specific heat capacity problem solving

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Specific Heat Example Problem

Specific heat is the amount of heat per unit mass needed to increase the temperature of a material by one degree Celsius or Kelvin. These three specific heat example problems will show how to find the specific heat of a material or other information involving the specific heat.

Specific Heat Equation

The equation most commonly associated with specific heat is

where Q = Heat energy m = mass c = specific heat ΔT = change in temperature = (T final – T initial )

A good way to remember this formula is Q = “em cat”

Basically, this equation is used to determine the amount of heat added to a material to raise the temperature some amount (or the amount lost as the material cools).

This equation only applies to materials that stay in the same state of matter (solid, liquid, or gas) as the temperature changes. Phase changes require additional energy considerations.

Specific Heat Example Problem – Find the Amount of Heat

Question : A 500 gram cube of lead is heated from 25 °C to 75 °C. How much energy was required to heat the lead? The specific heat of lead is 0.129 J/g°C.

Solution : First, let’s the variables we know.

m = 500 grams c = 0.129 J/g°C ΔT = (T final – T initial ) = (75 °C – 25 °C) = 50 °C

Plug these values into the specific heat equation from above.

Q = mcΔT

Q = (500 grams)·(0.129 J/g°C)·(50 °C)

Answer : It took 3225 Joules of energy to heat the lead cube from 25 °C to 75 °C.

Specific Heat Example Problem – Find the Specific Heat

Question: A 25-gram metal ball is heated 200 °C with 2330 Joules of energy. What is the specific heat of the metal?

Solution : List the information we know.

m = 25 grams ΔT = 200 °C Q = 2330 J

Place these into the specific heat equation.

2330 J = (25 g)c(200 °C)

2330 J = (5000 g°C)c

Divide both sides by 5000 g°C

c = 0.466 J/g°C

Answer : The specific heat of the metal is 0.466 J/g°C.

Specific Heat Example Problem – Find the Initial Temperature

Question: A hot 1 kg chunk of copper is allowed to cool to 100°C. If the copper gave off 231 kJ of energy, what was the initial temperature of the copper? The specific heat of copper is 0.385 J/g°C.

Solution : List our given variables:

m = 1 kg T final = 100 °C Q = -231 kJ (The negative sign is because the copper is cooling and losing energy.) c = 0.385 J/g°C

We need to make our units consistent with the specific heat units, so let’s convert the mass and energy units.

m = 1 kg = 1000 grams

1 kJ = 1000 J Q = -231 kJ · (1000 J/kJ) = -231000 J

Plug these values into the specific heat formula.

-231000 J = 1000 g · (0.385 J/g°C) · ΔT

-231000 J = 385 J/°C · ΔT

ΔT = -600 °C

ΔT = (T final – T initial )

Plug in the values for ΔT and T final .

-600 °C = (100 °C – T initial )

Subtract 100 °C from both sides of the equation.

-600 °C – 100 °C =  – T initial

-700 °C = – T initial

T initial  = 700 °C

Answer : The initial temperature of the copper chunk was 700 °C.

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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.

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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.

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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  

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Chemistry Steps

General Chemistry

Thermochemistry.

In the previous post , we talked about heat capacity, specific heat, and the formula for the relationship between these and the change in temperature.

Below are some additional practice examples of heat capacity problems. 

When solving a problem related to heat capacity and heat transfer, remember that most of the time, it is assumed the heat is not lost, and it only flows from the object with a higher temperature to the colder one:

The second expression adding the lost and gained heat is easier to use because you do not need to remember which one has a negative sign.

How much heat does it take to increase the temperature of a 540.6-g sample of Fe from 20.0 °C to 84.3 °C? The specific heat of iron = 0.450 J/g °C.

Calculate the specific heat capacity of a metal if a 17.0 g sample requires 481 J to change the temperature of the metal from 25.0 °C to 67.0 °C?

Calculate the energy of combustion for one mole of butane if burning a 0.367 g sample of butane (C 4 H 10 ) has increased the temperature of a bomb calorimeter by 7.73 °C. The heat capacity of the bomb calorimeter is 2.36 kJ/ °C.

How many joules of energy is required to melt 40.0 g of ice at 0 °C? The heat of fusion (Δ H fus ) for ice is 334.0 J/g.

