Solved Problems

The process of melting ice is as follows \begin{gather*} \underbrace{ice}_{\rm -78^\circ C}\xrightarrow{Q_1}\underbrace{ice}_{\rm 0^\circ C}\xrightarrow{Q_2}\underbrace{water}_{\rm 0^\circ C}\xrightarrow{Q_3}\underbrace{water}_{\rm T_f ^\circ C} \end{gather*} 
We can calculate the amount of heat absorbed by the ice as   
\begin{align*} Q_1&=m_ic_i\Delta T \\\\ &=(40\ {\rm g})\left(0.5\frac{{\rm cal}}{{\rm g.K}}\right)(0-(-78)) \\\\&=1560\, {\rm cal}\end{align*} 
\[Q_2=m_iL_f=(40\, {\rm g})\left(80\,\frac{{\rm cal}}{{\rm g}}\right){\rm =3200\, cal}\] 
\[Q_3=m_ic_w (T_f-0)=(40)(1)T_f\] 
Now calculate the amount of heat released by the calorimeter and the water 
\[\underbrace{water+Calo}_{\rm 25^\circ C}\xrightarrow{Q_4}\underbrace{water+Calo}_{\rm T_f ^\circ C} \]
\begin{align*} Q_4&=\left(m_{cal}\,c_{Cu}+m_wc_w\right)\Delta T\\\\&=\left(\underbrace{80\times 0.0923+560\times 1}_{567.384}\right)\left(T_f-25\right)\end{align*} 
Since ${\rm -78^\circ C}<T_f<{\rm 25^\circ C}$, so $Q_4<0$.

Set the negative of the heat released equal to the heats absorbed by the ice (by the first law of thermodynamics), to find the final temperature of the system \begin{gather*} \underbrace{Q_1+Q_2+Q_3}_{Q_{gain}}=-\underbrace{Q_4}_{lost}\\\\ 1560+3200+40\,T_f=-(567.384\,T_f-14184.6) \\\\ \Rightarrow T_f=\rm 15.51^\circ C\end{gather*}

Recall from the specific heat problems that the heat delivered or absorbed during a change in the temperature of an object is $Q=mc\Delta T$. If we have a phase change in the material, then the amount of heat absorbed or delivered is $Q=\pm mL$, where the plus sign is for the melting process and the minus is for condensation. $L$ is called the latent heat and $c$ is the specific heat capacity of that substance. 

In the following, The heat and temperature changes of this solid material are as follows 
\begin{gather*} \underbrace{\rm -10^\circ C}_{solid} \xrightarrow{Q_1} \underbrace{\rm 0^\circ C}_{Solid}\xrightarrow{Q_2} \underbrace{\rm 0^\circ C}_{fluid} \\\\ Q_1=mc\Delta T=20\times 2500\times (0-(-10))=50\times {10}^4\ {\rm J}\\\\ Q_2=+mL=20\times 40\times {10}^4=800\times {10}^4\quad {\rm J} \end{gather*} Thus, the total heat is the sum of them as \[Q_{total}=Q_1+Q_2=850\times {10}^4 \quad {\rm J}\] 

The heat released by the water is \begin{align*} Q_1&=m_wc_w(T_f-50) \\ &=(0.330)(4190)(T_f-50)\end{align*} And the heat absorbed by the aluminum is \begin{align*} Q_2&=m_{Al}c_{Al}(T_f-10)\\ &=(0.855)(900)(T_f-10)\end{align*} Duo to the conservation of energy the sum of the two heats must be zero, i.e., $\Sigma Q_{tot}=0$ or the negative of heat lost must be equal to the heat gained $-Q_{lost}=Q_{gained}$, therefore \begin{gather*} -(330)(4190)(T_f-50)=(0.855)(900)(T_f-10) \\ \Rightarrow T_f=\rm 35.67^\circ C \end{gather*}

(a) The process of heat lost by the water is \[ \underbrace{water}_{\rm 80^\circ C} \xrightarrow{Q_1} \underbrace{water}_{\rm T_f^\circ C} \] 
Since there is no phase transition, we have \[Q_1=m_wc_w\Delta T=5(4186)(T_f-80) \] And the process of melting ice is as follows \[ \underbrace{ice}_{\rm 0^\circ C}\xrightarrow{Q_2}\underbrace{water}_{\rm 0^\circ C}\xrightarrow{Q_3}\underbrace{water}_{\rm T_f^\circ C}\] As one can see, there is a phase change in this process, so as we learned from latent heat of fusion problems, we must use the formula $Q=mL_f$, which yields \begin{align*} Q_2&=m_iL_f\\ &=2(3.33\times {10}^5) \\ &=666\, {\rm kJ} \\\\ Q_3&=m_ic_w(T_f-0)\\ &=2(4186)(T_f-0) \end{align*} The conservation of energy states that, the amount of released and absorbed heat must be equal or the sum of them must be zero, i.e., $Q_{lost}+Q_{gained}=0$ \begin{gather*} Q_1+Q_2+Q_3=0 \\\\ 4186[5(T_f-80)+2(T_f-0)]+666\times {10}^3=0 \\\\ 4186(7T_f-400)+666\times {10}^3=0 \\\\ \Rightarrow \quad \boxed{T_f=\, \rm 34.41^\circ C}\end{gather*} 
(b) By definition, the change in entropy of a system is \[\Delta S=\int^f_i{\frac{dQ_{rev}}{T}}\], where $dQ_{rev}$ is the heat absorbed by the system in a reversible process. Therefore,  \begin{align*} \Delta S_i &= \int^{fin}_{ini}{\frac{dQ}{T}} \\ \\ &= \int^{water\ 0^\circ C}_{ice\ 0^\circ C}{\frac{dQ}{T}}+\int^{water\ 34.41^\circ C}_{water\ 0^\circ C}{\frac{dQ}{T}} \\ \\ &= \frac{m_iL_f}{273+0^\circ C}+\int^{273+\ 34.41^\circ C}_{273+\ 0^\circ C}{c_wm_i\frac{dT}{T}}\\ \\ &= \frac{m_iL_f}{273}+m_ic_w \ln \left(\frac{273+34.41}{273}\right) \\\\ &= \frac{2\times 333\times {10}^3}{273}+2(4186) \ln \left(\frac{307.41}{273}\right) \\ \\ &=3433.40\,\rm J/K \end{align*} Note: temperatures must be in Kelvins. 

(c) First calculate the change in entropy of water and then find the total change of it
\begin{align*} \Delta S_w&=\int^{water\ 34.41^\circ C}_{water\ 80^\circ C}{\frac{dQ}{T}} \\ \\ &=\int^{273+34.41^\circ C}_{273+80^\circ C}{m_wc_w\frac{dT}{T}} \\ \\ &=m_wc_w \ln \left(\frac{307.41}{353}\right) \\\\ &=5(4186) \ln \left(\frac{307.41}{353}\right) \\\\ &=-2894.31\, {\rm J} \end{align*} The minus sign indicates that the entropy of the water has decreased. \begin{align*} \Delta S_{tot}&=\Delta S_i+\Delta S_w \\&=3433.40-2894.31 \\ &=+538.68\,\rm J/K\end{align*} The entropy change of the universe (total system) is positive, as the second law of thermodynamics requires. 
 

(a) First, find the final temperature of the system. The heat released from water is 
\[ \underbrace{water}_{\rm 10^\circ C} \xrightarrow{Q_1} \underbrace{water}_{\rm T_f^\circ C}\]  And the amount of heat released is also calculated as follows \begin{align*}  Q_{lost}=Q_1&=m_wc_w\Delta T \\ &=0.032\times 4190(T_f-10)\end{align*} And the heat absorbed by the ice is  \[\underbrace{ice}_{\rm -5^\circ C} \xrightarrow{Q_2} \underbrace{ice}_{\rm 0^\circ C} \xrightarrow{Q_3} \underbrace{water}_{\rm 0^\circ C} \xrightarrow{Q_4} \underbrace{water}_{\rm T_f^\circ C} \] Where the heats are found as below \begin{align*} Q_2&=m_ic_i\Delta T\\&=0.430\times 2090(0-(-5))\\&=4493.5, {\rm J} \\\\ Q_3&=m_iL_f\\&=0.430\times 334000 \\&=143620\, {\rm J} \\\\ Q_4&=m_wc_w\Delta T \\ &=0.430\times 4190(T_f-0)\end{align*} Conservation of energy states that $-Q_{lost}=Q_{absorbed}$ \begin{gather*} -134.08(T_f-10)=4493.5+143620+1801.7\left(T_f-0\right) \\\\  \Rightarrow \quad \boxed{T_f<0} \end{gather*} Since the final temperature is negative, so all of the ice does not melt, and instead some of the water freezes (we have a mixture of ice and water), thus $\boxed{T_f=\rm 0^\circ C}$. 

To find the amount of water that freezes, equate the heat absorbed by the ice with the heat released by the $m_X\,{\rm kg}$ of water to reaches ice at $0^\circ{\rm C}$ as follows \[ \underbrace{m_X \, water}_{\rm 10^\circ C} \xrightarrow{Q_1} \underbrace{m_X water}_{\rm 0^\circ C} \xrightarrow{Q'_1} \underbrace{m_X\, ice}_{\rm 0^\circ C}\] And the process for the ice is \[\underbrace{ice}_{\rm -10^\circ C} \xrightarrow{Q_2}\underbrace{ice}_{\rm 0^\circ C}\] According to these processes, the heat absorbed by the ice and released by the water are computed as below \begin{align*} Q_{absorbed}&=Q_2=4493.5\, {\rm J} \\\\ Q_{released}&=m_Xc_w\Delta T+m_XL_f \\&=m_X(4190)(0-10)+m_X(-334000) \\ &=-375900m_X \end{align*} Therefore, applying the principle of the conservation of energy gives us \begin{gather*} Q_{absorbed}=-Q_{released} \\\\  4493.5=375900m_X \\\\ \Rightarrow \quad \boxed{m_X=0.0119\,\rm kg} \end{gather*} Thus, $11.9\ {\rm g}$ of the $32\ {\rm g}$ of water freezes.  

Note: the latent heat of freezing is $m\left(-L_f\right)$ since freezing is the opposite process of fusion.

(b) First, find the change in entropy of water and that of ice then combine them. \begin{align*} \Delta S_i &= \int^{fi}_{ini}{\frac{dQ}{T}} \\\\ &=\int^{\rm 0^\circ C+273}_{\rm -10^\circ C+273}{\frac{dQ}{T}}+\int^{water\ \rm 0^\circ C}_{ice\ \rm 0^\circ C}{\frac{dQ}{T}} \\\\ &= m_ic_i \int^{\rm 0^\circ C+273}_{\rm -10^\circ C+273}{\frac{dT}{T}}+\frac{m_iL_f}{273} \\\\ &=m_ic_i \ln T+\frac{m_iL_f}{273} \\\\ &= (0.430)(2090) \ln \left(\frac{273}{263}\right)+\frac{(0.430)(334000)}{273} \\\\ &= \ \boxed{5568.17\, \rm J/K} \end{align*} And the entropy change of ice water is  \begin{align*} \Delta S_w &= \int^{water\ \rm 0^\circ C}_{watre\ \rm 10^\circ C}{\frac{dQ}{T}} \\\\ &=\int^{273+\rm 0^\circ C}_{273+\rm 10^\circ C}{\frac{m_wc_wdT}{T}} \\\\ &=m_wc_w \ln \left(\frac{273}{283}\right) \\\\ &=0.032\times 4190\times \ln \left(\frac{273}{283}\right) \\\\ &= \boxed{-4.82\, \rm J/K} \end{align*} Therefore, the total change in entropy of the system is found to be \[ \Delta S_{tot}=\Delta S_w+\Delta S_i= +5563.35\, {\rm J/K}\] Since $\Delta S_{tot}>0$, so the second law of thermodynamics is satisfied. 

(a) In the section on the latent heat of fusion practice problems we learned the heat that a substance absorbs during a constant temperature process and a change in its phase occurs is calculated by the formula $Q=\pm mL$. In the melting process, we have $Q=+mL_f$. So 
\[Q_{ice}=mL_f=20\times (334\times {10}^3)=6.68\times {10}^6\, {\rm J}\] 
Now use the entropy definition and find the change in the entropy of the ice
\begin{align*} \Delta S_{ice}&=\frac{\Delta Q_{ice}}{T}\\\\ &=\frac{6.68\times {10}^6}{\rm 0^\circ C+273.15} \\\\ &=24.455\,\rm J/K \end{align*} Note: in the entropy definition, temperatures must be in Kelvins.
(b) Ice requires $Q_{ice}$ to totally melts and this energy has gained from the ocean water. So, by convention, the ocean water has lost $Q_{ocean}=-Q_{ice}$. Using the definition of the entropy change of the ocean water, we obtain
\begin{align*} \Delta S_{ocean}&=\frac{\Delta Q_{ocean}}{T}\\\\ &=\frac{-6.68\times {10}^6}{\rm 5^\circ C+273.15} \\\\ &=-24.016\,\rm J/K \end{align*} 

(c) The total change in the entropy of the system of ice + ocean water is the sum of them
\begin{align*} \Delta S_{tot}&=\Delta S_{ice}+\Delta S_{ocean} \\\\ &=24.455+(-24.016) \\\\ &=+439\,\rm J/K \end{align*} As required by the second law of thermodynamics.  

