Heat and Heat Transfer Methods
101 Temperature Change and Heat Capacity
Learning Objectives
- Observe how heat transfer causes changes in temperature and relates to mass.
- Calculate the final temperature of a system after thermal energy is exchanged between two objects.
One of the most common effects of heat transfer is a change in temperature. When a system is heated, its temperature typically increases; when it is cooled, its temperature drops. In this section, we assume there is no phase change and that no mechanical work is being done. The amount of heat transferred to or from a substance depends on three main factors:
- the magnitude of the temperature change,
- the mass of the substance, and
- the type of substance (and its phase).
These relationships are rooted in the idea that temperature reflects the average kinetic energy of atoms or molecules in a substance. A higher temperature means greater internal energy. So, increasing the temperature of a substance means transferring energy to increase that internal motion. Since internal energy depends on both temperature and the number of particles (related to mass), more heat is required for larger masses or larger temperature changes.
Different substances respond differently to heat because they store energy in unique ways. For example, heating 1 kg of copper and 1 kg of water by the same amount requires vastly different amounts of heat. These differences are described by a property called specific heat.
Heat Transfer and Temperature Change
The relationship between heat transfer and temperature change is given by:
Where:
- [latex]Q[/latex] is the heat transferred (in joules or kilocalories),
- [latex]m[/latex] is the mass of the substance (in kilograms),
- [latex]c[/latex] is the specific heat capacity (in [latex]\text{J/(kg·ºC)}[/latex] or [latex]\text{kcal/(kg·ºC)}[/latex]), and
- [latex]\Delta T[/latex] is the temperature change in degrees Celsius (or kelvins).
The units of [latex]c[/latex] match the units used for [latex]Q[/latex]. If you use joules, then [latex]c[/latex] is in [latex]\text{J/(kg·ºC)}[/latex]. If using kilocalories, [latex]c[/latex] is in [latex]\text{kcal/(kg·ºC)}[/latex]. Since the size of 1ºC and 1 K is the same, the temperature difference [latex]\Delta T[/latex] is numerically identical in either unit.
The specific heat [latex]c[/latex] tells us how resistant a material is to temperature change. A high specific heat means a substance requires more energy to heat up or cool down. Water, for instance, has one of the highest specific heats of any common substance. This is biologically important—it helps regulate body temperature and stabilizes climate by absorbing or releasing large amounts of heat with only small temperature changes.
In contrast, metals like iron or copper have much lower specific heat values and change temperature quickly with heat transfer. This is why metal objects often feel hotter or colder to the touch—they heat up or cool down rapidly compared to your skin or water.
While there is no easy formula to calculate specific heat for a given material, values are typically found in reference tables like Table 101.1. The specific heat may vary slightly with temperature, but for most solids and liquids (except gases), this variation is usually small.
Key point for health and life sciences: Understanding specific heat helps explain why water is such an effective temperature buffer in biological systems. It also underlies thermal regulation, fever response, and climate science. For example, the human body is over 60% water—its high specific heat helps prevent rapid body temperature changes even when external conditions fluctuate.
Example 101.1 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 liters of water from [latex]\text{20.0ºC}[/latex] to [latex]\text{80.0ºC}[/latex]. (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?
Strategy
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 the heat transfer for the given temperature change and mass of water and aluminum. The specific heat values for water and aluminum are given in Table 101.1.
Solution
Because water is in thermal contact with the aluminum, the pan and the water are at the same temperature.
- Calculate the temperature difference:
[latex]\text{Δ}T={T}_{\text{f}}-{T}_{\text{i}}=\text{60.0ºC.}[/latex]
- Calculate the mass of water. Because the density of water is [latex]\text{1000}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}[/latex], one liter of water has a mass of 1 kg, and the mass of 0.250 liters of water is [latex]{m}_{w}=0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex].
- Calculate the heat transferred to the water. Use the specific heat of water in Table 101.1:
[latex]{Q}_{w}={m}_{w}{c}_{w}\text{Δ}T=\left(0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(\text{4186}\phantom{\rule{0.25em}{0ex}}\text{J/kg}\text{ºC}\right)\left(\text{60.0}\text{ºC}\right)=\text{62}.8 kJ.[/latex]
- Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 101.1:
[latex]{Q}_{\text{Al}}={m}_{\text{Al}}{c}_{\text{Al}}\text{Δ}T=\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)\left(\text{60.0ºC}\right){\text{= 27.0 × 10}}^{4}\text{J = 27.0 kJ.}[/latex]
- Compare the percentage of heat going into the pan versus that going into the water. First, find the total transferred heat:
[latex]{Q}_{\text{Total}}={Q}_{\text{W}}+{Q}_{\text{Al}}=\text{62}\text{.}8\phantom{\rule{0.25em}{0ex}}\text{kJ}+\text{27}\text{.}0\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\text{kJ = 89}\text{.}8\phantom{\rule{0.25em}{0ex}}\text{kJ.}[/latex]
Thus, the amount of heat going into heating the pan is
and the amount going into heating the water is
Discussion
In this example, the heat transferred to the container is a significant fraction of the total transferred heat. Although the mass of the pan is twice that of the water, the specific heat of water is over four times greater than that of aluminum. Therefore, it takes a bit more than twice the heat to achieve the given temperature change for the water as compared to the aluminum pan.