How many kJ of energy does it take to change 36.0 g of ice at -15.0 °C to water at 0. °C ? The specific heat of ice is 2.10 J/g°C and the heat of fusion (Δ H fus ) for ice is 334.0 J/g.

• Energy Related to Heat and Work
• Endothermic and Exothermic Processes
• Heat Capacity and Specific Heat
• What is Enthalpy
• Constant-Pressure Calorimetry
• Bomb calorimeter – Constant Volume Calorimetry
• Stoichiometry and Enthalpy of Chemical Reactions
• Hess’s Law and Enthalpy of Reaction
• Hess’s Law Practice Problems
• Standard Enthalpies of Formation
• Enthalpy of Reaction from Enthalpies of Formation
• Thermochemistry Practice Problems

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11.2 Heat, Specific Heat, and Heat Transfer

Section learning objectives.

By the end of this section, you will be able to do the following:

• Explain heat, heat capacity, and specific heat
• Distinguish between conduction, convection, and radiation
• Solve problems involving specific heat and heat transfer

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (F) contrast and give examples of different processes of thermal energy transfer, including conduction, convection, and radiation.

Section Key Terms

[BL] [OL] [AL] Review concepts of heat, temperature, and mass.

[AL] Check prior knowledge of conduction and convection.

Heat Transfer, Specific Heat, and Heat Capacity

We learned in the previous section that temperature is proportional to the average kinetic energy of atoms and molecules in a substance, and that the average internal kinetic energy of a substance is higher when the substance’s temperature is higher.

If two objects at different temperatures are brought in contact with each other, energy is transferred from the hotter object (that is, the object with the greater temperature) to the colder (lower temperature) object, until both objects are at the same temperature. There is no net heat transfer once the temperatures are equal because the amount of heat transferred from one object to the other is the same as the amount of heat returned. One of the major effects of heat transfer is temperature change: Heating increases the temperature while cooling decreases it. Experiments show that the heat transferred to or from a substance depends on three factors—the change in the substance’s temperature, the mass of the substance, and certain physical properties related to the phase of the substance.

The equation for heat transfer Q is

where m is the mass of the substance and Δ T is the change in its temperature, in units of Celsius or Kelvin. The symbol c stands for specific heat , and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00 ºC. The specific heat c is a property of the substance; its SI unit is J/(kg ⋅ ⋅ K) or J/(kg ⋅ ⋅ °C °C ). The temperature change ( Δ T Δ T ) is the same in units of kelvins and degrees Celsius (but not degrees Fahrenheit). Specific heat is closely related to the concept of heat capacity . Heat capacity is the amount of heat necessary to change the temperature of a substance by 1.00 °C °C . In equation form, heat capacity C is C = m c C = m c , where m is mass and c is specific heat. Note that heat capacity is the same as specific heat, but without any dependence on mass. Consequently, two objects made up of the same material but with different masses will have different heat capacities. This is because the heat capacity is a property of an object, but specific heat is a property of any object made of the same material.

Values of specific heat must be looked up in tables, because there is no simple way to calculate them. Table 11.2 gives the values of specific heat for a few substances as a handy reference. We see from this table that the specific heat of water is five times that of glass, which means that it takes five times as much heat to raise the temperature of 1 kg of water than to raise the temperature of 1 kg of glass by the same number of degrees.

[BL] [OL] [AL] Explain that this formula only works when there is no change in phase of the substance. The transfer of thermal energy, heat, and phase change will be covered later in the chapter.

Misconception Alert

The units of specific heat are J/(kg ⋅ °C ⋅ °C ) and J/(kg ⋅ ⋅ K). However, degrees Celsius and Kelvins are not always interchangeable. The formula for specific heat uses a difference in temperature and not absolute temperature. This is the reason that degrees Celsius may be used in place of Kelvins.

Temperature Change of Land and Water

What heats faster, land or water? You will answer this question by taking measurements to study differences in specific heat capacity.