(a) The temperature of water and ice cube are $0{\rm{}^\circ\,\!C}$ so there is a change in phase and the heat absorbed by the ice is determined by 
\[Q=mL_f=0.012\times 333000=3996\ {\rm J}\] 
Now compute the entropy change of the system by the following relation
\[\Delta S=\frac{\Delta Q}{T}=\frac{3996}{0{\rm{}^\circ\,\!C}+273}=14.6\frac{{\rm J}}{{\rm K}}\] 
(b) Similarly, $Q=mL_V=0.005\times \left(2256\times {10}^3\right)=11.28\ {\rm kJ}$
\[\Delta S=\frac{\Delta Q}{T}=\frac{11280}{100{\rm{}^\circ\,\!C}+273}=30.2\frac{{\rm J}}{{\rm K}}\] 
 

(a) Let $T_f$ be the final temperature. First find the amount of heat released by the coffee 
\[coffee\ 70{\rm{}^\circ\,\!C}\xrightarrow{Q_1}coffee\ T_f\] 
\[Q_1=m_{co}c\Delta T=0.3\times 4186\left(70-T_f\right)=1255.8\left(T_f-70\right)\] 
Now compute the heat that the cream absorbed 
\[cream\ 5{\rm{}^\circ\,\!C}\xrightarrow{Q_1}cream\ T_f\] 
\[Q_2=m_{cr}c\Delta T=0.02\times 4186\left(T_f-5\right)=83.72(T_f-5)\] 
Equal these two heats, due to the first law of thermodynamics, to find the final temperature i.e. $Q_{abs}=-Q_{releas}$
\[\Rightarrow 83.72\left(T_f-5\right)=-1255.8\left(T_f-70\right)\] 
\[\Rightarrow T_f=65.9{\rm{}^\circ\,\!C}\] 

(b) First, compute the change in entropy of coffee and cream and then calculate the total change in entropy of the universe. 
\begin{align*} \Delta S_{cr} &= \int^{T_f}_{T_i}{\frac{{\rm d}{{\rm Q}}_{{\rm in}}}{T}}=\int^{T_f}_{T_i}{\frac{\left(m_{cr}cdT\right)}{T}} \\ &= m_{cr}c{\ln  \left(\frac{T_f}{T_i}\right)\ }=0.02\times 4186\times {\ln \left(\frac{65.9{\rm{}^\circ\,\!C}+273}{5{\rm{}^\circ\,\!C}+273}\right)\ }\end{align*} 
\[\Rightarrow \Delta S_{cr}=+16.58\,\frac{{\rm J}}{{\rm K}}\] 
\begin{align*}\Delta S_{co} &= \int^{T_f}_{T_i}{\frac{dQ_{in}}{T}}=\int^{T_f}_{T_i}{\frac{m_{co}cdT}{T}}\\&= m_{co}c{\ln  \left(\frac{T_f}{T_i}\right)\ }=0.3\times 4186\times {\ln \left(\frac{65.9{\rm{}^\circ\,\!C}+273}{70{\rm{}^\circ\,\!C}+273}\right)\ }\end{align*}
\[\Rightarrow \Delta S_{co}=-15.10\ \frac{{\rm J}}{{\rm K}}\] 
\[\Delta S_{universe}=\Delta S_{coffee}+\Delta S_{cream}=-15.10+16.58=+1.48\frac{{\rm J}}{{\rm K}}\] 
\[{\rm and\ }\Delta {{\rm S}}_{{\rm surrounding}}=0\ \] 
So $\Delta S_{universe}>0$ as it should be, since the second law of thermodynamics states that in an irreversible process the entropy change of the universe is always positive or increased. This process is irreversible since there is a spontaneous flow of heat.

(c) Setting the heat released by the coffee equal to $m_x\ {\rm g}$ of coffee absorbs to evaporate. 
\[Q=m_{co}c\Delta T=0.3\times 4186\times 5=6280\ {\rm J}\] 
\[Q^{'}=m_xL_V\] 
\[Q=Q^{'}\Rightarrow 6280=2.26\times {10}^6m_x\Rightarrow m_x=2.77\ {\rm g}\ \] 
 

Similar to previous problems
(a) $\Sigma Q_i=0\to Q_{water}+Q_{ice}=0$ When there is no phase change, so we have $Q=mc\Delta T$, in the case of a change in phase of matter, $Q=mL_f$ Thus, 
\begin{gather*} m_iL_f+m_ic_w(T_{final}-0)+m_wc_w(T_{final}-80)=0 \\\\ \Rightarrow T_{final}=\frac{m_wc_w80-m_iL_f}{\left(m_i+m_w\right)c_w}=34.4^\circ{\rm C}\end{gather*}

(b) The entropy is obtained as below \begin{align*} \Delta S &= \int^{final}_{initial}{\frac{dQ}{T}}\\\\ &=\int^{0{}^\circ\, C\ water}_{ice}{\frac{dQ}{273}}+\int^{273+34.4}_{273}{\frac{dQ}{T}}\\\\&=\frac{m_iL_f}{273}+\int^{273+34.4}_{273}{m_ic_w\frac{dT}{T}}=\frac{m_iL_f}{273}+m_ic_w{\ln  \frac{273+34.4}{273}}\\\\&=\frac{2\times 333\times {10}^3}{273}+2\times 4186{\ln  \frac{273+34.4}{273}} \end{align*} Therefore, \[\Delta S_{ice}=3433\quad {\rm J/K}\] 
Note: in the definition of entropy, $T$ should be in Kelvin.

(c) Now first determine the $\Delta S_{water}$
\begin{align*} \Delta S_{water} &=\int^{273+34.4}_{273+80}{\frac{dQ}{T}}\\\\&=\int^{307.4}_{353}{m_wC_w\frac{dT}{T}}\\\\&= m_wC_w{\ln  \left(\frac{307.4}{353}\right)}\\\\&=5\times 4186\,{\ln \left(\frac{307.4}{353}\right)}\\\\&=-2895\quad {\rm \ J/K} \end{align*} Thus, \[\Delta S_{total}=\Delta S_{ice}+\Delta S_{water}=3933-2895=+538\, {\rm J/K}\] 

(a) Determine the heat lost and gained during these processes. The process of heat lost by the water is as: $water\, 70^\circ {\rm C}\xrightarrow{Q_1}water\, T_f^\circ {\rm C}$
\begin{align*}Q_{lost}&=m_wc_w\Delta T\\\\&=200\times 4.186\times (t_f-70^\circ {\rm C})\end{align*} Heat gained by the ice follows the process: \[ice\, 0^\circ{\rm C}\xrightarrow{Q_2}water\ 0^\circ {\rm C}\xrightarrow{Q_3}water\, T_f^\circ {\rm C}\] So
\begin{align*} Q_{gain}&=\underbrace{m_iL_f}_{Q_2}+\underbrace{m_ic_w\Delta T}_{Q_3}\\\\&=100\times 80+100\times c_w\times (T_f-0)\end{align*}Now using the conservation of energy that is $Q_{gained}=-Q_{lost}$, we obtain \begin{gather*} -200\times 4.186\times (t_f-70^\circ {\rm C})=8000+100\times 4.186\times T_f \\\\ \Rightarrow T_f=40.3^\circ {\rm C} \end{gather*}
(b) The total change in the entropy of the mixing is the sum of the change in the entropy of each of them so $\Delta S_{mixing}=\Delta S_{water}+\Delta S_{ice}$.
\begin{align*} \Delta S_{mixing} &= \int^{T_f}_{T_i}{\frac{\Delta Q_{water}}{T}}+\int^{T_f}_{T_i}{\frac{\Delta Q_{ice}}{T}}\\\\&=\int^{40.3^\circ {\rm C}+273}_{70^\circ {\rm C}+273}{\frac{m_wc_wdT}{T}}+\underbrace{\frac{m_iL_f}{T(0^\circ {\rm C}+273)}+\int^{40.3^\circ {\rm C}+273}_{0^\circ {\rm C}+273}{\frac{m_ic_wdT}{T}}}_{\Delta S_{ice}} \\\\ &=m_wc_w{\left.{\ln  T}\right|}^{313.3}_{343}+\frac{m_iL_f}{273}+m_ic_w{\left.{\ln  T\ }\right|}^{313.3}_{273}\\\\& =+11.03\quad {\rm J/K}\end{align*}
(c) We want calculate the amount of energy needed to vaporize the $300\, {\rm g}$ of water, so the process is as \[water\, 40.3^\circ {\rm C}\xrightarrow{Q_1} water\, 100^\circ {\rm C}\xrightarrow{Q_2} vapor\, 100^\circ{\rm C}\] Thus,
\begin{align*} Q_{tot}&=\underbrace{m_wc_w\Delta T}_{Q_1}+\underbrace{m_wL_V}_{Q_2}\\\\&=300\times 4.186+\left(100-40.3\right)+300\times 540\\\\& =234\quad {\rm kJ}\end{align*}
(d) Use the ideal gas law, $PV=nRT$, and find the desired quantity.
\begin{align*} V&=\frac{nRT}{P}\\\\&=\left(\frac{300}{18}\right)\frac{8.314\times (100^\circ{\rm C}+273)}{{10}^5}\\\\&=0.517\quad {\rm m^3} \end{align*}

(a) 
\[{\rm ice-8.5}{\rm{}^\circ\,\!C}\xrightarrow{Q_1}{\rm ice\ 0}{\rm{}^\circ\,\!C}\xrightarrow{Q_2}{\rm water\ 0}{\rm{}^\circ\,\!C}\xrightarrow{Q_3}{\rm water\ 16.9}{\rm{}^\circ\,\!C}\] 
Heat lost or gained without change in phase is given by $Q=mc\Delta T$ and during phase change is given by $Q=mL$ where $L$ is the latent heat of the substance and $c$ is the specific heat of it. Therefore
\[Q_1=m_{ice}c_{ice}\Delta T_{ice}=\left(0.01\right)\left(2100\right)\left(0-\left(-8.5\right)\right)=178.5\ {\rm J}\] 
\[Q_2=m_{ice}L_f=\left(0.01\right)\left(3.335\right)=3330\ {\rm J}\] 
\[Q_3=m_{ice}c_w\Delta T=\left(0.01\right)\left(4186\right)\left(16.9\right)=707.4\ {\rm J}\] 
The total heat gained by the ice is $Q_{gained}=Q_1+Q_2+Q_3=4216\ {\rm J}$

(b) The total heat lost by the water and glass is
\[({\rm water+glass)\ 20}{\rm{}^\circ\,\!C}\xrightarrow{Q_4}{\rm \ (water+glass)\ }{\rm 16.9}{\rm{}^\circ\,\!C}\] 
\[Q_{lost}=Q_4=m_wc_w\Delta T_w+m_gc_g\Delta T_g=\left(m_wc_w+m_gc_g\right)\Delta T\] 
Note: the glass and the water experience a similar temperature change i.e. $\Delta T_w=\Delta T_g$

Note: conservation of energy states that the heat gained must be equal to the negative of the heat lost so: $-Q_{lost}=Q_{gained}$ 
\[\Rightarrow \ \left(m_wc_w+m_gc_g\right)\Delta T=4216\to -\left(m_w\left(4186\right)+0.100\left(840\right)\right)\left(16.9-20\right)=4216\] 
\[\therefore m_w=0.305\ {\rm kg}\] 
 

We can solve this question, from the principles that we learned from the heat practice problems page as below.

(a) The process is as follows \begin{gather*} \underbrace{\rm Pb_{liquid}}_{\rm 350^\circ C} \xrightarrow{Q_1} \underbrace{\rm Pb_{liquid}}_{\rm 327^\circ C} \xrightarrow{Q_2}\underbrace{\rm Pb_{solid}}_{\rm 327^\circ C} \\\\ Q_1=m_{Pb}c_{Pb}\Delta T=0.2\times 450\times (327-350)=-2070\, {\rm J} \\\\ Q_2=m_{Pb}(-L_f)=0.2\times 2.5\times {10}^4=-5000\ {\rm J} \end{gather*}  Note: the heat required to change the phase of a given pure substance is $Q=\pm mL$, the minus sign is used when it freezes (heat leaving), and plus sign (heat entering) is used when the material melts. In this case, the lead loses its heat and becomes solid (freeze) so we must use the minus sign.

In the above, the negative indicates that the lead has lost the heat and there is no other significance so the total heat lost going from the liquid lead at $\rm 350^\circ C$ to solid lead at $\rm 327^\circ C$ is \[Q_{total}=\left|Q_1+Q_2\right|=5000+2070=7070\, {\rm J}\]

(b) \begin{gather*} \underbrace{\rm Pb-liquid}_{\rm 350^\circ C} \xrightarrow{Q_1} \underbrace{\rm Pb-liquid}_{\rm 327^\circ C} \xrightarrow{Q_2} \underbrace{\rm Pb-solid}_{\rm 327^\circ C}\xrightarrow{Q_3}\underbrace{\rm Pb-solid}_{\rm T_f} \\\\ Q_3=m_{Pb}c_{Pb}\Delta T=0.2\times 450(T_f-327) \end{gather*} The total heat lost by lead is also found as below \[Q'_{tot}=7070+0.2\times 450(T_f-327)\]

Now calculate the heat that the cast has gained \begin{gather*} \underbrace{cast}_{\rm 25^\circ C}\xrightarrow{Q_{gained}}\underbrace{cast}_{T_f} \\\\ Q_{gained}=m_{Al}c_{Al}\Delta T=0.8\times 900(T_f-25) \end{gather*} 
Therefore, according to the conservation of the energy $Q_{gained}=-Q_{lost}$, we have \begin{gather*} 7070+0.2\times 450(T_f-327)=0.8\times 900(T_f-25) \\\\ \Rightarrow T_f=\rm 67^\circ C \end{gather*} 

The given information are $P_i=1.01\times {10}^5{\rm Pa\ ,\ }{{\rm V}}_{{\rm i}}{\rm =600}{{\rm m}}^{{\rm 3}}\ ,\ T_i=300{\rm k}$. 

