Heat Practice Problems With Detailed Answers

Struggling with heat problems in your physics class?

Our article is here to help! Packed with solved examples specifically designed for high school students, this guide will make mastering heat problems easy.

Get ready to boost your grades and deepen your understanding with these easy-to-follow solutions!

When heat energy ($Q$) causes a change in temperature $\Delta T=T_f-T_i$ in a sample with specific heat capacity ($c$) and mass ($m$), the relationship between these physical quantities is expressed by the following formula: \[Q=mc\Delta T=mc(T_f-T_i)\] where $T_f$ and $T_i$ are the initial and final temperatures.

In all these example problems, there is no change in the state of the substance. If there were a change in the phase of matter (solid $\Leftrightarrow$ liquid or to liquid$\Leftrightarrow$ gas) read the following page to learn more:

Solved problems on latent heat of fusion

Solved Problems on latent heat of vaporization

Heat Practice Problems

Problem (1): 5.0 g of copper was heated from 20°C to 80°C. How much energy was used to heat Cu? (Specific heat capacity of Cu is 0.092 cal/g. °C)

Solution: The energy required to change the temperature of a substance of mass $m$ from initial temperature $T_i$ to final temperature $T_f$ is obtained by the formula $Q=mc(T_f-T_i)$, where $c$ is the specific heat of the substance. Thus, we have \begin{align*} Q&=mc\Delta T\\ &= 5\times 0.092\times (80^\circ-20^\circ)\\&= 27.6 \quad {\rm cal} \end{align*} So, it would require 27.6 calories of heat energy to increase the temperature of this substance from 20°C to 80°C.

Problem (2): How much heat is released when 30 g of water at 96°C cools to 25°C? The specific heat of water is 1 cal/g.°C.

Solution: the amount of energy released is obtained by formula $Q=mc\Delta T$ as follows: \begin{align*} Q&=mc\Delta T\\&=30\times 1\times (25^\circ-96^\circ)\\&= -2130\quad {\rm cal}\end{align*} The negative sign in the result indicates that the energy is being released from the water. This is because the temperature of the water is decreasing, which means it is losing heat energy.

Therefore, 2130 calories of heat energy are released from the water when its temperature decreases from 96°C to 25°C. Depending on the specific circumstances, this energy could be transferred to the surrounding environment or used to do work.

Problem (3): If a 3.1 g ring is heated using 10.0 calories, its temperature rises 17.9°C. Calculate the ring's specific heat capacity.

Solution: In specific heat problems, we learned that specific heat is defined as the amount of heat energy required to change the temperature of a sample with mass $m$ by $\Delta T$.

By putting known values into the equation $Q=mc(T_f-T_i)$, and solving for the unknown value, the specific heat of the ring is calculated as follows: \begin{align*} c&=\frac{Q}{m(T_f-T_i)}\\ \\ &=\frac{10}{3.1\times 17.9^\circ}\\ \\&=0.18\quad {\rm cal/g\cdot ^\circ C}\end{align*} So, the specific heat of the ring is calculated to be $0.18\,{\rm cal/g\cdot ^\circ C}$. This value tells us how much heat is required to raise the temperature of $1$ gram of the ring by $1$ degree Celsius. Note that in this problem, the difference between temperatures is used, not the initial or final temperatures.

Problem (4): A 1.80 kg hammer strikes a nail with a velocity of $7.80\, \rm m/s$. If $60\%$ of the hammer's kinetic energy is converted into heat within the nail, and this heat is entirely absorbed by the nail, determine the temperature increase of an 8.00 g aluminum nail after it is struck 10 times. ($c_{Al}=910\,{\rm J/kg\cdot ^\circ C}$)

Solution: When $60\%$ of the hammer's kinetic energy ($K$) is converted into heat, we express this as $Q=0.6K$, where $Q$ is the heat transferred to the nail in each strike. Therefore, we have: \begin{align*} Q&=0.6\times \frac 12 mv^2 \\\\ &=(0.6)(0.5)(1.8)(7.8)^2 \\\\ &=32.853\,\rm J \end{align*} Thus, the total heat gained by the nail after 10 strikes is: \[Q_{tot}=10Q=328.53\,\rm J\] This heat increase causes an rise in the temperature of the nail. Using the heat equation, we get: \begin{align*} \Delta T &=\frac{Q_{tot}}{m_{Al}c_{Al}} \\\\ &=\frac{328.53}{(0.008)(910)} \\\\ &=\rm 45.14^\circ C \end{align*} So, the increase in temperature of the 8.00 g aluminum nail after it is strike 10 times is approximately $\rm 45^\circ C$.

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Problem (5): A 4.50 g coin of copper absorbed 54 calories of heat. What was the final temperature of the copper if the initial temperature was 25°C? The specific heat of copper is 0.092 cal/g.°C.