• Open flame—Tie back all loose hair and clothing before igniting an open flame. Follow all of your teacher's instructions on how to ignite the flame. Never leave an open flame unattended. Know the location of fire safety equipment in the laboratory.
• Sand or soil
• Oven or heat lamp
• Two small jars
• Two thermometers

Instructions

• Place equal masses of dry sand (or soil) and water at the same temperature into two small jars. (The average density of soil or sand is about 1.6 times that of water, so you can get equal masses by using 50 percent more water by volume.)
• Heat both substances (using an oven or a heat lamp) for the same amount of time.
• Record the final temperatures of the two masses.
• Now bring both jars to the same temperature by heating for a longer period of time.
• Remove the jars from the heat source and measure their temperature every 5 minutes for about 30 minutes.
• The pond will reach 0 °C first because of water’s greater specific heat.
• The field will reach 0 °C first because of soil’s lower specific heat.
• They will reach 0° C at the same time because they are exposed to the same weather.
• The water will take longer to heat as well as to cool. This tells us that the specific heat of water is greater than that of land.

Conduction, Convection, and Radiation

Whenever there is a temperature difference, heat transfer occurs. Heat transfer may happen rapidly, such as through a cooking pan, or slowly, such as through the walls of an insulated cooler.

There are three different heat transfer methods: conduction , convection , and radiation . At times, all three may happen simultaneously. See Figure 11.3 .

Conduction is heat transfer through direct physical contact. Heat transferred between the electric burner of a stove and the bottom of a pan is transferred by conduction. Sometimes, we try to control the conduction of heat to make ourselves more comfortable. Since the rate of heat transfer is different for different materials, we choose fabrics, such as a thick wool sweater, that slow down the transfer of heat away from our bodies in winter.

As you walk barefoot across the living room carpet, your feet feel relatively comfortable…until you step onto the kitchen’s tile floor. Since the carpet and tile floor are both at the same temperature, why does one feel colder than the other? This is explained by different rates of heat transfer: The tile material removes heat from your skin at a greater rate than the carpeting, which makes it feel colder.

[BL] [OL] [AL] Ask students what the current temperature in the classroom is. Ask them if all the objects in the room are at the same temperature. Once this is established, ask them to place their hand on their desk or on a metal object. Does it feel colder? Why? If their desk is Formica laminate, then it will feel cool to their hand because the laminate is a good conductor of heat and draws heat from their hand creating a sensation of “cold” due to heat leaving the body.

Some materials simply conduct thermal energy faster than others. In general, metals (like copper, aluminum, gold, and silver) are good heat conductors, whereas materials like wood, plastic, and rubber are poor heat conductors.

Figure 11.4 shows particles (either atoms or molecules) in two bodies at different temperatures. The (average) kinetic energy of a particle in the hot body is higher than in the colder body. If two particles collide, energy transfers from the particle with greater kinetic energy to the particle with less kinetic energy. When two bodies are in contact, many particle collisions occur, resulting in a net flux of heat from the higher-temperature body to the lower-temperature body. The heat flux depends on the temperature difference Δ T = T hot − T cold Δ T = T hot − T cold . Therefore, you will get a more severe burn from boiling water than from hot tap water.

Convection is heat transfer by the movement of a fluid. This type of heat transfer happens, for example, in a pot boiling on the stove, or in thunderstorms, where hot air rises up to the base of the clouds.

Tips For Success

In everyday language, the term fluid is usually taken to mean liquid. For example, when you are sick and the doctor tells you to “push fluids,” that only means to drink more beverages—not to breath more air. However, in physics, fluid means a liquid or a gas . Fluids move differently than solid material, and even have their own branch of physics, known as fluid dynamics , that studies how they move.

As the temperature of fluids increase, they expand and become less dense. For example, Figure 11.4 could represent the wall of a balloon with different temperature gases inside the balloon than outside in the environment. The hotter and thus faster moving gas particles inside the balloon strike the surface with more force than the cooler air outside, causing the balloon to expand. This decrease in density relative to its environment creates buoyancy (the tendency to rise). Convection is driven by buoyancy—hot air rises because it is less dense than the surrounding air.

Sometimes, we control the temperature of our homes or ourselves by controlling air movement. Sealing leaks around doors with weather stripping keeps out the cold wind in winter. The house in Figure 11.5 and the pot of water on the stove in Figure 11.6 are both examples of convection and buoyancy by human design. Ocean currents and large-scale atmospheric circulation transfer energy from one part of the globe to another, and are examples of natural convection.