The internal energy of an ideal gas is given by $E_{int}=C_V\Delta T$, where $C_V$ is the heat capacity at constant volume. For an monoatomic gasses ($He, Ne, Ar, Kr, Xe,\dots$) $C_V=\frac{3}{2}nR$ and for all diatomic gasses ($N_2, H_2, O_2,CO,\dots $) $C_V=\frac{5}{2}nR$. Therefore, the change in the internal energy of neon gas (monatomic) is \textit{}
\[\Delta E_{int}=\frac{3}{2}nR\Delta T\] 
Since the combination $nR$ is not given, we have used the ideal gas law and replaced it by the following 
\[P_iV_i=nR\,T_i\ \Longrightarrow nR=\frac{P_iV_i}{T_i}\] 
\[\Delta E_{int}=\frac{3}{2}\left(\frac{P_iV_i}{T_i}\right)\Delta T=\frac{3}{2}\frac{\left(1.01\times {10}^5\right)\left(600\right)}{300}\left(302-300\right)\] 
\[\Delta E_{int}=6.06\times {10}^5{\rm J}\] 
Important note: the internal energy of an ideal gas depends only on its absolute temperature, not on its pressure or volume.
 

By definition, the work done on the gas is $W_{on}=-P\Delta V$ and we know that $W_{on}=-W_{by}$ so the work done by the gas in an isothermal expanding is as follows
\[W_{by}=\int{PdV}=\int^{V_f}_{V_i}{nRT\frac{dV}{V}}\] 
\[W_{by}=nRT\,{\ln  \frac{V_f}{V_i}\ }=5\left(8.31\right)\left(200\right){\ln  \frac{2V_i}{V_i}\ }\] 
\[W_{by}=8.31\times 1000\times 0.7=5817\ {\rm J\ \ }\Longrightarrow {\rm W\sim \ 6000\ J}\] 
In above we have used the ideal gas law $P=nRT/V$.
 

The work done on a gas is defined as the negative of the product of pressure and the change in the volume of the gas i.e. 
\[W_{on}=-\int^{V_f}_{V_i}{PdV}\] 
By convention, the work done by the gas is negative of the work done on the gas so
\[W_{by}=-W_{on}\] 
When a $PV$ diagram has been given, the work done by the gas is the area under the diagram. In this case, the minimum of work is 
\[W_{min}=P_{min}\left(V_f-V_i\right)=9\left(12-8\right)=36\ {\rm J}\] 
 

(a) Use the equation of state for an ideal gas, $PV=nRT$
\[T_A=\frac{P_AV_A}{nR}=\frac{6000\times 1}{1\times 8.314}=721.6\ {\rm K}\] 
\[T_D=\frac{P_DV_D}{nR}=\frac{6000\times 4}{1\times 8.314}=962.2\ {\rm K}\] 

(b) The work done on the system is given by $W_{on}=-P\Delta V$ or its magnitude by the area under the $P-V$ diagram. So 
\[W_{AB}=-P\left(V_B-V_A\right)=-6000\left(2-1\right)=-6000\ {\rm J}\] 
\[\left|W_{BC}\right|=S_{trapezoid}=\frac{\left(2000+6000\right)\left(3-2\right)}{2}=4000\ {\rm J}\] 
Since $\Delta V_{BC}>0\ $so due to $W=-P\Delta V$, the work done on the gas is negative i.e. $W_{BC}=-4000\ {\rm J}$
\[W_{CD}=-P\Delta V_{CD}=-2000\left(4-3\right)=-2000\ {\rm J}\] 

(c) Use the first law of thermodynamics i.e. $\Delta U_{int}=W_{on}+Q_{in}$ and that the change in internal energy of an monoatomic ideal gas is ${\Delta U}_{int}=\frac{3}{2}nR\Delta T$, to find the desired heat 
\[\frac{3}{2}nR\left({{\rm T}}_{{\rm D}}{\rm -}{{\rm T}}_{{\rm A}}\right){\rm =}{\left({{\rm W}}_{{\rm AD}}\right)}_{on}+Q_{in}\] 
\[W_{AD}=W_{AB}+W_{BC}+W_{CD}=-6000-4000-2000=-12000\ {\rm J}\] 
\[\Rightarrow \ Q_{in}=\frac{3}{2}nR\left(T_D-T_A\right)-\left(-12000\right)=\frac{3}{2}\left(P_DV_D-P_AV_A\right)+12000\] 
\[\Rightarrow Q_{in}=\frac{3}{2}\left(2000\times 4-6000\times 1\right)+12000=+15000\ {\rm J}\] 
Where in the third line we have replaced the $nRT$ by $PV$ to avoid uncertainty in the result. Since $Q_{in}>0$ thus the total heat absorbed by the gas. 
 

Solution:

(a) First, find the temperatures of $a,b,c$ states. Because of $PV\sim T$ and maximum attainable temperature is $327^{\circ}$ and $bc$ is isothermal $T_b=T_c=327^{\circ}$ so we have \[P_aV_a<P_bV_b\ or\ P_cV_c\to T_a<T_b=T_c\]Applying the ideal gas law  get \begin{align*} \frac{P_a}{T_a}&=\frac{P_b}{T_b}\\ \to T_a&=T_b\frac{P_a}{P_b}\\&=\left(273+373\right)\frac{1\times {10}^5}{3\times {10}^5}\\&=200{\rm K}\end{align*}
Now applying the first law of thermodynamics as $\Delta U_{int}=\rm{\Delta W_{on}+ \Delta Q_{in}}$ we can get the desired heat in each process. 
In the process $a\to b$, we have
\[\Delta V_{a\to b}=0 \Rightarrow \Delta W_{a\to b}=-P\Delta V=0 \] Thus, the heat in the process $ab$ is \begin{align*} \Delta U_{a \to b}&=\Delta Q_{a\to b}\\&=nC_{V}\Delta T\\ &=2\left(\frac{3}{2}R\right)(T_b - T_a)\\&=3(8.314)(600-200)\\&=9.97\times10^{3}\,{\rm J}\end{align*} The bc process is isothermal so \begin{gather*}\Delta T_{b\to c}=0\\\Rightarrow \Delta U_{b\to c}=0\\\Rightarrow \Delta Q=-\Delta W\end{gather*} Recall that the work done on a gas in an isothermal process is as below \begin{align*} W_{isothermal}&=-\int{PdV}\\&=-\int^{V_f}_{V_i}{\frac{nRT}{V}dV}\\&=-nRT \ln \frac{V_f}{V_i} \end{align*}Therefore, the rquired heat in the $bc$ process is obtained as \begin{align*} \Delta Q_{b\to c}&=nRT{\ln  \frac{V_c}{V_b}\ }\\&=2(8.314)(600){\ln \left(\frac{T_c}{T_b}\frac{P_b}{P_c}\right)\ }\\&=9976.8\,{\ln  3\ }\end{align*} And in the process $ac$ which is an isobaric, we have $\Delta P_{c\to a}=0$, therefore the heat gained or lost in this process is \begin{align*} \Delta Q_{c\to a} &= \Delta U_{c\to a}-\Delta W_{c\to a}\\&=\frac{3}{2}nR\left(T_a-T_c\right)+\underbrace{P\left(V_a-V_c\right)}_{nR\left(T_a-T_c\right)}\\&=\frac{5}{2}nR\left(T_a-T_c\right)\\ &=\frac{5}{2}\left(2\right)\left(8.134\right)\left(200-600\right)\\&=-1.66\times {10}^4\ {\rm J} \end{align*}Now the signs of heats determine whether heat absorbed or lost by the system. $\Delta Q_{a\to b}>0$ and $\Delta Q_{b\to c}>0$ so heat enters the gas and $\Delta Q_{c\to a}<0$ so heat leaves the gas.
Note: Change in the internal energy of a monoatomic gas is $\Delta U_{int}=\frac{3}{2}nRT$. Helium is such a gas.
(b) In a cycle $\Delta T=0$ then $\Delta U=\frac{3}{2}nR\Delta T=0$ so $\Delta Q_{cycle}=-\Delta W_{cycle}$. The amount of heat enters the system (gas) is \begin{align*}Q_{in}&=Q_{ab}+Q_{bc}\\&=9.97\times {10}^3+1.1\times {10}^4\\&=2.097\times {10}^4\, {\rm J}\end{align*} and the lost heat is $Q_{out}=Q_{ca}=1.66\times {10}^4\,{\rm J}$Therefore, the change in the heat of the system is as below \begin{align*} \Delta Q_{cycle}&=Q_{net}\\&=Q_{in}-Q_{out}\\&=\left(2.097-1.66\right)\times {10}^4\\&=4.4\times {10}^3\ {\rm J}\end{align*}Thus, the magnitude of the work done on or by the system is obtained as $ \left|\Delta W\right|=4.4\times {10}^3\ {\rm J}$ and consequently, the efficiency of the cycle is 
\[e=\frac{W}{Q_{in}}=0.21=21\%\] 
(c) Recall that the most efficient heat engine is the Carnot cycle which is obtained as $e_{Carnot}=1-\frac{T_C}{T_H}$
So in a cycle, we have \begin{align*} e_{max}&=e_{Carnot}\\ \\&=1-\frac{T_C}{T_H}\\\\&=1-\frac{200}{600}\\ \\&=\frac{2}{3}\\&=66\%\end{align*}

(a) By definition, the work done on the gas is $W_{on}\ =-P\Delta V$, so
\begin{align*} W&=-P\left(\frac{nRT_2}{P}-\frac{nRT_1}{P}\right)\\ \\&=-nR\left(T_2-T_1\right)\\ \\&=-\left(7.1\right)\left(8.314\right)\left(300-370\right)\\ \\&=+4132.058\quad {\rm J}\end{align*} Since the volumes of the gas is unknown, we have used the ideal gas law $V=nRT/P$ .

(b) The change in internal energy of an ideal gas in general is given by \begin{align*}\Delta U_{int}&=nC_V\Delta T\\ &=7.1\times 12.24\left(300-370\right)\\&=-6083.28\quad{\rm J}\end{align*} Note: for ideal gases, the equation above is always valid, even when the volume isn't constant as this case!

(c) Use the first law of thermodynamics to find $Q$ \begin{align*}\Delta U_{int}&=Q_{trans}+W_{on}\\ \Rightarrow \quad Q_{trans}&=-6083.28-\left(+4132.058\right)\\ &=-10215.338\quad {\rm J}\end{align*} Since $Q_{trans}<0$ so the heat lost by the gas. 
 

(a) Given data: $Q_{in}=1000\ {\rm J}$ , $T_i=300{\rm \ K}$, $W_{on}=?$

 By definition, $W_{on}=-P\Delta V=0\ $since $\Delta V=0$ 
(b) Use the first law of thermodynamics as $\Delta U_{int}=Q_{in}+W_{on}$
\[\Delta U_{int}=+1000+0=+1000\ {\rm J}\] 

(c) The heat transferred in a constant volume by a system is given by $Q=nC_V\Delta T$
\[\Rightarrow \ +1000=\left(1\right)\frac{3}{2}\left(8.314\right)\left(T_f-300\right)\Rightarrow T_f=380.18\ {\rm K}\] 
 

(a) Use the first law of thermodynamics to find the heat exchanged of each process. 
BC is isothermal i.e.$\Delta T=0$, thus \begin{gather*} \Delta U_{int}=nC_V\Delta T=0 \\ \\ {\left(\Delta U_{int}\right)}_{BC}={\left(Q_{in}+W_{on}\right)}_{BC}\\ \\ \Rightarrow 0=Q_{in}+W_{on}\end{gather*}
\begin{align*}Q_{BC}=-W_{BC}&=-\left(-\int{PdV}\right)\\ \\&=\int^{V_f}_{V_i}{\frac{nRT}{V}dV}\\ \\&=\underbrace{nRT}_{PV}{\ln \frac{V_f}{V_i}\ }\end{align*}Thus, the heat in the BC part is computed as below \begin{align*} Q_{BC}&=P_BV_B{\ln  \frac{V_C}{V_B}\ }\\ \\&=\left(3\times 1.01\times {10}^5{\rm Pa}\right)\left(5\times {10}^{-3}{{\rm m}}^{{\rm 3}}\right){\ln  \frac{15}{5}}\\ \\&=+1664.39\ {\rm J}\end{align*} For the $CA$ process, we have \begin{gather*}\Delta P=0 \\ \\  Q_{CA}=nC_P\Delta T_{CA}=\left(\frac{7}{2}R\right)n\left(T_A-T_C\right)\end{gather*} Therefore,\begin{align*} Q_{CA}&=\frac{7}{2}\left(P_AV_A-P_CV_C\right)\\ \\ &=\frac{7}{2}\left(5\times 1-15\times 1\right)\left({\rm atm.L}\right)\\ \\&=-35\left({\rm 1.01\times }{{\rm 10}}^{{\rm 5}}{\rm Pa.}(10^{-3}\,{\rm m^3})\right)\\ \\&=-3535\ {\rm J} \end{align*} For the isochoric process $AB$, we have \begin{gather*} \Delta V=0 \\ \\ Q_{AB}=nC_V\Delta T_{AB}=n\left(\frac{5}{2}R\right)\left(T_B-T_A\right)\end{gather*} Therefore, \begin{align*} Q_{AB}&=\frac{5}{2}\left(P_BV_B-P_AV_A\right)\\ \\&=\frac{5}{2}\left(3-1\right){\rm atm\times 5L=25\quad atm.L} \\ \\ &=25\left(1.01\times {10}^5\right)\left({10}^{-3}\right) \\ \\&=+2525\quad {\rm J}\end{align*}
By convention, if $Q_{in}>0$ then heat transferred into the system (gas).Therefore, the total heat absorbed by the gas is $Q_h=Q_{AB}+Q_{BC}=2525+1664.39=4189.39\ {\rm J}$

Note: in above, instead of calculating temperatures of each point, we have used the ideal gas law to convert $nRT$ into $PV$. (Since the data is given in terms of $P$ and $V$)

 Note: for an diatomic gas $C_V=\frac{5}{2}R$ and in general $C_P=C_V+R=\frac{7}{2}R$.

