Specific Heat Problems with Answers for AP Physics
Several specific heat practice problems with solutions are provided for high school students. Each problem discusses the definition and formula of specific heat.
Specific heat capacity is defined as the amount of heat energy, Q, required to raise the temperature of a unit mass (one kilogram or one gram) of a substance by one degree Celsius or one degree Kelvin.
If $m$ kilogram of a substance absorbs $Q$ joules of heat energy from a heat source and its temperature changes by $\Delta T=T_{final}-T_{initial}$, then its specific heat capacity is found by the following formula \[c=\frac{Q}{m\Delta T}\] This equation is only applicable when there is a temperature change in substance.
Problems involving phase change are excluded from using this equation. Refer to the following pages for them:
Heat of fusion: formula and solved problems
Heat of vaporization: formula and solved problems
With this information in mind, we can now solve some specific heat problems to further understand its definition.
Specific Heat Problems
Problem (1): An chunk of steel with a mass of 1.57 kg absorbs a net thermal energy of $2.5\times 10^{5}$ J and increases its temperature by 355°C. What is the specific heat of the steel?
Solution: the specific heat of a substance is defined as the energy transferred ($Q$) to a system divided by the mass ($m$) of the system and the change in its temperature ($\Delta T$). Its formula is \[c\equiv \frac {Q}{m\Delta T}\] Substituting the given numerical values into the above formula, we have \begin{align*} c&=\frac {Q}{m\Delta T}\\ \\ &=\frac{2.5\times 10^{5}}{(1.57)(273+355)}\\ \\&=4485.51\quad{\rm J/kg\cdot\,^{\circ}C}\end{align*} Where ${\rm J/kg\cdot\,^{\circ}C}$ is the SI unit of specific heat.
Problem (2): To one kilogram of water, 4190 J thermal energy is added so that its temperature raises by one Kelvin degree. Find the specific heat capacity of water.
Solution: The specific heat capacity is the amount of heat energy required to raise the temperature of a system with a mass of $m$ by $\Delta T$ so applying its formula we get \begin{align*} c&=\frac {Q}{m\Delta T}\\ \\&=\frac{4190\,J}{(1\,kg)(1\,K)}\\ \\&=4190 \quad{\rm J/kg\cdot K}\end{align*}
Problem (3): m kilogram of ice is heated, and its temperature increases from -10°C to -5°C. A larger amount of energy is added to the same mass of water, and the change in its temperature is from 15°C to 20°C. From these observations, compare the specific heat capacities of ice and water.
Solution: The heats added to the ice and water are $Q_i=m_ic_i \Delta T_i$ and $Q_w=m_wc_w \Delta T_w$, respectively. Now, solving for mass m and equating them (since it is said in the question that the mass of both samples is the same), we can get \begin{align*} m_i &= m_w\\\frac{Q_i}{c_i\Delta T_i}&=\frac{Q_w}{c_w \Delta T_w}\\ \\ \Rightarrow \frac{c_i}{c_w} &= \Big(\frac{Q_i}{Q_w}\Big)\Big(\frac{\Delta T_w}{\Delta T_i}\Big)\\ \\ &=\Big(\frac{Q_i}{Q_w}\Big) \Big(\frac{20-15}{-5-(-10)}\Big)\\ \\&=\Big(\frac{Q_i}{Q_w}\Big)\end{align*}Since $Q_i < Q_w$, we conclude that the specific heat capacity of ice is less than that of water.
A look at the specific heat table can confirm this. $c_i= 2090\,{\rm J/kg.K}$ and $c_w=4190\,{\rm J/kg.K}$.
Problem (4): The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 47 calories of heat. What is the specific heat of iron?
Solution: Initial and final temperatures are $T_i=50.4^\circ$ and $T_f=25^\circ$, respectively. Applying the specific heat equation and substituting the numerical values, we get \begin{align*} c&=\frac{Q}{m\Delta T}\\ \\&=\frac{47\, {\rm cal}}{1\,{\rm g}(50.4-25)}\\ \\ &=1.850\quad {\rm cal/g\cdot K}\end{align*}
Problem (5): the temperature of a 200-g sample of an unknown substance changed from 40°C to 25°C. In the process, the substance released 569 calories of energy. What is the specific heat capacity of the substance?
Solution: again applying the specific heat definition, we have \begin{align*} c&=\frac{Q}{m\Delta T}\\ \\ &=\frac{-569\, {\rm cal}}{200\,{\rm g}\,(25-40)}\\ \\&=0.19 \quad {\rm cal/g\cdot K}\end{align*} since thermal energy has left the substance so a negative sign is placed.
Problem (6): A 200-g sample of an unknown object is heated using 100 J such that its temperature rises by 2°C. What is the specific heat of this unknown object?
Solution: in this problem, the transferred heat $Q$ and the change in temperature $\Delta T$ are given. Using specific heat capacity formula we have \begin{align*} c&=\frac{Q}{m\Delta T}\\ \\ &=\frac{100\, {\rm J}}{0.2\,{\rm kg}\,(2\,^\circ {\rm C})}\\ \\&=500 \quad {\rm J/kg\cdot ^\circ C}\end{align*}
Problem (7): What is the specific heat of metal if its mass is 27 g and it requires 420 J of heat energy to increase its temperature from 25°C to 50°C?