Radiation is a form of heat transfer that occurs when electromagnetic radiation is emitted or absorbed. Electromagnetic radiation includes radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays, all of which have different wavelengths and amounts of energy (shorter wavelengths have higher frequency and more energy).

[BL] [OL] Electromagnetic waves are also often referred to as EM waves. We perceive EM waves of different frequencies differently. Just as we are able to see certain frequencies as visible light, we perceive certain others as heat.

You can feel the heat transfer from a fire and from the sun. Similarly, you can sometimes tell that the oven is hot without touching its door or looking inside—it may just warm you as you walk by. Another example is thermal radiation from the human body; people are constantly emitting infrared radiation, which is not visible to the human eye, but is felt as heat.

Radiation is the only method of heat transfer where no medium is required, meaning that the heat doesn’t need to come into direct contact with or be transported by any matter. The space between Earth and the sun is largely empty, without any possibility of heat transfer by convection or conduction. Instead, heat is transferred by radiation, and Earth is warmed as it absorbs electromagnetic radiation emitted by the sun.

All objects absorb and emit electromagnetic radiation (see Figure 11.7 ). The rate of heat transfer by radiation depends mainly on the color of the object. Black is the most effective absorber and radiator, and white is the least effective. People living in hot climates generally avoid wearing black clothing, for instance. Similarly, black asphalt in a parking lot will be hotter than adjacent patches of grass on a summer day, because black absorbs better than green. The reverse is also true—black radiates better than green. On a clear summer night, the black asphalt will be colder than the green patch of grass, because black radiates energy faster than green. In contrast, white is a poor absorber and also a poor radiator. A white object reflects nearly all radiation, like a mirror.

Ask students to give examples of conduction, convection, and radiation.

Virtual Physics

Energy forms and changes.

In this animation, you will explore heat transfer with different materials. Experiment with heating and cooling the iron, brick, and water. This is done by dragging and dropping the object onto the pedestal and then holding the lever either to Heat or Cool. Drag a thermometer beside each object to measure its temperature—you can watch how quickly it heats or cools in real time.

Now let’s try transferring heat between objects. Heat the brick and then place it in the cool water. Now heat the brick again, but then place it on top of the iron. What do you notice?

Selecting the fast forward option lets you speed up the heat transfers, to save time.

• Water will take the longest, and iron will take the shortest time to heat, as well as to cool. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body.
• Water will take the shortest, and iron will take the longest time to heat, as well as to cool. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body.
• Brick will take shortest and iron will take longest time to heat up as well as to cool down. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body.
• Water will take shortest and brick will take longest time to heat up as well as to cool down. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body.

Have students consider the differences in the interactive exercise results if different materials were used. For example, ask them whether the temperature change would be greater or smaller if the brick were replaced with a block of iron with the same mass as the brick. Ask students to consider identical masses of the metals aluminum, gold, and copper. After they have stated whether the temperature change is greater or less for each metal, have them refer to Table 11.2 and check whether their predictions were correct.

Solving Heat Transfer Problems

Worked example, calculating the required heat: heating water in an aluminum pan.

A 0.500 kg aluminum pan on a stove is used to heat 0.250 L of water from 20.0 °C °C to 80.0 °C °C . (a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water?

The pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and the pan is increased by the same amount. We use the equation for heat transfer for the given temperature change and masses of water and aluminum. The specific heat values for water and aluminum are given in the previous table.

Because the water is in thermal contact with the aluminum, the pan and the water are at the same temperature.

• Calculate the temperature difference. Δ T = T f − T i = 60.0 °C Δ T = T f − T i = 60.0 °C 11.8
• Calculate the mass of water using the relationship between density, mass, and volume. Density is mass per unit volume, or ρ = m V ρ = m V . Rearranging this equation, solve for the mass of water. m w = ρ ⋅ V = 1000  kg/m 3 × ( 0 .250 L× 0 .001 m 3 1 L ) =0 .250 kg m w = ρ ⋅ V = 1000  kg/m 3 × ( 0 .250 L× 0 .001 m 3 1 L ) =0 .250 kg 11.9
• Calculate the heat transferred to the water. Use the specific heat of water in the previous table. Q w = m w c w Δ T =   ( 0.250  kg ) ( 4186 J/kg °C ) ( 60 .0 °C )  = 62 .8 kJ Q w = m w c w Δ T =   ( 0.250  kg ) ( 4186 J/kg °C ) ( 60 .0 °C )  = 62 .8 kJ 11.10
• Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in the previous table. Q A l = m A l c A l Δ T =   ( 0.500  kg ) ( 900 J/kg °C ) ( 60 .0 °C )  = 27 .0 ×10 3 J = 27 .0 kJ Q A l = m A l c A l Δ T =   ( 0.500  kg ) ( 900 J/kg °C ) ( 60 .0 °C )  = 27 .0 ×10 3 J = 27 .0 kJ 11.11
• Find the total transferred heat. Q T o t a l = Q w + Q A l = 62 .8 kJ + 27 .0 kJ = 89 .8 kJ Q T o t a l = Q w + Q A l = 62 .8 kJ + 27 .0 kJ = 89 .8 kJ 11.12