(b) By convention, if $Q_{in}<0$ then the heat released from the gas into the surrounding thus in this case $Q_c=Q_{CA}=-3535\ {\rm J}$

(c) Gas absorbed the $Q_h$ from hot reservoir, does work $W_{net}$ and gives up $Q_c$ to the cold reservoir so due to the 2${}^{nd}$ law of thermodynamics we have \begin{align*} \left|W_{net}\right|&=\left|Q_h\right|-\left|Q_c\right|\\&=4189.39-3535\\&=654.39\quad {\rm J}\end{align*}

(d) The thermal efficiency of a heat engine is defined as the work done by the engine $W_{net}$ divided by the heat absorbed during one cycle i.e. \begin{align*} e&=\frac{W_{net}}{\left|Q_h\right|}\\ \\&=1-\frac{\left|Q_c\right|}{\left|Q_h\right|}\\ \\&=1-\frac{3535}{4189.39}\\ \\&=15.6\,\% \end{align*}
 

Given data: $T_1=273\ {\rm K\ ,\ }{{\rm P}}_{{\rm 1}}=1\ {\rm atm\ ,\ \ }{{\rm V}}_{{\rm 1}}{\rm =1\ lit\ ,\ }{{\rm V}}_{{\rm 2}}{\rm =0.5\ lit\ ,\ \ }{{\rm P}}_{{\rm 2}}{\rm =?,\ \ }{{\rm T}}_{{\rm 2}}{\rm =?\ \ }$

(a) In the adiabatic process the equation of state is $PV^\gamma={\rm constant}$ or $TV^{\gamma-1}={\rm constant}$. So 
\[P_1V^\gamma_1=P_2V^\gamma_2\Rightarrow P_2=P_1{\left(\frac{V_1}{V_2}\right)}^\gamma=\left(1\right){\left(\frac{1}{0.5}\right)}^{1.3}=2.46\ {\rm atm}\] 
\[T_1V^{\gamma-1}_1=T_2V^{\gamma-1}_2\Rightarrow T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma-1}=273{\left(\frac{1}{0.5}\right)}^{1.3-1}=336.1\ {\rm K}\] 

(b) Use the ideal gas law $PV=nRT$ between these points to find the final volume
\[\frac{P_2V_2}{T_2}=\frac{P_fV_f}{T_f}\Rightarrow V_f=V_2\left(\frac{T_f}{T_2}\right)=\left(0.5\right)\left(\frac{273}{336.1}\right)=0.406\ {\rm lit}\] 
 

(a) Use the definition of the internal energy as $U_{int}=nC_V\Delta T$.  Recall that for an ideal diatomic gas $C_V=\frac{5}{2}R$ so 
\[\Delta U_{int}=\frac{5}{2}nR\left(T_c-T_a\right)\xrightarrow{ideal\ gas\ law}\Delta U_{int}=\frac{5}{2}\left(P_cV_c-P_aV_a\right)\] 
\[\Rightarrow \Delta U_{int}\left(a\to c\right)=\frac{5}{2}\left(2000\times 4-5000\times 2\right)=-5000\ {\rm J}\] 
Note: the change in internal energy of a system is independent of the path and only dependent on the initial and final temperatures of the system, as the formula shows this fact.
(b) See part (d), $Q\left(a\to c\right)=+2000\ {\rm J}$
(c) See part (d), $Q\left(a\to b\to c\right)=Q_1+Q_2=35000-30000=+5000\ {\rm J}$
(d) To solve this part, calculate work and heat of each process individually.
\[a\to b\ :isobaric\ \left(\Delta P=0\right)\] 
\[\ \ W_1=area\ under\ curve=\left(4-2\right)5000=10000\ {\rm J}\] 
\begin{align*} Q_1 &=nC_P\Delta T_{ab}=n\ \left(C_V+R\right)\left(T_b-T_a\right)\\ &=\frac{7}{2}\left(P_bV_b-P_aV_a\right)=\frac{7}{2}\left(5000\right)\left(4-2\right)=35000\ {\rm J} \end{align*}
\[b\to c:isochoric\left(\Delta V=0\right),\ \ W_2=-P\Delta V=0\] 
\begin{align*} Q_2 &=nC_V\Delta T_{bc}=\frac{3}{2}nR\left(T_c-T_b\right)\\ &=\frac{5}{2}\left(P_cV_c-P_bV_b\right)=\frac{5}{2}\left(2000-5000\right)\left(4\right)=-30000\ {\rm J}\end{align*}
\begin{align*} a\to c\ :W_3=area\ under\ curve(trapezoid)&=\frac{1}{2}(2000+5000)(4-2)\\ &=7000\ {\rm J}\end{align*}
\[Q_3=\Delta U_{int}\left(a\to c\right)-W_3=-5000-\left(-7000\right)=+2000\ {\rm J}\] 
Note: since in this case $\Delta V_{ac}>0$ and $\Delta V_{ab}>0\ $so by definition of work done on the gas i.e. $W_{on}=-P\Delta V$, we have $W_3=-7000\ {\rm J}$ and $W_1=-10000\ {\rm J}$
Therefore the total heat added to the system, by convention (i.e. $Q_{in}>0$), is 
\begin{gather*} Q_{in}=Q_1=35000{\rm \ J}\\ \\ W_{on}=-10000-7000=-17000\ {\rm J}\end{gather*}
Note: the area under $P-V$ diagram, gives the magnitude of the work done on the gas. To find the correct sign ($\pm $), we must use the formal definition of the work done on the gas that is $W_{on}=-P\Delta V$. If $\Delta V>0$, then $W_{on}<0$ and vice versa. 
Note: the area of a trapezoid is as follows \[S=\frac{a+b}{2}h\] 

(a) $20.9\ {\rm J}$ heat was added to the gas that is $Q_{in}=+20.9\ {\rm J}$ . First from the definition of the work done on the gas find $W_{on}$, then use the 2${}^{nd}$ law of thermodynamics as $\Delta U_{int}=Q_{in}+W_{on}$ to determine the $\Delta U_{int}$. 
\[W_{on}=-P\left(V_2-V_1\right)=-\left(1\times 1.01\times {10}^5{\rm Pa}\right)\left({\rm 100-50}\right)\left({{\rm 10}}^{{\rm -}{\rm 6}}{{\rm m}}^{{\rm 3}}\right){\rm =-5.05\ J}\] 

\[\Delta U_{int}=+20.9+\left(-5.05\right)=+15.85\] 
(b) Molar specific heat of the gas at constant pressure is defined by $C_P=\frac{1}{n}\frac{\Delta Q}{\Delta T}$ so first find the temperatures of the gas at the given points as
\[T_1=\frac{P_1V_1}{nR}=\frac{\left(1.01\times {10}^5\right)\left(50\times {10}^{-6}\right)}{\left(2\times {10}^{-3}\right)\left(8.314\right)}=303.7\ {\rm K}\] 
\[T_2=\frac{P_2V_2}{nR}=\frac{\left(1.01\times {10}^5\right)\left(100\times {10}^{-6}\right)}{\left(2\times {10}^{-3}\right)\left(8.314\right)}=607.4\ {\rm K}\] 
Therefore, 
\[C_P=\frac{1}{n}\frac{Q_{in}}{T_2-T_1}=\frac{1}{0.002}\left(\frac{20.9}{607.4-303.7}\right)=34.4\frac{{\rm J}}{{\rm mol.K}}\] 
(c) Similarly, Molar specific heat of the gas at constant volume is defined by 
\[C_V=\frac{1}{n}\frac{\Delta U_{int}}{\Delta T}\] 
\[\therefore C_V=\frac{1}{0.002}\left(\frac{15.85}{607.4-303.7}\right)=26.09\ \frac{{\rm J}}{{\rm mol.K}}\] 
 

(a) Recall that the efficiency of a Carnot engine is given by 
\begin{gather*} e=\frac{W_{net}}{Q_h} \\  {\rm or\ \ } \\  e=1-\frac{\left|Q_c\right|}{\left|Q_h\right|}\ \ \ \ \ {\rm or} \\ e=1-\frac{T_c}{T_h}\end{gather*} 
Here, engine takes $Q_h$ from the hot reservoir at temperature $T_h$, does work $W_{net}$ and release $Q_c$ to the cold reservoir at $T_c$.  

Therefore, \begin{align*} e&=1-\frac{115{\rm{}^\circ\,\!C}+273}{235{\rm{}^\circ\,\!C}+273} \\& =1-\frac{388}{508}\\&=23.6 \% \end{align*}

 Note: temperatures must be in Kelvins.

(b) Use the first equation to find the work done by the engine 
\begin{align*}W_{net}&=eQ_h\\&=23.6\times {10}^{-2}\times 6.3\times {10}^4\\ &=14.9\quad {\rm kJ}\end{align*} 
 

The efficiency of a heat engine is defined as the ratio of the work done by the engine to the heat absorbed from the high-temperature reservoir i.e. $e=W/Q_h$. So 

Given data: $T_c=17{\rm{}^\circ\,\!C}+273=290{\rm K\ ,\ \ }{{\rm e}}_{{\rm 1}}{\rm =40\%\ \ \ ,\ \ }{{\rm e}}_{{\rm 2}}{\rm =50\%}$
\[e_1=1-\frac{T_c}{T_h}\Rightarrow T_h=\frac{T_c}{1-e_1}=\frac{290}{1-0.40}=483.3\ {\rm K}\] 
\[T^{'}_h=\frac{T_c}{1-e_2}=\frac{290}{1-0.50}=580\ {\rm K}\] 

\[\Delta T=T^{'}_h-T_h=96.7\ {\rm K}\] 

Note: the work done on the gas in an isothermal expansion is given by $W_{on}=-nRT{\ln  \left(\frac{V_f}{V_i}\right)\ }$ and its change in internal energy ($\Delta U_{int}=nC_V\Delta T$) is zero since $\Delta T=0$. So use the first law of thermodynamics to find the heat absorbed by the gas and then use the definition of entropy change to find its value. 
\[\Delta U_{int}=W_{on}+Q_{in}\Rightarrow Q_{in}=-W_{on}=nRT\,{\ln  \left(\frac{V_f}{V_i}\right)\ }\] 
The change in entropy of a system as it goes from one state to another is defined as $\Delta S=\Delta Q_{in}/T$ where $\Delta Q_{in}$ is the heat absorbed by the system in a reversible process. 
\[\Rightarrow \Delta S=\frac{nRT\,{\ln  \left(\frac{V_f}{V_i}\right)\ }}{T}=nR\,{\ln  \frac{V_f}{V_i}\ }\]
\[\Rightarrow \ \ n=\frac{\Delta S}{R\left({\ln  \frac{V_f}{V_i}\ }\right)}=\frac{22}{8.314\times {\ln  \left(\frac{3.4}{1.3}\right)\ }}=2.75\ {\rm mol}\] 
 

(a) The work done by the gas in an isothermal process is $W=nRT\,{\ln  \frac{\displaystyle V_f}{\displaystyle V_i}\ }$, so 
\[W_{by}=4\left(8.314\right)\left(400\right){\ln  \left(\frac{2V_1}{V_1}\right)\ }\ \sim \ 9220\ {\rm J}\] 

(b) The entropy change of a system is $\Delta S=\Delta Q_{in}/T$, so use the first law of thermodynamics to find the heat absorbed by the gas as 
\[{\left(\Delta U_{int}\right)}_{isothermal}=W_{on}+Q_{in}=0\Rightarrow Q_{in}=-W_{on}\] 
Recall that the work done on the gas is equal to the minus of the work done by the gas i.e. $W_{on}=-W_{by}$, thus $Q_{in}=-W_{on}=W_{by}$
\[\Delta S=\frac{W_{by}}{T}=\frac{nRT\,{\ln  2\ }}{T}=nR\,{\ln  2\ }=\left(4\right)\left(8.314\right){\ln  2\ }=+23.05\frac{{\rm J}}{{\rm K}}\] 
(c) In adiabatic processes the heat exchange is zero i.e. no heat transfers into or out of the system, $Q_{in}=Q_{out}=0$ so using the definition of the entropy change we have
\[\Delta S=\frac{\Delta Q_{in}}{T}=0\] 
This is always true for any reversible adiabatic process.
 