Solution: applying the specific heat formula, we get \begin{align*} c&=\frac{Q}{m\Delta T}\\ \\&=\frac{420\,{\rm J}}{0.27\,{\rm kg}\times (50-25)}\\ \\&=62.2\quad {\rm \frac{J}{kg\cdot\,^\circ C}}\end{align*} Be sure to convert grams to kilograms.
In another type of problem, the specific heat capacity is determined by calorimetry.
In calorimetry problems, the specific heat of a sample is determined by inserting it into a certain amount of water with a known temperature.
This system is isolated so the sum of all energy gains or losses by all objects in the system is zero i.e. $\Sigma Q_i=0$ where $Q_i$ is the energy of the ith object.
If we measure the temperature of the equilibrium point, then we can solve the equation above for the unknown specific heat to find it.
Below, some problems for finding the heat capacity using this method are presented.
Problem (8): A 125-g block of a substance with unknown specific heat at a temperature of 90°C is placed in an isolated box containing 0.326 kg of water at 20°C. The equilibrium temperature of the system is 22.4°C. What is the block's specific heat?
Solution: since the temperature of the block is more than the water so the block loses thermal energy and water gains it. Equate these, $Q_{gain}=-Q_{lose}$ and solve for the unknown specific heat. \begin{align*} m_w c_w (T-T_w) &= -m_x c_x (T-T_x)\\ \\ \Rightarrow c_x&=\frac{m_w c_w (T-T_w)}{m_x (T_x-T)}\\ \\ &=\frac{(0.326)(4190)(22.4-20)}{(0.125)(90-22.4)}\\ \\&=388\quad {\rm J/kg\cdot\, ^\circ C}\end{align*}
In calorimetry problems, thermal energy is transferred from the warmer substance to the cooler object.
Therefore, a negative is placed in front of the warmer object to ensure that both sides of the above equality become positive.
For more practice about this, check out the page on thermodynamics problems and solutions.
Problem (9): A piece of metal of unknown specific heat, weighing 25 g and temperature 90°C, is dropped into 150 g of water at 10°C. They finally reach thermal equilibrium at a temperature of 24°C. Calculate the unknown specific heat capacity. (Specific heat of water is 1.0 cal/g°C)
Solution: Since the equilibrium temperature $T_f$ is lower than the metal one, it loses as much heat as $Q_m=-m_m c_m (T_f-T_m)$ and the water gains heat as much as $Q_w=m_w c_w (T_f-T_w)$.
As a rule of thumb keep in mind that, once any objects lose heat we should place a negative in front of it.
Conservation of energy says that $Q_w=Q_m$, therefore we have \begin{align*} m_w c_w (T_f-T_w)&= -m_m c_m (T_f-T_m) \\ \\ \Rightarrow c_m &= \frac{m_w c_w (T_f-T_w)}{m_m \,(T_m-T_f)}\\ \\ &=\frac{(150)(1)(24-10)}{25\times(90-24)}\\ \\&=1.27\quad {\rm cal/g\cdot\,^\circ C}\end{align*}
Problem (10): a piece of unknown substance with a mass of 2 kg and a temperature of 280 Kelvins is brought in thermally contact with a 5-kg block of copper initially at 320 K.
This system is in isolation. After the system reaches thermal equilibrium, its temperature becomes 290 K.
What is the specific heat of the unknown substance in cal/g.°C? (Specific heat of the copper is 0.093 cal/g.°C).
Solution: copper loses heat and the unknown substance gains that amount of lost heat, by considering complete isolation. As previous questions, we get \begin{align*} m_s c_s (T_f-T_s)&= -m_c c_c (T_f-T_c) \\ \\ \Rightarrow c_s &= \frac{m_c c_c (T_c-T_f)}{m_s \,(T_f-T_s)}\\ \\&=\frac{(5000)(0.093)(320-290)}{2000\times(290-280)}\\ \\ &=0.6975 \quad {\rm cal/g\cdot K}\end{align*} In above, the subscripts $s$ and $c$ denote the substance and copper, respectively.
Note: in calorimetry equations where a temperature change appears, it's possible to use either Celsius or Kelvin temperatures because a change in Celsius equals a change in Kelvin.
Problem (11): a 20-g block of a solid initially at 70°C is placed in 100 g of a fluid with a temperature of 20°C. After a while, the system reaches thermal equilibrium at a temperature of 30°C. What is the ratio of the specific heat of the solid to that of the fluid?
Solution: solid loses heat and fluid absorbs it so $Q_f=-Q_s$ where f and s denote the fluid and solid, respectively. From the definition of specific heat, we can find the heat transferred during a process as below \begin{align*} m_f c_f (T-T_f) &= -m_s c_s (T-T_s)\\ \\ \Rightarrow \frac{c_f}{c_s} &=\frac{m_s (T_s-T)}{m_f (T-T_f)}\\ \\ &= \frac{(20)(70-30)}{(100)(30-20)}\\ \\ &=0.8\end{align*} Therefore, the specific heat capacity of fluid is less than that of solid.