The percentage of heat going into heating the pan is

The percentage of heat going into heating the water is

In this example, most of the total heat transferred is used to heat the water, even though the pan has twice as much mass. This is because the specific heat of water is over four times greater than the specific heat of aluminum. Therefore, it takes a bit more than twice as much heat to achieve the given temperature change for the water than for the aluminum pan.

Water can absorb a tremendous amount of energy with very little resulting temperature change. This property of water allows for life on Earth because it stabilizes temperatures. Other planets are less habitable because wild temperature swings make for a harsh environment. You may have noticed that climates closer to large bodies of water, such as oceans, are milder than climates landlocked in the middle of a large continent. This is due to the climate-moderating effect of water’s large heat capacity—water stores large amounts of heat during hot weather and releases heat gradually when it’s cold outside.

Calculating Temperature Increase: Truck Brakes Overheat on Downhill Runs

When a truck headed downhill brakes, the brakes must do work to convert the gravitational potential energy of the truck to internal energy of the brakes. This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck, and keeps the truck from speeding up and losing control. The increased internal energy of the brakes raises their temperature. When the hill is especially steep, the temperature increase may happen too quickly and cause the brakes to overheat.

Calculate the temperature increase of 100 kg of brake material with an average specific heat of 800 J/kg ⋅ °C ⋅ °C from a 10,000 kg truck descending 75.0 m (in vertical displacement) at a constant speed.

We first calculate the gravitational potential energy ( Mgh ) of the truck, and then find the temperature increase produced in the brakes.

• Calculate the change in gravitational potential energy as the truck goes downhill. M g h = ( 10 , 000  kg ) (9 .80 m/s 2 ) ( 75 .0 m ) = 7.35 × 10 6 J M g h = ( 10 , 000  kg ) (9 .80 m/s 2 ) ( 75 .0 m ) = 7.35 × 10 6 J 11.15

where m is the mass of the brake material (not the entire truck). Insert the values Q = 7.35×10 6 J (since the heat transfer is equal to the change in gravitational potential energy), m = = 100 kg, and c = = 800 J/kg ⋅ ⋅ °C °C to find

This temperature is close to the boiling point of water. If the truck had been traveling for some time, then just before the descent, the brake temperature would likely be higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of the brake material above the boiling point of water, which would be hard on the brakes. This is why truck drivers sometimes use a different technique for called “engine braking” to avoid burning their brakes during steep descents. Engine braking is using the slowing forces of an engine in low gear rather than brakes to slow down.

Practice Problems

How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 °C ?

Calculate the change in temperature of 1.0 kg of water that is initially at room temperature if 3.0 kJ of heat is added.

Check Your Understanding

Use these questions to assess student achievement of the section’s learning objectives. If students are struggling with a specific objective, these questions will help identify which and direct students to the relevant content.

• The mass difference between two objects causes heat transfer.
• The density difference between two objects causes heat transfer.
• The temperature difference between two systems causes heat transfer.
• The pressure difference between two objects causes heat transfer.
• The overall direction of heat transfer is from the higher-temperature object to the lower-temperature object.
• The overall direction of heat transfer is from the lower-temperature object to the higher-temperature object.
• The direction of heat transfer is first from the lower-temperature object to the higher-temperature object, then back again to the lower-temperature object, and so-forth, until the objects are in thermal equilibrium.
• The direction of heat transfer is first from the higher-temperature object to the lower-temperature object, then back again to the higher-temperature object, and so-forth, until the objects are in thermal equilibrium.
• conduction, radiation, and reflection
• conduction, reflection, and convection
• convection, radiation, and reflection
• conduction, radiation, and convection

True or false—Conduction and convection cannot happen simultaneously

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Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute Texas Education Agency (TEA). The original material is available at: https://www.texasgateway.org/book/tea-physics . Changes were made to the original material, including updates to art, structure, and other content updates.