From the equation of state of an ideal gas, we have $P\Delta V=nR\Delta T$ combine this with the definition of the work done on the gas i.e. $W_{on}=-P\Delta V$ to find the desired temperature.  
\[W_{on}=-P\Delta V=-1590\ {\rm J}\] 
Since the volume is contracted so the work done on the gas, by convention, is negative i.e. $-1590\ {\rm J}$.
\[P\Delta V=nR\Delta T\to \ -W=nR\left(T_f-T_i\right)\ \]
\begin{align*}\Rightarrow T_f&=-\frac{W}{nR}+T_i\\ &=-\frac{-1590}{\left(84.6\right)\left(8.314\right)}+\left(273+12.6{\rm{}^\circ\,\!C}\right)\\ &=287.86\ {\rm K}\end{align*} Use the relation $T\left(K\right)=273+T\left({\rm{}^\circ\,\!C}\right)$ to convert Kelvins to Celsius as \[ \Rightarrow T\left({\rm{}^\circ\,\!C}\right)=287.86-273=14.86\ {\rm{}^\circ\,\!C} \]

(a) In a adiabatic process, we have $PV^\gamma={\rm Constant}$. Where $\gamma$ is the ratio of the heat capacities i.e. $\gamma=C_P/C_V$. So between initial and final states we have $P_iV^\gamma_i=P_fV^\gamma_f$ 
\begin{align*} \Rightarrow P_f&=P_i{\left(\frac{V_i}{V_f}\right)}^\gamma\\ &=\left(2\times {10}^5\right){\left(\frac{0.2}{0.2}\right)}^{\frac{5}{3}}\\ &=6.34\times {10}^5\ {\rm Pa}\end{align*}Recall that in the monatomic case $C_V=\frac{3}{2}R\ $and $C_P=C_V+R=\frac{5}{2}R$.
(b) One of the other relations for an adiabatic process is $TV^{\gamma-1}={\rm Constant}$. Using this between initial and final points $T_iV^{\gamma-1}_i=T_fV^{\gamma-1}_f$ we have
\begin{align*} \Rightarrow T_i&=T_f{\left(\frac{V_f}{V_i}\right)}^{\gamma-1}\\ &=305{\left(\frac{0.1}{0.2}\right)}^{\frac{5}{3}-1}\\ &=192.13\ {\rm K} \end{align*} 
(c) The rms speed of a molecule of a gas is related to the absolute temperature by 
\[v_{rms}=\sqrt{\frac{3RT}{M}}\quad {\rm or}\quad v_{rms}=\sqrt{\frac{3kT}{m}}\] 
Where $M$ is the molar mass of the molecule and $m$ is the mass of the molecule.
\[v_{rms}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3\times 1.381\times {10}^{-23}\times 305}{3.35\times {10}^{-26}}}=614.16\frac{{\rm m}}{{\rm s}}\] 
(d) The change in the internal energy of a system is given by $\Delta U_{int}=nC_V\Delta T$ that for an ideal monatomic gas is $\Delta U_{int}=\frac{3}{2}nR\Delta T$. Therefore, first compute the number of moles i.e. $n$ as
\begin{gather*} P_iV_i=nRT_i\\ \Rightarrow \left(2\times  {10}^5\right)\left(0.2\right)=n\left(8.314\right)\left(192.13\right)\\ \Rightarrow n=25.04\ {\rm mole} \end{gather*} Now substitute $n$ in the definition of the internal energy
\begin{align*} \Delta U_{int}&=\frac{3}{2}nR\left(T_f-T_i\right)\\ &=\frac{3}{2}\left(25.04\right)\left(8.314\right)\left(305-192.13\right)\\ &=3.52\times {10}^4\ {\rm J} \end{align*} 
(e) Since the heat exchange in an adiabatic process is zero i.e. $Q_{in}=0$ use first law of thermodynamics to find the work done by the gas
\[\Delta U_{int}=W_{on}+Q_{in}\Rightarrow \Delta U_{int}=W_{on}\] 
By convention, $W_{on}=-W_{by}$, thus 
\[\Delta U_{int}=-W_{by}\] 
\[\therefore \ W_{by}=-3.52\times {10}^4{\rm \ J}\] 
The negative sign indicates that the work is done by an external agent on the gas to compress it. 
 

(a) We have an important relation for Carnot engine that relate temperatures and heat exchanges as 
\[\frac{\left|Q_c\right|}{Q_h}=\frac{T_c}{T_h}\] Use this relation to find the heat $Q_c$ released to the cold reservoir.
\begin{align*} \frac{\left|Q_c\right|}{Q_h}=\frac{T_c}{T_h} \\ \Rightarrow \left|Q_c\right|&=Q_h\frac{T_c}{T_h}\\&=2000\, \left(\frac{100+273}{500+273}\right)\\&=965.1\quad {\rm J\ }\end{align*} \[\Rightarrow Q_c=-965.1\quad {\rm J}\] We have inserted, by convention, a negative to indicate that the heat leaves the cycle.

(b) In the Carnot engines (cycles) from first law of thermodynamics, we have
\begin{gather*}W_{net}+\left|Q_c\right|=Q_h \\ \\ \Rightarrow W_{net}=2000-965.1=1034.9\quad{\rm J}\end{gather*}
Where $W_{net}$ is the work done by the engine.

(c) The efficiency of an Carnot engine is given by 
\[e=1-\frac{T_c}{T_h}=1-\frac{373}{773}=0.517=51.7\%\] 

(a) To piston fall freely, its weight must be greater than the force provided by the pressure $P_1$ that is $F_1=P_1A$. Therefore,
\[W>F_1\Rightarrow M_Pg>P_1A\] 
\[\ \therefore \ P_1<\frac{M_pg}{A}\] 

(b) When the force produced ($F_2=P_2A$) by the pressure at later time balanced with the weight of the piston, the system reaches the final equilibrium state, that is  $M_pg=P_2A$. Use the ideal gas law $PV=nRT$ between initial and final states to find the final height $h_2$.
\[P_1V_1=Nk_BT=P_2V_2\ {\rm since\ temperature\ is\ constant}\] 
\[P_1Ah=P_2Ah_2\] 
\[\therefore h_2=\frac{P_1A}{P_2A}h=\frac{P_1A}{M_Pg}h\] 
In the last line, we have used the final equilibrium condition. 
 

(a) Average kinetic energy ($\bar{K}$) of a molecule with temperature $T$ is $\bar{K}=\frac{3}{2}k_BT$ so for $N_A$ molecules have: \[N_A\underbrace{\left({\frac{1}{2}mv^2}\right)}_{\bar{K}}=\frac{3}{2}N_Ak_BT=\frac{3}{2}RT\] where $mN_A=M$ is the molar mass. 

The criteria for the escape of $N_A$ molecules from the surface of a planet is stated as the kinetic energy $\ge$ potential energy
\[ \frac{3}{2}RT\ge \left(N_Am\right)gR_P\Rightarrow T>\frac{2N_AmgR_P}{3R}\] 
Thus the minimum temperature achieved by $N_2$ is \[T_{N_2}=\frac{2(N_Am)gR_E}{3R}=\frac{2MgR_E}{3R}\] Substituting the numerical values into it, gives \begin{align*} T_{N_2}&=\frac{2\times 28\times {10}^{-3}\times 9.8\times 6.38\times {10}^6}{3\times 8.31} \\\\ &=1.4\times {10}^5\,\rm K \\\\ \Rightarrow \quad T_{H_2}&=\frac{2}{28}T_{N_2}\\\\& =1\times {10}^4{\rm K} \end{align*}
(b) First, find the ratio of $gR$ on the moon to the earth, and then substitute it into the minimum temperature achieved by a molecule to escape the surface.\begin{align*} \frac{g_MR_M}{g_ER_E}&=\frac{(1.63)(1740)}{(9.8)(6380)} \\\\ &=4.53\times {10}^{-2} \\\\ T_{N_2}&=\frac{2Mg_MR_M}{3R} \\\\&=\left(4.53\times {10}^{-2}\right)\frac{2Mg_ER_E}{3R} \end{align*} Therefore, the temperature required to escape $N_2$ molecules off the surface of the moon is \[T_{N_2}=4.53\times {10}^{-2}\left(1.40\times {10}^5\right)=6.35\times {10}^3\, {\rm K}\] and with similar calculation for $H_2$ is \[T_{H_2}=4.53\times {10}^{-2}\left(1\times {10}^4\right)=453\ {\rm K}\] 
At a given temperature, the escape temperatures of $N_2$ and $H_2$ for the moon are much less than for the earth. 

(c) Similar to the previous, \[\frac{g_sR_s}{g_ER_E}=\frac{\left(274.2\right)\left(6.96\times {10}^5\right)}{\left(9.8\right)\left(6380\right)}=3.06\times {10}^3\] 
\[T_{N_2}=\left(3.06\times {10}^3\right)\left(1.4\times {10}^5\right)=4.28\times {10}^8\ {\rm K}\] 
\[T_{H_2}=\left(3.06\times {10}^3\right)\left(1\times {10}^4\right)=3.06\times {10}^7\ {\rm K}\] 
Sun is too big and not hot enough to provide temperature for escaping: 

Ratio is $\frac{3.06\times {10}^7}{5800}=5.27\times {10}^3$. Almost nothing escapes. 

Note: in an diatomic gas, we have $C_V=\frac{5}{2}R$ and $C_P=C_V+R=\frac{7}{2}$
(a) First, use the equation of state of ideal gas to find the volume of point 1:
\[PV=nRT\to V_1=\frac{nRT_1}{P_1}=\frac{\left(0.350\right)\left(8.31\right)\left(300\right)}{1.01\times {10}^5}=8.62\times {10}^{-3}\ {{\rm m}}^{{\rm 3}}\] 
Since in diagram $V_1=V_2$ use the ideal gas law between theses points to find the $P_2$.
\[\frac{P_2V_2}{T_2}=\frac{P_1V_1}{T_1}\to \frac{P_2}{T_2}=\frac{P_1}{T_1}\ \Longrightarrow P_2=\frac{600}{300}\times 1.01\times {10}^5=2\ {\rm atm}\] 
Process $1\to 3$ is isobar that is $P_1=P_3$, use again the ideal gas law between them
\[\frac{{P_3V}_3}{T_3}=\frac{P_1V_1}{T_1}\ \ \Longrightarrow V_3=\frac{492}{300}\times \left(8.62\times {10}^{-3}\right)=14.1\times {10}^{-3}\ {{\rm m}}^{{\rm 3}}\] 
(b) We know, by definition, the work done on the system is $W_{on}=-P\Delta V$ , 
In $1\to 2$ : $\Delta V=0\ \Longrightarrow W_{1\to 2}=0$
$Q_{1\to 2}=nC_V\Delta T=0.35\times 20.79\times \left(600-300\right)=+2180\ {\rm J}$. Since its value is positive, so by convention, heat flows into the engine. Now use the first law of thermodynamics to find the change in internal energy of the gas.
\[{\left(\Delta U_{int}\right)}_{1\to 2}={\left(Q_{in}\right)}_{1\to 2}+{\left(W_{on}\right)}_{1\to 2}=+2180\ {\rm J}\] 
In $2\to 3$: 
In an adiabatic process heat exchange is zero so, $Q_{in-out}=0$
\[\Delta U_{int}=nC_V\Delta T=\left(0.35\right)\left(20.79\right)\left(492-600\right)=-780{\rm J}\] 
\[\Delta U_{int}=Q_{in}+W_{on}\to W_{on}=\Delta U_{int}-Q_{in}=-780\ {\rm J}\] 
In $3\to 1$:
Isobar process so,
\[W_{on}=-P\Delta V=-\left(1.01\times {10}^5\right)\left(8.62-14.1\right)\times {10}^{-3}=+560\ {\rm J}\] 
\[Q_{in}=nC_P\Delta T=\left(0.35\right)\left(29.11\right)\left(300-492\right)=-1960\ {\rm J}\] 
\[\Delta U_{int}=Q_{in}+W_{on}=-1960+\left(+560\right)=-1400\ {\rm J}\] 
Since $Q_{in}<0$, so some heat transferred into the surrounding. 
(c) In each cycle the total internal energy of a system is zero since $\Delta U_{int}=nC_V\Delta T=\Delta Q_{in}{\rm +}\Delta W_{on}$ And $\Delta {\rm T=0}$ (we have come back to the initial position!), then $\Delta U_{int}=0$and $\Delta Q_{in}=-\Delta W_{on}$. Therefore, the work done on the gas and heat transferred into the system is as follows
\[W_{net\ on}=W_{1\to 2}+W_{2\to 3}+W_{3\to 1}=0-780+560=-220\ {\rm J}\] 
\[Q_{net\ in}=Q_{1\to 2}+Q_{2\to 3}+Q_{3\to 1}=2180+0-1960=+220\ {\rm J}\] 
In general, the work done by the gas is the negative of the work done on the gas so
\[W_{on}=-W_{by}\Rightarrow W_{by}=+220\ {\rm J}\] 
(d) By convention, if $Q>0$ , then heat transferred into the system. In this problem, only $Q\left(1\to 3\right)>0$, so the net heat flows into the engine is $Q_{in}=2180\ {\rm J}$. The thermal efficiency of a heat engine is defined as the ratio of the net work done by the engine during one cycle to the heat transferred into it at higher temperature during the cycle. Therefore,
\[e=\frac{W_{net}}{Q_H}=\frac{+220}{2180}=0.101\ \sim \ 10.1\%\] 
The Carnot engine is the most efficient heat engine that works between $T_h$ and $T_c$ and its efficiency is defined as 
\[e_{Carnot}=1-\frac{T_c}{T_H}=1-\frac{300}{600}=0.5=50\%\] 
As expected $e_{Carnot}>e$.
 

