Specific Heat Problems with Answers

Specific heat capacity measures the heat energy needed to raise the temperature of one unit of mass by one degree Celsius or Kelvin. This guide includes practice problems with solutions to help students strengthen their understanding of heat transfer calculations.

This guide provides several practice problems with solutions to help the students grasp this concept.

Definition:

If a substance of mass $m$ (in kilogram) 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 given by: \[c=\frac{Q}{m\Delta T}\] This equation is only applicable a temperature change occurs.

However, for the phase change process, this equation does not apply because the temperature remains constant, even though energy is being absorbed or released. Since specific heat capacity depends on temperature change, a different approach is required for such processes.

For a deeper understanding of these thermal processes and how energy calculations work during phase changes, refer to the following resources:

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 understand its definition further.

Specific Heat Problems

Problem (1): A 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): 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, a negative sign is included to indicate heat loss.

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 heat added to the ice and water is given by: $Q_i=m_ic_i \Delta T_i$ and $Q_w=m_wc_w \Delta T_w$, respectively. Since the problem states that both samples have the same mass, we set $m_i=m_w$, which allows us to equate their respective heat expressions. Solving for the ratio of specific heat capacities, gives us: \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 lower than that of water.

A quick check of a specific heat table confirms this. $c_i= 2090\,{\rm J/kg \cdot K}$ and $c_w=4190\,{\rm J/kg \cdot K}$.

Problem (4): A 1.5-kg copper block is initially at 25°C. If 45 kJ of energy are added to the block, what is its final temperature?

Solution: Using the specific heat formula and rearranging and solving for $T_f$, we have: \[T_f=T_i+ \frac{Q}{mc}\] Substituting the numerical values into this gives: \begin{align*} T_f &=25+ \frac{45000}{1.5 \times 385} \\\\ \Rightarrow T_f & \approx \boxed{\rm 103 ^\circ C} \end{align*} Thus, the final temperature of the copper block is approximately 103°C.

Problem (5): (a) How much energy is required to heat 1.5 L of water from 15°C to 95°C? (b) For how long could this amount of energy power a 75-W lightbulb?

Solution: (a) Since water has a density of 1 kg/L, the mass of 1.5 L of water is 1.5 kg. Substituting the given values into the heat transfer energy equation, we have: \begin{align*} Q&=mc\Delta T \\\\ &=(1.5)(4190)(95-15) \\\\ &=\boxed{502800\,\rm J} \end{align*} Thus, 502.8 kJ of energy is required to heat the water.

(b) Power is defined as energy per unit time, $P=\frac{Q}{t}$. Rearranging to solve for $t$, and substituting the known values, we get: \begin{align*} t &=\frac{Q}{P} \\\\ &=\frac{502800}{75} \\\\ &=\boxed{6704\,\rm s} \end{align*} Converting to minutes: \[t \approx 112\quad\rm minutes\] Thus, the same amount of energy could power a 75-W lightbulb for about 112 minutes.

In another type of problem, specific heat capacity is determined through calorimetry.

In calorimetry problems, the specific heat of a sample is measured by placing it into a known quantity of water at a specific temperature.

Since this system is isolated, the total energy gained or lost by all objects within it sums to zero, expressed as $\Sigma Q_i=0$, where $Q_i$ represents the energy of the ith object.

By measuring the equilibrium temperature, we can solve the equation above to determine the unknown specific heat.

Below, some problems for finding the heat capacity using this method are presented.

Problem (6): 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 block's temperature is higher than the water's, the block loses thermal energy while the water absorbs it. Applying the principle of energy conservation ($Q_{gain}=-Q_{lose}$), and solving for the unknown specific heat, gives us: \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 one.

A negative sign is placed in front of the heat lost by the warmer object to ensure both sides of the equation remain positive.

For more practice about this, check out the page on thermodynamics problems and solutions.

Problem (7): A piece of metal of unknown specific heat, weighing 25 g and temperature of 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 that of the metal, the metal loses heat given by: $Q_m=-m_m c_m (T_f-T_m)$ while the water gains heat: $Q_w=m_w c_w (T_f-T_w)$.

A useful rule of thumb is that when an object loses heat, a negative sign should be placed in front of its heat expression to indicate energy loss.

Conservation of energy tells us $Q_w=Q_m$, thus 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 (8): A piece of unknown substance with a mass of 2 kg and a temperature of 280 Kelvin is brought into thermal 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: To solve for the specific heat of the unknown substance, we will use the principle of conservation of energy. In an isolated system, the heat lost by the copper is equal to the heat gained by the unknown substance. 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 the 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 (9): 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: The solid loses heat while the fluid absorbs it, so we express this as $Q_f=-Q_s$ where f and s denote the fluid and solid, respectively. Using the definition of specific heat, we can find the heat transferred during this process as follows: \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*} Thus, the specific heat capacity of fluid is lower than that of the solid.

Problem (10): 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 surroundings, 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 surroundings, the conservation of energy principle gives $Q_{water}=-Q_{metal}$. Using the definition of specific heat, the heat transfer equation is as follows. Solving this for $c_{meta}$ and substituting the given numerical values into it, we have: \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.

Problem (11): 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 lose 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*}

Problem (12): 200 g of water at 22.5°C is mixed with 150 g 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.

Problem (13): 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: The minimum mass of the metal ensures that it has just enough heat energy to completely melt the ice without raising the temperature of the system. The amount of heat energy released by $m$ grams of the metal as it cools from 250°C to 0°C is given by: $Q_{lost}=mc(T_f-T_i)$ where $T_f$ is the equilibrium temperature, 0°C.

Meanwhile, the energy required to melt 200 g of ice is: \[Q_{gain}=m_{ice} L_f=200\times 336\times 10^{3}\,{\rm \frac{J}{kg}}\] where $L_f$ is the latent heat of fusion for ice. By the conservation of energy principle, equating the heat lostby the metal to the heat gained by the ice, we have: \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*} Thus, the minumum mass of the metal required to fully melt the ice is 672 g. If the metal's mass exceeds this value, the excess heat will cause the final system temperature to rise above 0°C.

For more practice, you can 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: April 17, 2025