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• Book title: Physics
• Publication date: Mar 26, 2020
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Heat Capacity Example Problem

Calculate the heat needed to raise water from freezing to boiling

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Heat capacity is the amount of heat energy required to change the temperature of a substance. This example problem demonstrates how to calculate heat capacity .

Problem: Heat Capacity of Water From Freezing to Boiling Point

What is the heat in joules required to raise the temperature of 25 grams of water from 0 degrees C to 100 degrees C? What is the heat in calories?

Useful information: specific heat of water = 4.18 J/g·°C Solution:

Use the formula

q = mcΔT where q = heat energy m = mass c = specific heat ΔT = change in temperature q = (25 g)x(4.18 J/g·°C)[(100 C - 0 C)] q = (25 g)x(4.18 J/g·°C)x(100 C) q = 10450 J Part II 4.18 J = 1 calorie x calories = 10450 J x (1 cal/4.18 J) x calories = 10450/4.18 calories x calories = 2500 calories Answer: 10450 J or 2500 calories of heat energy are required to raise the temperature of 25 grams of water from 0 degrees C to 100 degrees C.

Tips for Success

• The most common mistake people make with this calculation is using incorrect units. Make certain temperatures are in Celsius. Convert kilograms to grams.
• Be mindful of significant figures, particularly when working problems for homework or an exam.
• Specific Heat Example Problem
• Heat of Formation Worked Problem
• Heat Capacity Definition
• Calorimetry and Heat Flow: Worked Chemistry Problems
• Specific Heat Capacity in Chemistry
• Molar Heat Capacity Definition and Examples
• Heat of Vaporization Example Problem
• Heat of Fusion Example Problem: Melting Ice
• Enthalpy Change Example Problem
• Calculate Energy Required to Turn Ice Into Steam
• Enthalpy Definition in Chemistry and Physics
• Calculate the Change in Entropy From Heat of Reaction
• Converting Kilometers to Meters
• Calorimeter Definition in Chemistry
• What Happens When You Touch Dry Ice?
• A List of Common General Chemistry Problems

• Why Does Water Expand When It Freezes
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• Faraday Cage
• Oil Drop Experiment
• Magnetic Monopole
• Why Do Fireflies Light Up
• Types of Blood Cells With Their Structure, and Functions
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• Why Does Ice Float on Water
• Why Does Oil Float on Water
• How Do Clouds Form
• What Causes Lightning
• How are Diamonds Made
• Types of Meteorites
• Types of Volcanoes
• Types of Rocks

Specific Heat and Heat Capacity

Specific heat is defined as the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. It plays a crucial role in understanding how different materials respond to heating and cooling and describes their ability to store and release thermal energy . For example, water has a higher specific heat than metals. It means that more heat is required to raise water’s temperature by one degree than metals’, as shown in the image below.

On the other hand, heat capacity is the amount of heat required to increase the temperature of the entire substance by one degree Celsius.

Specific Heat and Heat Capacity Formula

The specific heat can be calculated from the amount of heat transferred into and out of a substance. The heat transfer equation provides a quantitative relationship between heat transfer, substance’s mass, specific heat, and temperature change. It is shown as follows:

• Q represents the heat transfer in Joules (J) or calories (cal).
• m denotes the mass of the substance in grams (g) or kilograms (kg) being heated or cooled.
• c is the specific heat
• ΔT represents the temperature change in degrees Celsius ( ∘ C) of the substance

This formula assumes no phase change occurs during heating or cooling processes and applies specifically to substances with constant specific heat within a given temperature range.

Rearranging the above equation, one can find the expression for specific heat.

The heat capacity (C) can be calculated by multiplying the specific heat with the mass. Therefore,

The unit of specific heat is Joules per gram per degree Celsius or J/g∙ ∘ C. Another unit of specific heat is calories per gram per degree Celsius or J/cal∙ ∘ C. The temperature change (∆T) in the Celsius (C) scale is the same as that in the Kelvin (K) scale, although the temperature values differ. Therefore, one can replace ∘ C with K. In that case, the SI unit of specific heat is Joules per kilogram per degree Kelvin or J/kg∙K.