(a) In an isothermal process, temperatures between two points are the same. So use the ideal gas law $PV=nRT$ to find $T_a$ and $T_b$.
\[T_a=\frac{P_aV_a}{nR}=\frac{\left(2\times {10}^5\right)\left(0.01\right)}{\left(1\right)\left(8.314\right)}=240.55\ {\rm K}\] 
\[T_b=\frac{P_bV_b}{nR}=\frac{\left(4\times {10}^5\right)\left(0.005\right)}{\left(1\right)\left(8.314\right)}=240.55\ {\rm K}\] 
Since $T_a=T_b$ so the path $b\to a$ is an isothermal process.
(b) Use the definitions of heat transferred $Q=nC_{V,P}\Delta T$ and work done on the system $W_{on}=-P\Delta V$ to find desired values. Path $ab$ is isothermal. Using first law of thermodynamics $\Delta U_{int}=W_{on}+Q_{in}$ we know that $\Delta U_{int}=0$ so $Q_{in}=-W_{on}=nRT{\ln  \frac{V_b}{V_a}\ }\ \ ,\ \ V_a>V_b\ \Longrightarrow \ \ Q_{in}<0$ so heat is rejected. 

In $bc$ process $P$ is constant, so $=nC_P\Delta T$ , use the ideal gas law to substitute $\frac{P\Delta V}{R}$ for $n\Delta T$ so  $Q=\frac{C_P}{R}P\Delta V$. Since $\Delta V=\left(V_c-V_a\right)>0$ so $Q>0$ and subsequently heat is absorbed.

In $ca$ process $V$ is constant (isochoric process), so $Q=nC_V\Delta T=\frac{C_V}{R}V\Delta P\ ,\ \Delta P<0\ \Longrightarrow Q<0\ $so heat is rejected.

(c) \[T_a=T_b=\frac{P_aV_a}{nR}=\frac{2\times {10}^3}{1\times 8.314}=240.55\ {\rm K}\] 
\[T_c=\frac{P_cV_c}{nR}=\frac{4\times {10}^3}{1\times 8.314}=481.11\ {\rm K}\] 

(d) In part (c) we have find the temperatures of points, so 
\[Q_{bc}=nC_P\Delta {{\rm T}}_{{\rm bc}}{\rm =}\left({\rm 1.0}\right)\left(\frac{{\rm 7}}{{\rm 2}}\right)\left({\rm 8.134}\right)\left({\rm 481.11-240.55}\right){\rm =7.00\times }{{\rm 10}}^{{\rm 3}}\ {\rm J}\] 
\[Q_{ca}=nC_V\Delta T_{ca}=\left(1.0\right)\left(\frac{5}{2}\right)\left(8.314\right)\left(240.55-481.11\right)=-5.00\times {10}^3{\rm J}\] 
\[Q_{ab}=nRT{\ln  \left(\frac{V_b}{V_a}\right)\ }=\left(1.0\right)\left(8.314\right)\left(240.55\right){\ln  \left(\frac{0.005}{0.01}\right)\ }=-1.38\times {10}^3\ {\rm J}\] 
\[\therefore Q_{total}=+620\ {\rm J}\] 
The path $ab$ is isothermal that is $\Delta U_{int}=\frac{3}{2}nR\Delta T={{\rm Q}}_{{\rm in}}{\rm +}{{\rm W}}_{{\rm on}}{\rm =0}$ ,thus the work done on the gas is ${\left(W_{on}\right)}_{ab}=-{\left(Q_{in}\right)}_{ab}=+1.38\times {10}^3\ {\rm J}$. By convention, the work done by the gas is ${(W_{by})}_{ab}=-1.38\times {10}^3{\rm J}$.
\[{\left(W_{on}\right)}_{bc}=-P\Delta V=-\left(4\times {10}^5\right)\left(0.01-0.005\right)=-2000\ {\rm J}\] 
\[{\left(W_{on}\right)}_{ca}=-P\Delta V=0\] 
\[\therefore {\left(W_{on}\right)}_{net}=1380{\rm -}{\rm 2000=-620}\ {\rm J\ and\ }{\left(W_{by}\right)}_{net}{\rm =+620\ J}\] 
\[e=\frac{{\left(W_{by}\right)}_{net}}{Q_H}=\frac{620}{7.01\times {10}^3}=8.84\ \%\] 
 

(a) The translational kinetic energy of a monoatomic molecule such as ($H_2,N_2,\dots )$ is $K_{tr}=\frac{1}{2}m\overline{v^2}$ where $\overline{v}$ is the average velocity of the it. This molecule can move in three direction ($x,y,z$) or have three degrees of freedom (DOF). By Equipartition theorem, each of the DOF contributes $\frac{1}{2}k_BT$ to the kinetic energy. therefore, the total translational kinetic energy of $n$ moles of a gas containing $N$ molecules is
\[K_{tr-ave}=N\left(\frac{1}{2}m\overline{v^2}\right)=N\left(\frac{3}{2}k_BT\right)=\frac{3}{2}nRT\] 
Where we have used $Nk_B=nN_Ak_B=nR$. ($N_A$ is Avogadro's number). In other words, the absolute temperature is a measure of the translational kinetic energy of the molecules. Since container A and B have the same $T$ so $K_{tr-A}=K_{tr-B}$. In the above equation, the square root of $\overline{v^2}$  is an estimation of the order of magnitude of the speeds of the molecules in the gas, that can be determined by $v_{rms}=\sqrt{\overline{v^2}}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3k_BT}{m}}$, where $M=N_Am$ is the molar mass of the molecules. because $v_{rms}\propto \ 1/\sqrt{m}$ and $m_A>m_B$ therefore, the lightest molecule (A) has the greatest $rms$ speed. 
(b) Since B is heavier, we need to raise its energy to get the same $v_{rms}$ as A.
(c) We want ${\left(v_{rms}\right)}_A={\left(v_{rms}\right)}_B$
\[\sqrt{\frac{3k_BT_A}{m_A}}=\sqrt{\frac{3k_BT_B}{m_B}}\Rightarrow T_B=\frac{m_B}{m_A}T_A=\frac{5.34\times {10}^{-26}}{3.34\times {10}^{-27}}\left(283\right)=4531\ {\rm K}\] 
Therefore, we must raise the temperature of the container B to $4531\ {\rm K}$ so that both of them have the same rms speed. Since translational kinetic energy only depends on the absolute temperature, so the B molecules have the greatest translational energy. 
(d) \[v_{rms}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\right)\left(4531\right)}{5.34\times {10}^{-26}}}=1874\frac{{\rm m}}{{\rm s}}\]
\[v_{mp}=\sqrt{\frac{2}{3}}v_{rms}=0.816\times 1874=1529\frac{{\rm m}}{{\rm s}}\] 
\[v_{ave}=\sqrt{\frac{8}{3p}}v_{rms}=0.92\times 1874=1724\frac{{\rm m}}{{\rm s}}\] 
 

(a) Use the ideal gas law, $PV=nRT$ between a and b. 
\[\frac{P_aV_a}{T_a}=\frac{P_bV_b}{T_b}\ \xrightarrow{a\to b\, \text{is isobar process}}\frac{V_a}{T_a}=\frac{V_b}{T_b}\] 
\[\Rightarrow V_b=\frac{T_b}{T_a}\ V_a=\frac{1}{4}V_a\] 
(b) External work done on the gas is defined as $W_{on}=-P\Delta V=-P\left(V_b-V_a\right)$
\begin{align*} W_{on}&=-P\left(\frac{1}{4}V_a-V_a\right)\\&=\frac{3}{4}PV_a\\&=\frac{3}{4}\ \left(1.5\ {\rm atm}\right)\left(0.5\ {\rm L}\right)\\&=+0.5625{\rm \ (lit.atm)}\end{align*}
In the SI units work must be in Joules, we know that ${\rm Pa.}{{\rm m}}^{{\rm 3}}={\rm J}$ so convert the above value as follows
\[+0.5625{\rm \ }\left({\rm lit.atm}\right){\rm =0.5625}\left({{\rm 10}}^{{\rm -}{\rm 3}}{\rm m^3}\right)\left(1.01\times {10}^5{\rm Pa}\right){\rm =56.81\ J}\] 
Since $W>0$ so  work done on the gas.
(c) 
\begin{align*}U_{int}&=nC_V\Delta T=nC_V\left(T_b-T_a\right)\\&=nC_V\left(\frac{1}{4}-1\right)T_a\\&=-\frac{3}{4}nC_VT_a\\&=-\frac{3}{4}C_V\left(\frac{P_aV_a}{R}\right)\\ &=-\frac{3}{4}\left(\frac{3}{2}R\right)\left(1.5\times \left(1.01\times {10}^5{\rm Pa}\right)\right)\frac{\left(0.5\times {10}^{-3}\ {{\rm m}}^{{\rm 3}}\right)}{R}\\&=-85.85\ {\rm J}\end{align*}
In the third line, we have substituted $PV/R$ for $nT$ (from ideal gas law).
Note: for monoatomic gasses $C_V=\frac{3}{2}R$ and in general $C_P=C_V+R$
(d) From the first law of thermodynamics we have
\[\Delta U_{int}=Q_{in}+W_{on}\to Q_{in}=-85.85-56.81=-142.66\ {\rm J}\] 

(a) In any process, the work done by the gas is the area under the $P-V$ diagram.
So we must find the sum of the area of triangle and rectangular.
\begin{align*} W_{by} &= S=S_{tri}+S_{rec}\\&=\frac{1}{2}\left(2\times {10}^5\right)\left(0.01-0.002\right)+\left(1\times {10}^5\right)\left(0.01-0.002\right)\\&=1000+800=1800\, {\rm J}\end{align*}
(b) The change in internal energy of a substance is independent of path and only depends on the initial and final temperature of the path. In general, we have $\Delta U_{int}=nC_V\Delta T$ where $C_V$ is the heat capacity at constant volume and for monoatomic molecules is equals to $\frac{3}{2}R$ per mole. So we must first determine the $T_a$ and $T_b$ by using the ideal gas law $PV=nRT$
\begin{align*}\Delta U_{int}&=\frac{3}{2}nR\left(T_c-T_a\right)\\&=\frac{3}{2}nR\left(\frac{P_cV_c}{nR}-\frac{P_aV_a}{nR}\right)\\&=\frac{3}{2}\left(P_cV_c-P_aV_a\right)\\ &=\frac{3}{2}\left(1\times {10}^5\right)\underbrace{\left(0.01-0.002\right)}_{0.008}\\&=1200\quad {\rm J}\end{align*}
(c) Use the first law of thermodynamics as $\Delta U_{int}=W_{on}+Q_{in}$ where $W_{on}$ is the work done on the gas and $Q_{in}$ is the heat transferred into it. 

Note: recall that $W_{on}=-W_{by}$
\[\Rightarrow \Delta U_{int}=-W_{by}+Q_{in}\Rightarrow 1200=-1800+Q_{in}\] 
\[\Rightarrow Q_{in}=+3000\ {\rm J}\] Since $Q_{in}>0$ so heat absorbed by the gas. 

(a)

(b) The work done by an ideal gas is $W_{by}=P\Delta V$ where $P$ and $V$ must be in Pa and ${{\rm m}}^{{\rm 3}}$, respectively. ($1\ {\rm atm=1.01\times }{{\rm 10}}^{{\rm 5}}{\rm Pa\ ,\ 1lit=}{{\rm 10}}^{{\rm -}{\rm 3}}\ {{\rm m}}^{{\rm 3}}$)
\begin{align*}
W_{a\to b}&=5{\rm atm}\times \left(0.8-0.4\right){\rm lit}\\
&=\left(5\times 1.01\times {10}^5\right)\left(0.4\times {10}^{-3}\right)=202.6\ {\rm J}
\end{align*}
\[W_{b\to c}=0\ {\rm since}\ \Delta V=0\] 
\[\therefore \ \ {\left(W_{by}\right)}_{tot}=W_{a\to b}+W_{b\to c}=202.6\ {\rm J}\] 
Note that the work done on the gas is ${\left(W_{on}\right)}_{tot}=-P\Delta V=-202.6\ {\rm J}$
(c) First law of thermodynamics: 
The change in the internal energy of a system equals the heat transfer into the system plus the work done on the system i.e. $\Delta U_{int}=Q_{in}+W_{on}$.
Note: the internal energy of an ideal gas is $U_{int}=\frac{3}{2}nRT$ i.e. depends only on the absolute temperature of the gas. Since at the final stage $c$, temperature is the same as initial stage $a$, Thus 
\[\Delta U_{int}=\frac{3}{2}nR\left(T_f-T_i\right)=\frac{3}{2}nR\left(0\right){\rm =0}\] 
\[{\left(\Delta U_{int}\right)}_{tot}={\left(Q_{in}\right)}_{tot}+{\left(W_{on}\right)}_{tot}=0\Rightarrow {\left(Q_{in}\right)}_{tot}=-{\left(W_{on}\right)}_{tot}\] 
\[\therefore \ \ {\left(Q_{in}\right)}_{tot}=+202.6\ {\rm J}\] 
Because $Q>0$ then heat transferred into the system.
 

(a) Recall that the equation of state for an ideal gas is given by $PV=nRT$ where $n$ is the total mole of the substance and $R$ is the universal gas constant.

Note: if in a problem there are two states then we must use the equation below:
\[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\] 
Now we have
\[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\ {\rm since\ T\ does\ not\ change\ so\ }P_1V_1=P_2V_2\] 
\[110\times \left(1.8\times 0.9\times 0.15\right)=P_2\left(1.8\times 0.9\times 0.13\right)\] 
\[P_2=\frac{110\times 0.15}{0.13}=126.92\ {\rm kPa}\] 

(b) Using the equation of state 
\[n=\frac{PV}{RT}=\frac{110\times {10}^3\times \left(1.8\times 0.9\times 0.15\right)}{8.315\times \left(273+30\right)}=10.6\ {\rm mol}\] 
Note: in the equation of state $T$ is in kelvins.
 

