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 $m$, then we can relate all these physical quantities as following formula \[Q=mc\Delta T=mc(T_f-T_i)\] where $T_f$ and $T_i$ are the initial and final temperatures.
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.
If this page helped you with homework or anything else, please support me here.
Problem (2): How much heat is absorbed by a 20 g granite boulder as energy from the sun causes its temperature to change from 10°C to 29°C? (Specific heat capacity of granite is 0.1 cal/g.°C)
Solution: to raise the temperature of the granite boulder from 10°C to 29°C, we must add $Q=mc\Delta T$ energy to the granite as below \begin{align*} Q&=mc\Delta T\\ &=20 \times 0.1\times (29^\circ-10^\circ)\\&=38\quad {\rm cal}\end{align*} So, it would require 38 calories of heat energy to increase the temperature of this granite boulder from 10°C to 29°C. This is the amount of energy we must add to the boulder.
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
Problem (3): 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 below \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. This energy could be transferred to the surrounding environment or used to do work, depending on the specific circumstances.
Were these not helpful? My income comes from your generous donations. Please support me.
Problem (4): If a 3.1 g ring is heated using 10.0 calories, its temperature rises 17.9°C. Calculate the specific heat capacity of the ring.
Solution: Since the given heat causes a change in the temperature of the ring, the amount of heat is obtained by the formula $Q=mc(T_f-T_i)$. By putting known values into it and solving for the unknown value, the specific heat of the ring is calculated as below, \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 not the initial or final temperatures.
Problem (5): The temperature of a sample of water increases from 20°C to 46.6°C as it absorbs 5650 calories of heat. What is the mass of the sample? (Specific heat of water is 1.0 cal/g.°C)
Solution: As before, using heat formula and solving for mass $m$, we get \begin{align*} m&=\frac{Q}{c\Delta T}\\\\ &=\frac{5650}{1\times (46.6^\circ-20^\circ)}\\ \\&=212.4\quad {\rm g}\end{align*}
Problem (6): 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: In the specific heat problems, we learned that the specific heat is defined as the amount of heat energy required to change the temperature of a sample with mass $m$ by $\Delta T$.
Here, energy is released by as much as 47 calories, so we must put it with a negative sign into the equation. Thus, we have \begin{align*} c&=\frac{Q}{m\Delta T}\\ \\&=\frac{-47}{10\times (25^\circ-50.4^\circ)}\\ \\&= 0.185\quad {\rm cal/g\cdot ^\circ \! C}\end{align*}
Problem (7): 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 (8): A 155 g sample of an unknown substance was heated from 25°C to 40°C. In the process, the substance absorbed 569 calories of energy. What is the specific heat of the substance?
Solution: In the heat formula $Q=mc\Delta T$, the specific heat of any substance is denoted by $c$. Putting known values into this formula and solving for unknown specific heat, we get \begin{align*} c&=\frac{Q}{m\Delta T}\\ \\ &=\frac{569}{155\times (40^\circ-25^\circ)}\\ \\&=0.244\quad {\rm cal/g\cdot^\circ\! C} \end{align*}
Problem (9): What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its temperature changes from 25°C to 20°C?
Solution: same as above, we have \begin{align*} c&=\frac{Q}{m(T_f-T_i)}\\ \\&=\frac{12}{2.5\times (20^\circ-25^\circ)}\\\\&=0.96\quad {\rm cal/g\cdot ^\circ \! C}\end{align*}
Problem (10): 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 value of specific heat is given in grams, and the weight of water is in kilograms. Therefore, first convert them into grams or kilograms and then continue to solve the problem. Here, we converted 3 kg to 3000 g.
The negative sign indicates that the heat is released from the water.
Problem (11): How much heat is needed to raise a 0.30 kg piece of aluminum from 30°C to 150°C? ($c_{Al}=0.9\,{\rm J/g\cdot ^\circ C}$)
Solution: Let $T_f$ and $T_i$ be the initial and final temperatures of the aluminum so the required heat is computed as below \begin{align*} Q&=mc(T_f-T_i)\\&=0.3\times 900\times (150^\circ-30^\circ)\\&=-32400\quad {\rm J}\\ or &=-32.4\quad {\rm kJ}\end{align*} Here, we converted specific heat in SI units.
Problem (12): 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) 1.96 kJ of heat is added to 500 g of copper.($c_{Cu}=0.385\,{\rm J/g\cdot ^\circ C}$)
Solution: In both parts, we use the heat formula when temperature changes, $Q=mc(T_f-T_i)$.
(a) Substituting known values $m=10\,{\rm kg}$ and $Q=232\,{\rm kJ}$ into the above equation and solving for the change in temperature $\Delta T=T_f-T_i$, we get: \begin{align*} \Delta T&= \frac{Q}{mc}\\ \\&=\frac{-232000}{10\times 4179}\\ \\&=-5.55\,{\rm ^\circ C}\end{align*} Since water loses heat energy (which justifies why we inserted a minus sign for Q), its temperature must be decreasing. In the above, kJ means 1000 J of energy.
(b) Heat is added to the water, so $Q>0$ must be inserted into the formula, \begin{align*}\Delta T&=\frac{Q}{mc}\\ \\&=\frac{1960}{0.5\times 385}\\ \\&=10.18\,{\rm ^\circ C}\end{align*}
Problem (13): When heated, the temperature of a water sample increased from 15°C to 39°C. It absorbed 4300 joules of heat. What is the mass of the sample?
Solution: putting known values into the equation $Q=mc(T_f-T_i)$ and solving for unknown mass, we get \begin{align*} m&=\frac{Q}{c(T_f-T_i)}\\ \\ &=\frac{4300}{4179\times (15^\circ-39^\circ)}\\ \\&=0.0428\quad {\rm kg}\\ \\ or &=42.8\quad {\rm g} \end{align*}
Problem (14): 5.0 g of copper was heated from 20°C to 80°C. How much energy was used to heat Cu?
Solution: the necessary energy is calculated as below: \begin{align*} Q&=mc(T_f-T_i)\\&=5\times 0.385\times (20^\circ-80^\circ)\\&=115.5\quad {\rm J}\end{align*} So, the necessary energy or heat absorbed by the object is calculated to be $\rm 115.5\, J$. This value tells us how much heat energy is required to change the temperature of the $\rm 5\, g$ object by $60$ degrees Celsius.
Problem (15): The temperature of a sample of water increases from 20°C to 46.6°C as it absorbs 5650 Joules of heat. What is the mass of the sample?
Solution: known values are $T_i={\rm 20^\circ C}$, $T_f={\rm 46.6^\circ C}$ and $Q=5650\,{\rm J}$. We can rearrange the formula to solve for ($m$): \begin{align*} m&=\frac{Q}{c(T_f-T_i)}\\ \\&=\frac{5650}{4179\times (46.6^\circ-20^\circ)}\\ \\ &=0.0508\quad {\rm kg} \\ \\ or &=50.8\quad {\rm g} \end{align*} So, the mass of the water sample is approximately ($50.7$) grams.
Author: Dr. Ali Nemati
Page Created: 3/9/2021
© 2015 All rights reserved. by Physexams.com