The unit of heat capacity is J/ ∘ C or cal/ ∘ C.

Molar Specific Heat

The specific heat of a substance can also be described in terms of its molar amount. In that case, we use a term called molar specific heat. It is defined as the amount of heat required to change the temperature of one mole of a substance by one degree. It is represented in the unit of Joules per mole per degree Celsius or J/mol ∘ C.

Specific Heat Table

The specific heat values can be calculated for different substances from the above formula. Below is a table that shows the values for a few common substances:

Water has a particularly high specific heat compared to many other substances. Its specific heat capacity is 4.184 J/g°C, which means it takes 4.184 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius. Let us discuss the significance of this remarkable property of water.

Specific Heat of Water

Water has an exceptionally high capacity to absorb and retain heat energy without undergoing large temperature changes. This property is significant in thermoregulation in nature and human-made systems.

Thermoregulation is the process by which living organisms maintain their internal body temperature within a narrow range despite fluctuations in the external environment. In nature, bodies of water such as oceans, lakes, and rivers act as thermal regulators by absorbing excess heat during the daytime and slowly releasing it at night. This helps maintain stable temperatures in surrounding areas and supports diverse ecosystems.

Understanding water’s specific heat is crucial for designing efficient heating and cooling artificial systems like car radiators and hot water pipes. Water’s ability to store large amounts of thermal energy makes it an ideal medium for transferring heat between different areas while minimizing temperature fluctuations.

Difference Between Specific Heat and Heat Capacity

Below is a table summarizing the difference between specific heat and heat capacity.

Example Problems with Solutions

Problem 1 :  A 200 g piece of aluminum initially at 80°C is dropped into 400 mL of water at 20°C. If the final temperature of the system is 40°C, calculate the heat transferred to the water. The specific heat of aluminum is 0.9 J/g∙°C, and the specific heat of water is 4.18 J/g∙°C.

The heat transfer equation is given by:

Heat lost by aluminum is:

\[ q_{\text{Al}} = m_{\text{Al}} \cdot c_{\text{Al}} \cdot \Delta T_{\text{Al}} \]

\[ q_{\text{Al}} = (200 \, \text{g}) \cdot (0.9 \, \text{J/g°C}) \cdot (40 \, \text{°C} – 80 \, \text{°C}) \]

\[ q_{\text{Al}} = -7200 \, \text{J} \]

The negative sign implies that aluminum loses heat to water. Therefore, the heat transferred to water is 7200 J .

Problem 2 : A 150 g piece of copper is heated from 20°C to 100°C. Calculate the heat energy absorbed by the copper. The specific heat of copper is 0.39 J/g∙°C.

\[ q = m \cdot c \cdot \Delta T \]

\[ q_{\text{copper}} = m_{\text{copper}} \cdot c_{\text{copper}} \cdot \Delta T_{\text{copper}} \]

\[ q_{\text{copper}} = (150 \, \text{g}) \cdot (0.39 \, \text{J/g°C}) \cdot (100 \, \text{°C} – 20 \, \text{°C}) \]

\[ q_{\text{copper}} = 4680 \, \text{J} \]

Therefore, the heat absorbed by copper is 4680 J .

• Specific Heat – Hyperphysics.phy-astr.gsu.edu
• Specific Heat Capacity – Study.com
• What is Specific Heat? – Chemistrytalk.org
• Specific Heat –  Phys.libretexts.org
• Specific Heats: the relation between temperature change and heat – Web.mit.edu
• Heat – Chemed.chem.purdue.edu
• Heat Capacity and Specific Heat – Chem.libretexts.org
• Specific Heat Definition – Thoughtco.com
• Specific Heat Capacity – Energyeducation.ca
• Specific heat and Heat Capacity – Cpanhd.sitehost.iu.edu

Article was last reviewed on Sunday, August 11, 2024

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Heat Capacity: Problem-Solving

Previous video 19.10: mean free path and mean free time, next video 19.12: dalton’s law of partial pressure.

q lost, metal = q gained, water

(mass) (Δt) (C p, metal ) = (mass) (Δt) (C p, water )

(21.5) (100 − x) (0.449) = (132.0) (x − 20) (4.184)

a) 100 − x is the Δt for the metal; it starts at 100.0 °C and drops to some unknown, final value. b) x − 20 is the Δt for the water; it starts at 20.0 °C and rises to some unknown, final value. c) Since both metal and water wind up at the same ending value, we need to use only one unknown for the two Δt expressions.