(a) Use the ideal gas equation of state $PV=nRT$
\[\Rightarrow T_2=\frac{P_2V_2}{nR}=\frac{\left(1.31\times {10}^5\right)\left(24\times {10}^{-3}\right)}{1\left(8.314\right)}=378.15{\rm K}\] 

(b) For adiabatic expansion and compression the ideal gas equation of state is as follows
\[PV^\gamma=Constant\quad {\rm or}\quad TV^{\gamma-1}=Constant\] 
Where $\gamma$ is the ratio of the heat capacities $\gamma=C_P/C_V\ $

Therefore for the process ($2\to 3$) we have $T_2V^{\gamma-1}_2=T_3V^{\gamma-1}_3\Rightarrow T_3=T_2{\left(\frac{V_2}{V_3}\right)}^{\gamma-1}$
\[\Rightarrow \ \ T_3=378.15{\left(\frac{24\times {10}^{-3}}{16\times {10}^{-3}}\right)}^{\frac{5}{3}-1}=491.58\ {\rm K}\] 
Note: the heat capacity at constant pressure and constant volume for an ideal monoatomic gas is $C_P=C_V+R=\frac{5}{2}nR$ and $C_V=\frac{3}{2}nR$, respectively.

(c) The amount of heat absorbed by the gas at constant volume or pressure is $Q_V=C_V\Delta T$ or $Q_P=C_P\Delta T$, respectively. First, find the temperature of point 1.
\[T_1=\frac{P_1V_1}{nR}=\frac{\left(1.31\times {10}^5\right)\left(16\times {10}^{-3}\right)}{\left(1\right)\left(8.314\right)}=252.1\] 
\[Q_{1\to 2}=C_P\Delta T=\frac{5}{2}nR\Delta T=\frac{5}{2}\left(1\right)\left(8.314\right)\left(T_2-T_1\right)=2619.84\ {\rm J}\] 

(d) Use the first law of thermodynamics as $\Delta U_{int}=W_{on}+Q_{in}$, where $W_{on}$ is the work done on the system and $Q_{in}$ stand for the heat transfer into the system. 
\[\left(1\to 2\right):isobaric:\ \left\{ \begin{array}{c} W_{on}=-P\Delta V=-\left(1.31\times {10}^5\right)\left(24-16\right)\times {10}^3=-1048\ {\rm J} \\ Q_{in}=2619.84\ {\rm J} \end{array}\right.\] 
\[\left(2\to 3\right):adiabatic\left\{ \begin{array}{c}W_{on}=C_V\Delta T=\frac{3}{2}nR\Delta T=\frac{3}{2}(1)(8.314)(491.58-378.15)=1414.65)\ {\rm J} \\ Q_{in}=0 \end{array}\right.\] 

\[\left(3\to 1\right):isochoric\left\{ \begin{array}{c}W_{on}=-P\Delta V=0 \\ Q_{in}=C_V\Delta T=\frac{3}{2}\left(1\right)\left(8.314\right)\left(252.1-491.58\right)=-2986.55\ {\rm J} \end{array}\right.\] 

\[\Rightarrow \ W_{tot}=W_{1\to 2}+W_{2\to 3}+W_{3\to 1}=-1048+141465+0=\ +366.35\ {\rm J}\ \ \ \] 
 

(a) Because there is no heat flow in this piston that is $Q=0$ thus we have an adiabatic process. Using the equation of state for an adiabatic process i.e. $PV^\gamma=Constant$. Between initial and final states, we obtain 
\[P_1V^\gamma_1=P_2V^\gamma_2\Rightarrow V_2={\left(\frac{P_1}{P_2}\right)}^{\frac{1}{\gamma}}V_1\]
\begin{align*}\therefore\ \ V_2&={\left(\frac{1}{2}\right)}^{\frac{1}{\frac{5}{3}}}\left({10}^{-3}\right)\\&=6.6\times {10}^{-4}\ {{\rm m}}^{{\rm 3}}\end{align*} Where $\gamma=\frac{C_P}{C_V}=\frac{\frac{5}{2}nR}{\frac{3}{2}nR}=\frac{5}{3}$
(b) By definition, the pressure due to an external force is $P=\frac{F}{A}$, where $A$ is the cross-sectional area of the object.

In this problem, two forces acting on the piston: external force $F$ and air pressure $P_0$. Since we want the head of the piston maintains fixed so the net force on the piston must be zero.\begin{gather*} \Sigma F_{ex}=0\Rightarrow F+P_0A=PA\Rightarrow F=\left(P-P_0\right)A\\\therefore F=\left(2-1\right)\left(1.013\times {10}^5\right)\left(0.01\right)=1.013\times {10}^3\ {\rm N}\end{gather*}

First, draw a free body diagram and then use the equilibrium condition that is $\Sigma F_y=0$ to find the pressure of the piston. 
\[\Sigma F_y=0\Rightarrow P_0A+mg-PA=0\] 
\[\Rightarrow P=P_0+\frac{mg}{A}=P_0+\frac{mg}{\pi r^2}\] 
\[P=\left(1.01\times {10}^5\right)+\frac{50\times 9.8}{\pi{\left(0.05\right)}^2}=1.633\times {10}^5\ {\rm Pa}\] 
Now use the equation of state for the ideal gas confined in the piston i.e. $PV=nRT$
\[\Rightarrow V_{cylinder}=\frac{nRT}{P}=\left(\pi r^2\right)h\Rightarrow h=\frac{nRT}{P\left(\pi r^2\right)}\] 
\[\Rightarrow h=\frac{\left(0.89\right)\left(8.314\right)\left(273+30\right)}{\left(1.633\times {10}^5\right)\pi{\left(0.05\right)}^2}=1.748\ {\rm m=174.8\ cm}\] 
Note: in the equation of state, the temperature of the gas must be in Kelvins i.e. $T\left(K\right)=273+T({\rm{}^\circ\,\!C})$
 

First, draw its $PV$ diagram as 

The work done on the gas is defined as $W_{on}=-P\Delta V$ . The work done by the gas is the negative of the work done on the gas i.e.
\[W_{by}=-W_{on}=P\Delta V\] 
Or the area under a PV diagram represents the work done by the gas. Therefore, we must calculate the area of the shaded trapezoid in the figure. 
\[W_{by}=S_{under}=\frac{\left(P_2+P_1\right)\Delta V}{2}\] 
Find the final volume of the gas using $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\ $
\[\Rightarrow V_2=\frac{T_2}{T_1}\frac{P_1}{P_2}V_1=\frac{300}{390}\frac{300}{80}\left(0.090\right)=0.260\ {{\rm m}}^{{\rm 3}}\] 
\[W_{by}=\frac{\left(80+300\right)\times {10}^3}{2}\ \left(0.260-0.09\right)=32300\ {\rm J}\] 
 

The thermal energy is a form of internal energy or the amount of energy needed to increase the temperature of an object which is given by $Q=nC_VT$, $C_V$ is the heat capacity at constant volume.  For monoatomic ideal gasses $C_V=\frac{3}{2}R$. Therefore, using equation above, find the initial temperature of the gas
\[Q_1=n_1C_{V1}T_1\Rightarrow T_1=\frac{9000}{\left(4.8\right)\frac{3}{2}\left(8.314\right)}=150.3\ {\rm K}\] 
The temperature must be in Kelvins and $R=8.314\ {\rm J/(mol.K)}$. 
After mixing the two gasses the combined gas has the thermal energy 
\[Q_f=Q_1+Q_2=9000+5000=14000\ {\rm J}\] 
Now compute the final temperature of the system as 
\[T_f=\frac{Q_f}{\left(n_1+n_2\right)C_V}=\frac{14000}{\left(4.8+3.4\right)\frac{3}{2}(8.314)}=136.9\ {\rm K}\] 
Thus, the change in the thermal energy of the gas $\#1$ is 
\begin{align*}\Delta Q_1&=n_1C_V\left(T_f-T_1\right)\\ &=\left(4.8\right)\left(\frac{3}{2}\right)\left(8.314\right)\left(136.9-150.3\right)\\&=-802.1\ {\rm J}\end{align*} The negative indicates that the heat is transferred out of the gas $\#1$. 

Note: In a monoatomic molecule, there are three degrees of freedom (DOF) due to the translational motion. 

In a diatomic molecule, in addition to translational symmetry, we have two extra degrees of freedom due to the rotational symmetry, and the two DOF for vibrational symmetry. Therefore, in a diatomic molecule, we have 7 degrees of freedom (DOF). 

Equipartition of energy states that each of the DOF contributes $\frac{1}{2}kT$ per molecule (or $\frac{1}{2}RT$ per mol) to the internal energy of the system. Thus ${\left(E_{int}\right)}_{di}=\frac{7}{2}kT$ but in moderate temperatures that the molecule do not vibrate, the internal energy of a diatomic gas is defined by ${\left(E_{int}\right)}_{di}=\frac{5}{2}kT$.

The total internal energy of the combined gas is the sum of each of the internal energies
\begin{align*} {\left(E_{int}\right)}_{tot}&={(E_{int})}_{mono}+{\left(E_{int}\right)}_{di} \\\\ &=\frac{3}{2}n_1RT_1+\frac{5}{2}n_2RT_2 \\\\ {\left(E_{int}\right)}_{tot}&=\left[\frac{3}{2}(2)(300)+\frac{5}{2}(4)(600)\right]\left(8.314\right) \\\\&=5.74\times {10}^4\, {\rm J}\end{align*} Now use the definition of the internal energy to find the final temperature as \begin{gather*} {\left(E_{int}\right)}_{tot}=C_VT_f=\left(\frac{3}{2}n_{He}R+\frac{5}{2}n_{O_2}R\right)T_f\\ \Rightarrow 5.74\times {10}^4=\left(\frac{3}{2}(2)+\frac{5}{2}(4)\right)(8.314)T_f\\ \\ \Rightarrow \quad \boxed{T_f=531\,\rm K}\end{gather*}

The thermal efficiency of a heat engine is defined as the ratio of the magnitude of the work done by the engine to the magnitude of the heat absorbed from the high-temperature reservoir 
\[e=\frac{\left|W\right|}{\left|Q_H\right|}\] 
Recall that the area under a PV diagram represents the work done by the gas so 
\begin{align*} W_{by}&=S_{triangle}\\ &=\frac{1}{2}\left(200\times {10}^{-6}-100\times {10}^{-6}\right)\left(400\times {10}^3-200\times {10}^3\right)\\ &=10\ {\rm J} \end{align*}
So $e=\frac{\left|W\right|}{\left|Q_H\right|}=\frac{10}{50}=0.2=20\%$
 

The amount of heat that a system absorbs or release at a constant volume is determined by 
\[Q=nC_V\Delta T\xrightarrow{mono}\ Q=\frac{3}{2}nR\Delta T\] 
Where $C_V$is the heat capacity at constant volume and $n$ is the number of moles. So 
\[n=\frac{2Q}{3R\Delta T}=\frac{2\times (-831)}{3\times 8.314\times \left(405-455\right)}=1.33\ {\rm mole}\] 
Note: the minus sign is due to the heat removed (the gas is cooled) from the sample!
 

The mean free path $\lambda$ of a molecule with diameter $d$ and number of molecules per unit volume $n_V$ is determined by $\lambda=1/\sqrt{2}\left(n_V\pi d^2\right)$. First find the $n_V$ then use the $PV=Nk_BT$ to find the desired pressure. 
\[\lambda=\frac{1}{\sqrt{2}\left(n_V\pi d^2\right)}=\frac{1}{\sqrt{2}\left(\frac{N}{V}\right)\pi{\left(2r\right)}^2}\Rightarrow n_V=\frac{N}{V}=\frac{1}{4\sqrt{2}\pi r^2\lambda}\] 
\[\left\{ \begin{array}{c} \frac{N}{V}=\frac{1}{4\sqrt{2}\pi r^2\lambda} \\ P=\frac{N}{V}k_BT \end{array}\right.\]
\begin{align*}\Rightarrow P&=\frac{1}{4\sqrt{2}\pi r^2\lambda}k_BT\\&=\frac{\left(1.381\times {10}^{-23}\right)\left(273+20\right)}{4\sqrt{2}\pi{\left({10}^{-10}\right)}^2\left(4.6\right)}\\&=4.95\times {10}^{-3}{\rm Pa}\end{align*} 
\[or\ \ 4.95\times {10}^{-3}{\rm Pa}\frac{{\rm 1\ atm}}{{\rm 1.01\times }{{\rm 10}}^{{\rm 5}}{\rm Pa}}{\rm =4.9\times }{{\rm 10}}^{{\rm -}{\rm 8}}\ {\rm atm}\] 

The rms of a molecule of a gas is dependent only on the absolute temperature of the gas and is found by the following relation
\[v_{rms}=\sqrt{\frac{3k_BT}{m}}=\sqrt{\frac{3RT}{M}}\] 
Where $M$ is the molar mass of the sample. Since this process is isothermal, the temperature of the gas does not change, so $v_{rms}$ stays the same as the initial speed. 
 