Solution: Let $T_i$ and $T_f$ be the initial and final temperatures of the copper coin. Again using formula $Q=mc(T_f-T_i)$ and solving for final temperature $T_f$, we have \begin{align*} T_f&=\frac{Q}{mc}+T_i \\ \\ &=\frac{54}{0.092\times 4.5}+25^\circ\\ \\ &=155.43\,{\rm ^\circ C}\end{align*}

Problem (6): When 3 kg of water is cooled from 80°C to 10°C, how much heat energy is lost? (specific heat of water is $c_W=4.179\,{\rm J/g\cdot ^\circ C}$)

Solution: the heat has led to a change in temperature, so we must use the formula $Q=mc\Delta T$ to find the lost heat as shown below: \begin{align*} Q&=mc(T_f-T_i)\\&=3000\times 4.179\times (10^\circ-80^\circ)\\&=-877590\quad {\rm J} \\ or &=-877.590\quad {\rm kJ}\end{align*} Note that in the above calculation, the specific heat capacity is given in $\rm J/g\cdot ^\circ C$, and the mass of water is in kilograms. Therefore, first convert mass into grams or use consistent units throughout the calculation. Here, we converted 3 kg to 3000 g.

The negative sign indicates that the heat is released from the water.

Problem (7): Calculate the temperature change when:
(a) 10.0 kg of water loses 232 kJ of heat. ($c_w=4.179\,{\rm J/g\cdot ^\circ C}$)
(b) 8954 J of heat is added to 3 moles of copper. ($c_{Cu}=0.385\,{\rm J/g\cdot ^\circ C}$)

Solution: In both parts, we use the heat formula for temperature changes, $Q=mc(T_f-T_i)$.
(a) Substituting known values $m=10\,{\rm kg}$ and $Q=-232\,{\rm kJ}$ (note the negative sign for heat loss) into the equation and solving for the change in temperature $\Delta T=T_f-T_i$: \begin{align*} \Delta T&= \frac{Q}{mc}\\ \\&=\frac{-232000}{10\times 4179}\\ \\&=-5.55\,{\rm ^\circ C}\end{align*} Since the water loses heat energy (hence the negative sign for Q), its temperature decreases. Here, $\rm kJ$ (kilojoules) is converted to $\rm J$ by multiplying by 1000.

(b) The molar mass of copper $M$ is $63.5\,\rm g/mol$. Therefore, the mass of 3 moles of copper is calculated as follows: \begin{align*} m&=\text{moles}\times M \\ &= 3\times 63.5 \\ &=190.5\,\rm g \end{align*} Next, we find the temperature increase using the heat equation: \begin{align*} \Delta T &=\frac{Q}{mc} \\\\ &=\frac{8954}{190.5\times 0.385} \\\\ &=\boxed{\rm 119^\circ C} \end{align*} Thus, the temperature increase of the copper is $\rm 119^\circ C$.

Problem (8): A 180-gram sample of an unknown material is heated to 280°C. This hot sample is then immediately submerged into a 95-gram copper calorimeter containing 150 grams of water and a 12-gram glass thermometer. The initial temperature of the calorimeter, water, and thermometer is 20.0°C. After reaching thermal equilibrium, the final temperature of the system is 32.5°C. Determine the specific heat of the unknown material. (Assume no heat is lost to the surroundings.)

Solution: These types of problems fall under calorimetry. In such cases, some objects lose heat while others gain it until they reach thermal equilibrium at a specific final temperature.

In this question, the unknown material ($x$) at a higher temperature of $\rm 280^\circ C$ loses its heat energy to reach the equilibrium temperature of $\rm 20^\circ C$. The other components at a lower temperature gain this heat.

According to the energy conservation principle (or in this case, the principle of calorimetry), we have \[Q_{gain}=-Q_{lost}\] The negative sign ensures that the right side has a positive value. Therefore, \begin{align*} Q_{lost}&=mc\Delta T \\ &=(0.180)c_x(32.5^\circ-280^\circ) \\ &=-44.55 c_x \end{align*} And the heat gained by other components is given by: \[Q_{gain}=(m_w c_w+m_{calo}c_{calo}+m_{ther}c_{ther}) \Delta T \] Substituting the given values, we find the heat gained by these objects: \[Q_{gain}=8420.7\,\rm J\] Therefore, using energy conservation: \begin{gather*} Q_{gain}=-Q_{lost} \\\\ 8420.70=44.55c_x \\\\ \Rightarrow c_x \approx \rm 189\,J/g\cdot ^\circ C \end{gather*} Therefore, the specific heat of the unknown material is approximately 189 J/g°C.

Author: Dr. Ali Nemati
Page Created: 3/9/2021