(2150 − 21.5x) (0.449) = (132x − 2640) (4.184) 965.35 − 9.6535x = 552.288x − 11045.76 561.9415x = 12011.11

1. How many joules of energy did the water absorb? 2. How many joules of energy did the metal lose? 3. What is the specific heat of metal?

1) q = (50.0 g) (3.1 °C) (4.184 J g¯ 1 °C¯ 1 ) = 648.52 J 2) 648.52 J 3) 648.52 J / 70.9 °C = 9.147 J/°C 4) 9.147 J/°C / 12.48 g = 0.733 J g¯ 1 °C¯ 1

(9.15 g) (334.166 J g¯ 1 ) = 3057.62 J

q = (mass) (Δt) (C p, metal ) 3057.62 J = (43.2 g) (89.0 °C) (x) x = 0.795 J g¯ 1 °C¯ 1

q = (15.0 g) (334.166 J g¯ 1 ) = 5012.49 J Note that the 100 g of water is not mentioned yet.

q = (115 g) (10.0 °C) (4.184 J g¯ 1 °C¯ 1 ) = 4811.6 J Note the inclusion of the melted 15 g of ice. Also, notice that the water was at zero °C. We know this from the presence of the ice.

(5012.49 J + 4811.6 J) = (35.0 g) (70.0 °C) (x) x = 4.01 J g¯ 1 °C¯ 1

(6.02 kJ/mol) (109.5 g / 18.015 g/mol) = 36.5912 kJ

36591.2 J = (500.0 g) (153.0 °C) (x) x = 0.4783 J/g-°C

(0.4783 J/g °C) (x) = 26.31 J/mol °C x = 55.0 g/mol The element is manganese.

(a) How many joules of energy did the water absorb? (b) How many joules of energy did the metal lose? (c) What is the specific heat capacity of the metal?

q = (50.0 g) (3.1 °C) (4.181 J g¯ 1 °C¯ 1 ) = 648.52 J

q = 648.52 J

(50.0 g) (3.1 °C) (4.181 J g¯ 1 °C¯ 1 ) = (12.48 g) (74.0 °C) (x) Solve for x.

Comment: notice that the two metals are being added to each other. Imagine a situation where each sample is composed of dust or very small pellets. Then, the two dry samples are rapidly mixed together. (90.0 g) (111.4 °C) (x) = (210.0 g) (24.6 °C) (0.385 J/g °C) x = 0.198 J/g °C

(12.0 g) (334.166 J/g) = 4009.992 J 4009.992 J = (31.0 g) (88.0 °C) (x) x = 1.47 J / g °C

the heat lost by the methanol goes to (1) heating the ethanol and (2) heating the calorimeter (25.95 g) (6.95 °C) (x) = (38.65 g) (3.95 °C) (2.44 J g -1 °C -1 ) + (3.95 °C) (19.3 J/C) x = 2.49 J g -1 °C -1

0.451 J/g °C

(250.0 mL) (1.261 g/mL) = 315.25 g

(135 g) (180.3 °C) (0.451 J/g °C) = (315.25 g) (21.2 °C) (x) Comment: 180.3 comes from 225 − 44.7 and 21.1 comes from 44.7 − 23.5 10977.5655 = 6651.986x x = 1.65 J/g °C

a) Calculate the specific heat of the metal. b) What is the atomic weight? c) Identify the metal.

q = (55.7 g) (8.4 °C) (4.184 J g¯ 1 °C¯ 1 = 1957.61 J 1957.61 J = (130 g) (67.8 °C) (x) sp ht. = 0.222 J g¯ 1 °C¯ 1 Use the Dulong–Petit law : M = 3R / sp. ht M = (3) (8.31446 J mol¯ 1 K¯ 1 ) / 0.222 J g¯ 1 K¯ 1 M = 112 g /mol Cadmium