First, draw a schematic representation of a heat engine as below.
The engine absorbs heat $Q_H$ from the hot reservoir at $T_H$, does work $W$, and releases heat $Q_c$ to the cold reservoir at $T_c$.
Note: the efficiency of the Carnot cycle is defined as
\[e_{Carnot}=1-\frac{Q_c}{Q_H}=1-\frac{T_c}{T_H}\] 
Therefore, first calculate the Carnot efficiency of the system then use $e=1-\frac{Q_c}{Q_H}$ to find the heat releases to the cold reservoir $Q_C$.
\begin{gather*} e=1-\frac{T_c}{T_H}=1-\frac{300}{700}=0.571\\ \frac{Q_c}{Q_H}=1-e\Rightarrow Q_c=Q_H\left(1-e\right)\\ \therefore\ \  Q_c=\left(60\times {10}^3\right)\left(1-0.571\right)=25.71\ {\rm J} \end{gather*}

(a) A refrigerator is a heat engine run backward that is its engine absorbs heat from the cold reservoir (interior medium) and releases heat to the hot reservoir (surrounding). For any refrigerator, there is a number called Coefficient of Performance (COP) as the ratio of the magnitudes of heat absorbed from the cold reservoir to the work done on the refrigerator i.e. ${\rm COP}={\left|Q\right|}_c/\left|W\right|$. Another version of the COP is, by using the conservation of energy, ${\rm COP}=T_c/(T_H-T_c)$. Therefore, first convert all temperatures into the Kelvins then substitute in above equation 
\begin{gather*} T_H\left({\rm K}\right)=273+T\left({\rm{}^\circ\,\!C}\right)=273+20=293\ {\rm K}\\T_c\left({\rm K}\right)=273+T\left({\rm{}^\circ\,\!C}\right)=273-10=263\ {\rm K}\\\Rightarrow {\rm COP=}\frac{293}{293-263}=8.77\end{gather*}
(b) All of the processes are as follows
\[Water\ 30{\rm{}^\circ\,\!C}\xrightarrow{Q_1}water\ 0{\rm{}^\circ\,\!C}\xrightarrow{Q_2}ice\ 0{\rm{}^\circ\,\!C}\xrightarrow{Q_3}ice-10{\rm{}^\circ\,\!C}\] 
Now calculate the heats 
\begin{gather*} Q_{lost}=\underbrace{m_wc_c\left(0-30{\rm{}^\circ\,\!C}\right)}_{Q_1}+\underbrace{m_wL_f}_{Q_2}+\underbrace{m_wc_{ice}\left(-10{\rm{}^\circ\,\!C}-0\right)}_{Q_3}\\=0.5\times 4190\left(-30\right)+0.5\times \left(-334\times {10}^3\right)+0.5\times \left(2.01\times {10}^3\right)\left(-10\right)=-239900{\rm J} \end{gather*}
Note: freezing of the water is a reverse process of melting of the ice so $Q=-mL_f$.
(c)Use the definition of the COP and find the work done by the refrigerator's engine
\[COP=\frac{\left|Q_c\right|}{\left|W\right|}\Rightarrow W=\frac{239900}{8.77}=27400\ {\rm J}\ \] 
(d) During a refrigeration process energy is conserved so using the conservation of energy, we obtain 
\begin{gather*} \left|W\right|+\left|Q_c\right|=\left|Q_H\right|\\ \Rightarrow \left|Q_H\right|=27400+239900=267000\ {\rm J} \end{gather*}

Since this process is adiabatic, so its equation of state is
$PV^\gamma=Constant$ or $TV^{\gamma-1}=Constant$. Therefore, use the second equation between points 1 and 2.

\begin{gather*} T_1V^{\gamma-1}_1=T_2V^{\gamma-1}_2\\\Rightarrow {\left(\frac{V_2}{V_1}\right)}^{\gamma-1}=\frac{T_1}{T_2}\\ \Rightarrow V_2={\left(\frac{T_1}{T_2}\right)}^{\frac{1}{\gamma-1}}V_1={\left(\frac{300}{400}\right)}^{\frac{1}{1.5-1}}\left(20\right)\\ =11.25\ {\rm Lit} \end{gather*}
Note: $\gamma$ is the ratio of the heat capacities at constant pressure and volume i.e. $\gamma=C_P/C_V$, which is a dimensionless quantity. 

(a) In adiabatic process, the heat exchange is always zero. So $Q=0$

(b) Using the first law of thermodynamics (energy conservation), we get
\[\Delta U_{int}={{\rm Q}}_{{\rm in}}+W_{on}\] 
Where $W_{on}$ is the work done on the system. By convention, the negative of the work done on the system equals the work done by the system. Therefore, 
\[\Delta U_{int}=0+W_{on}=-120\ {\rm J}\] 

(c) Apply the ideal gas law at point $b$ as follows
\[PV=nRT\ \] 
\[\Rightarrow T_b=\frac{P_bV_b}{nR}=\frac{\left(6\times {10}^{-3}\right)\left(1\times 1.013\times {10}^5\right)}{\left(0.12\right)\left(8.314\right)}=609\ {\rm K}\] 
 

The change in length for a given change in temperature is $\Delta L=L_0a\Delta T$ where $a$ is the coefficient of linear expansion and  $T$ is absolute temperature (must be in Kelvins). 
In this problem, the temperature has given in Fahrenheit so first must be converted to Celsius as $T\left({}^\circ\, F\right)=\frac{9}{5}T\left({\rm{}^\circ\,\!C}\right)+32$. Since an $1\ {\rm K}$ increase in temperature equals a $1{\rm{}^\circ\,\!C}$ that is $\Delta T_c=\Delta T_K$ use this note to convert the change in Fahrenheit to the change in Celsius as
$\Delta T_c=\frac{5}{9}{\Delta T}_F=\frac{5}{9}(412-55.6)=198{\rm{}^\circ\,\!C}$  since $\Delta T_c=\Delta T_K$ we have $\Delta L=L_0a\Delta T$
\begin{align*} \Rightarrow \ \ L_0&=\frac{\Delta L}{a\Delta T}\\&=\frac{3.89\ {\rm mm}}{\left({\rm 6.44\times }{{\rm 10}}^{{\rm -}{\rm 5}}\right)\left(198\right)}\\&={\rm 305\ mm\times }\left(\frac{{\rm 1cm}}{{\rm 10mm}}\right)\\&={\rm 30.5\ cm}\end{align*}

By definition, the coefficient of linear expansion is $a=\Delta L/L_0\Delta T\ $so 
\begin{align*} a&=\frac{L-L_0}{L_0\left(T-T_0\right)}\\\\&=\frac{94.6-92.2}{\left(92.2\right)\left(96.6-22.2\right)}\\\\&=3.50\times {10}^{-4}\,^\circ C\end{align*}

One of the methods of heat transfer is conduction. Let us consider a slab with length $L$ and cross-sectional area $A$ which is located at different temperatures $T_H$ and $T_C$. The rate of heat transfer is computed by the following equation
\[\frac{Q}{t}=kA\frac{T_H-T_C}{L}\] 
The given data are $A_2=\pi{\left(2R\right)}^2=4\pi R^2=4A_1\ ,\ \ L_2=2L_1$ so using the conduction equation, we obtain 
\[\frac{Q_2}{t}=k\frac{A_2\Delta T}{L_2}=k\frac{4A_1\Delta T}{2L_1}\ \] 
\[\Longrightarrow \frac{Q_2}{t}=2\ kA_1\frac{\Delta T}{L_1}=2\frac{Q_1}{t}=2\times 20=40\ {\rm cal/s}\] 
 

(a) The thermal current $dQ/dt$ is the same at the interface so use the following relation for the thermal current through a substance with thermal conductivity $k$ 

\begin{gather*}I=\frac{dQ}{dt}=-kA\frac{\Delta T}{\Delta x}\\ {\rm where}\ \Delta T=T_{cooler}-T_{warmer}\end{gather*}
Since $I_{B\to I}=I_{I\to B}$ therefore 
\begin{gather*} k_B\frac{A\left(T-T_B\right)}{L_B}=K_I\frac{A\left(T_I-T\right)}{L_I}\\\Rightarrow T=\frac{\frac{k_IT_I}{L_B}+\frac{k_BT_B}{L_B}}{\frac{k_B}{L_B}+\frac{k_I}{L_I}}=10.04{\rm{}^\circ\,\!C}\end{gather*} 
(b) Since the inner and outer temperature is constant, so no heat accumulated in the wall. Therefore, the heat conducted through the brick must equal to the insulation, that is ${\left(\frac{Q}{t}\right)}_B={\left(\frac{Q}{t}\right)}_I$. Choosing ${\left(\frac{Q}{t}\right)}_I$, we have
\begin{align*} {\left(\frac{Q}{t}\right)}_I&=-k_I\frac{A\left({T-30}{\rm{}^\circ\,\!C}\right)}{l_{I}}\\&=-9\times {10}^{-2}\frac{\left(10\right)\left(10.04{\rm{}^\circ\,\!C}-30{\rm{}^\circ\,\!C}\right)}{0.4}\\&=44.91\ {\rm J}\end{align*} 
(c) When we have a temperature drop $\Delta T$ across a length $L_0$, due to the thermal expansion, the change in the length is given by $\Delta L=L_0a\Delta T$. Therefore,
\[\Delta L=\left(1\right)\left(1.7\times {10}^{-5}\right)\left(30-10\right)=3.4\times {10}^{-4}\ {\rm m}\] 

The heat $Q$ conducted through a bar of length $L$ and cross sectional area $A$ during a time $t$ is given by the
\[Q=k\frac{A\Delta T}{L}t\] 
where $k$ is the thermal conductivity of the material and $\Delta T=T_H-T_C$.\begin{align*} Q&=k\frac{A\left(T_H-T_C\right)}{L}t\\ \\&=\left(0.8\right)\frac{\left(1.7\times 1.4\right)\left(16-4\right)}{0.002}\left(60\times 60\right)\\ \\&=4.112\times {10}^7\ {\rm J} \end{align*}

Note: the net power radiated by an object with surface area $A$ at temperature  $T$ to its environment at $T_0$ is determined by the Stefan-Boltzmann law as $P_{net}=e\sigma A(T^4-T^4_0)$ where $e$ is the emissivity of the object and $\sigma$ is Stefan-Boltzmann constant. Therefore,
\[P_{net}=\left(0.895\right)\left(1.2\right)\left(5.6703\times {10}^{-8}\right)\left({300}^4-{290}^4\right)=62.55{\rm W}\] 
Note: $T$ must be in Kelvins. ($T\left(K\right)=273+17{\rm{}^\circ\,\!C}=290{\rm K}$)
 

(a) The power radiated by an object with surface area $A$ at absolute temperature $T$ is given by 
\[P=Ae\sigma T^4\] 
Where $e$ is emissivity and $\sigma$ is the Stefan- Boltzmann constant.
Therefore, we must first find the absolute temperatures of skin and surroundings
\begin{align*} T_{skin}&=273+T(\rm ^\circ C) \\ &=273+27=300\quad \rm K \\\\ P_{emit}&=Ae\sigma T^4 \\&=1.4\times 1\times (5.67\times {10}^{-8}) \times (300)^4 \\ &=643\quad \rm W \end{align*}
(b) Same as the previous part
\begin{align*}T_{surr}&=273+T(\rm ^\circ C) \\&=273+17=290\quad \rm K \\\\ P_{surr}&=AesT^4\\ &=1.4\times 1\times (5.67\times {10}^{-8})\times (290)^4 \\ &=562\quad \rm W \end{align*}
(c) The net power radiated is the difference of them as follows
\begin{align*} P_{net}&=P_{skin}-P_{surr} \\ &=643-562 \\& =81\quad \rm W \end{align*}

(a) By definition, mass is equals to the density times the volume.
\[m=\rho V=1.2\times \left(10\times 20\times 50\right)=12000\ {\rm kg}\] 
(b) All of the kinetic energy of the train is converted to the heat and is absorbed by the air so 
\begin{gather*} K=\frac{1}{2}mv^2=Q\\\Rightarrow Q=K=\frac{1}{2}\left(30000\right){\left(20\right)}^2=6\times {10}^6\ {\rm J} \end{gather*}
(c) Use the equation $Q=mc\Delta T$ to find the change in the temperature due to the heat absorbed by the air.
\[\Delta T=\frac{Q}{m_{air}c}=\frac{6\times {10}^6}{\left(12000\right)\left(1020\right)}=0.49\ {\rm K}\] 
Note: since the temperature increment of $1\ {\rm{}^\circ\,\!C}$ is the same as an increment of $1{\rm K}$ so the temperature increase of the air is equal to $0.49\ {\rm{}^\circ\,\!C}$. 
 

In this question, there is a change in the phase of water, so we must use the well-known formula $Q=mc\Delta T$, as we learned on the specific heat practice problems page. First, find the heat required to change the temperature of the water from ${\rm 10^\circ C}$ to ${\rm 30^\circ C}$ and then using the definition of the power determine the time required as follows
Since there is no change in the phase of water and only its temperature is changed so the heat required is calculated as $Q=mc\mathrm{\Delta }T$, where $c$ is the specific heat of the matter. Thus \begin{align*}Q&={\rm (1000 \,g)\left(1\,\frac{cal}{g.K}\right)(30-10)} \\ &={\rm 20\times 10^3 \, cal} \\ &={\rm 20\quad kcal} \end{align*} 
Note: the temperature difference of one Kelvin exactly coincides with one Celsius degree. 
Power is defined as the ratio of energy to time. In the definition of power, all quantities must be in SI units so we must convert cal to Joules as \begin{gather*} 20000\ \rm{cal\times 4.18\ J=83600\ J} \\\\ P=\frac{Q}{t}\Rightarrow t=\frac{Q}{P}=\frac{83600}{400}=209\rm{s=3.48\ min} \end{gather*} 

The correct answer is C.