Problem (12): A chunk of metal with a mass of 245.7 g at 75.2°C is placed in 115.43 g of water initially at a temperature of 22.6°C. The metal and water reach the final equilibrium temperature of 34.6°C. If no heat is exchanged between the system and its surrounding, what is the specific heat of the metal? (specific heat of water is 1 cal/g °C).
Solution: Since no heat is lost to the surrounding, so by conservation of energy we have $Q_{water}=-Q_{metal}$
\begin{align*}Q_{water}&=-Q_{metal}\\ \\m_w c_w (T_f-T_w) &=-m_{metal} \, c_{metal} (T_f-T_{metal})\\ &\\ \Rightarrow c_{metal}&=\frac{m_w c_w (T_f-T_w)}{m_{metal}(T_{metal}-T_f)}\\&\\ &=\frac{(115.43)(1)(34.6-22.6)}{(245.7)(75.2-34.6)}\\ \\&=0.13 \quad {\rm cal/g\cdot^\circ C}\end{align*} where $T_f$ is the equilibrium temperature.
Now, we solve some example problems with specific heat capacity using calorimetry.
Example Problem (13): how many liters of water at 80°C should be mixed with 40 liters of water at 10°C to have a mixture with a final temperature of 40°C?
Solution: conservation of energy tells us that $\Sigma Q=0$ that is $Q_{lost}+Q_{gain}=0$. By considering no heat exchange with the environment, hotter objects lost heat and colder ones gain it. Thus, the amount of heat lost by 80°C - water is \begin{align*} Q_{lost}&=mc(T_f-T_i)\\&=mc(40-80)\\&=-40mc\end{align*} and also the heat gained by 40°C - water is \begin{align*} Q_{gain}&=m'c(T_f-T_i)\\&=40\times c\times (40-10)\\&=1200c\end{align*}Therefore, by equating those two heats, we get the unknown mass of the water \begin{align*} Q_{gain}&=-Q_{lost}\\1200c&=-(-40mc)\\1200&=40m \Rightarrow m&=30,{\rm Lit}\end{align*}
Example Problem (14): 200 g of water at 22.5°C is mixed with 150 grams of water at 40°C, after reaching the thermal equilibrium, what is the final temperature of the water?
Solution: using the principle of conservation of energy, we have $Q_{lost}+Q_{gain}=0$. Substituting the specific heat equation into it, we get \begin{gather*} m_1 c_w (T_f-T_1)+m_2 c_w (T_f-T_2)=0\\ \\ 200 (T_f-22.5)+150 (T_f-40)=0\\ \\ \Rightarrow 350T_f=10500 \\ \\ \Rightarrow T_f= 30^{\circ}C \end{gather*} where $T_f$ is the equilibrium (final) temperature of the mixture. Also, the specific heat capacity of the water is dropped from both sides.
Example Problem (15): in a container, 1000 g of water and 200 g of ice are in thermal equilibrium. A piece of metal with a specific heat capacity of $c_m=400\,{\rm J/kg.K}$ and a temperature of 250°C is dropped into the mixture.
How much should the minimum mass of the metal be to melt down all of the ice? (heat of fusion of ice is $L_f=336\,{\rm \frac{kJ}{kg}}$.
Solution: minimum mass means that the metal has enough heat energy to only melt down the ice completely and not raise the temperature of the system. Thus, the amount of heat energy released by $m$ gram of the metal to reach from 250°C to 0°C is $Q_{lost}=mc(T_f-T_i)$ where $T_f$ is the equilibrium temperature which is 0°C.
On the other hand, the required energy for melting 200 g of ice is also obtained as below \[Q_{gain}=m_{ice} L_f=200\times 336\times 10^{3}\,{\rm \frac{J}{kg}}\] where $L_f$ is the latent heat of the ice. Now, conservation of energy tells us that equate them and find the unknown metal's mass \begin{gather*}m_{metal}\,c(T_f-T_i)+m_{ice}L_f=0 \\ \\ m_{metal} \times 400\times (0-250)+200\times 336 \times 10^{3}=0 \\ \\ \Rightarrow m_{metal}=672 \quad {\rm g} \end{gather*} If the metal's mass is greater than this value, the excess heat will cause the final temperature of the system goes higher.
For more exercise, you can also practice these heat problems.
In the end, we provided some specific heat capacities at room temperature.
| Substance |
Specific heat J/kg.K |
Specific heat cal/g.K |
|
Brass |
380 | 0.092 |
| Copper | 386 | 0.0923 |
| Glass | 840 | 0.20 |
| Aluminum | 900 | 0.215 |
| Mercury | 140 | 0.033 |
| Water | 4187 | 1.00 |
| Sea Water | 3900 | 0.93 |
| Ice (-10°C) | 2220 | 0.530 |
| Ethyl Alcohol | 2430 | 0.58 |
Author: Dr. Ali Nemati
Last Update: 1/9/2021