Example 101.2 Calculating the Temperature Increase from the Work Done on a Substance: Truck Brakes Overheat on Downhill Runs
Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into increased internal energy (higher temperature) of the brake material. This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck. The problem is that the mass of the truck is large compared with that of the brake material absorbing the energy, and the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment.
Calculate the temperature increase of 100 kg of brake material with an average specific heat of [latex]\text{800 J/kg}\cdot \text{ºC}[/latex] if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at a constant speed.
Strategy
If the brakes are not applied, gravitational potential energy is converted into kinetic energy. When brakes are applied, gravitational potential energy is converted into internal energy of the brake material. We first calculate the gravitational potential energy [latex]\left(\text{Mgh}\right)[/latex] that the entire truck loses in its descent and then find the temperature increase produced in the brake material alone.
Solution
- Calculate the change in gravitational potential energy as the truck goes downhill
[latex]\text{Mgh}=\left(\text{10,000 kg}\right)\left(9\text{.}\text{80 m}{\text{/s}}^{2}\right)\left(\text{75.0 m}\right)=\text{7.}\text{35}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{J.}[/latex]
- Calculate the temperature from the heat transferred using [latex]Q\text{=}\text{Mgh}[/latex] and
[latex]\text{Δ}T=\frac{Q}{\text{mc}}\text{,}[/latex]
where [latex]m[/latex] is the mass of the brake material. Insert the values [latex]m=\text{100 kg}[/latex] and [latex]c=\text{800 J/kg}\cdot \text{ºC}[/latex] to find
[latex]\text{Δ}T=\frac{\left(7\text{.35}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{J}\right)}{\left(\text{100 kg}\right)\left(\text{800 J/kgºC}\right)}=\text{92ºC.}[/latex]
Discussion
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, so this technique is not practical. However, the same idea underlies the recent hybrid technology of cars, where mechanical energy (gravitational potential energy) is converted by the brakes into electrical energy (battery).
Note that Example 101.2 is an illustration of the mechanical equivalent of heat. Alternatively, the temperature increase could be produced by a blow torch instead of mechanically.
| Substances | Specific heat (c) | |
|---|---|---|
| Solids | J/kg⋅ºC | kcal/kg⋅ºC2 |
| Aluminum | 900 | 0.215 |
| Asbestos | 800 | 0.19 |
| Concrete, granite (average) | 840 | 0.20 |
| Copper | 387 | 0.0924 |
| Glass | 840 | 0.20 |
| Gold | 129 | 0.0308 |
| Human body (average at 37 °C) | 3500 | 0.83 |
| Ice (average, -50°C to 0°C) | 2090 | 0.50 |
| Iron, steel | 452 | 0.108 |
| Lead | 128 | 0.0305 |
| Silver | 235 | 0.0562 |
| Wood | 1700 | 0.4 |
| Liquids | ||
| Benzene | 1740 | 0.415 |
| Ethanol | 2450 | 0.586 |
| Glycerin | 2410 | 0.576 |
| Mercury | 139 | 0.0333 |
| Water (15.0 °C) | 4186 | 1.000 |
| Gases3 | ||
| Air (dry) | 721 (1015) | 0.172 (0.242) |
| Ammonia | 1670 (2190) | 0.399 (0.523) |
| Carbon dioxide | 638 (833) | 0.152 (0.199) |
| Nitrogen | 739 (1040) | 0.177 (0.248) |
| Oxygen | 651 (913) | 0.156 (0.218) |
| Steam (100°C) | 1520 (2020) | 0.363 (0.482) |
Example 101.3 Calculating the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan
Suppose you pour 0.250 kg of [latex]\text{20}\text{.0ºC}[/latex] water (about a cup) into a 0.500-kg aluminum pan off the stove with a temperature of [latex]\text{150ºC}[/latex]. Assume that the pan is placed on an insulated pad and that a negligible amount of water boils off. What is the temperature when the water and pan reach thermal equilibrium a short time later?
Strategy
The pan is placed on an insulated pad so that little heat transfer occurs with the surroundings. Originally the pan and water are not in thermal equilibrium: the pan is at a higher temperature than the water. Heat transfer then restores thermal equilibrium once the water and pan are in contact. Because heat transfer between the pan and water takes place rapidly, the mass of evaporated water is negligible and the magnitude of the heat lost by the pan is equal to the heat gained by the water. The exchange of heat stops once a thermal equilibrium between the pan and the water is achieved. The heat exchange can be written as [latex]\mid {Q}_{\text{hot}}\mid ={Q}_{\text{cold}}[/latex].
Solution
- Use the equation for heat transfer [latex]Q=\text{mc}\text{Δ}T[/latex] to express the heat lost by the aluminum pan in terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final temperature:
[latex]{Q}_{\text{hot}}={m}_{\text{Al}}{c}_{\text{Al}}\left({T}_{\text{f}}-\text{150ºC}\right)\text{.}[/latex]
- Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water and the final temperature:
[latex]{Q}_{\text{cold}}={m}_{W}{c}_{W}\left({T}_{\text{f}}-\text{20.0ºC}\right)\text{.}[/latex]
- Note that [latex]{Q}_{\text{hot}}0[/latex] and [latex]{Q}_{\text{cold}}>0[/latex] and that they must sum to zero because the heat lost by the hot pan must be the same as the heat gained by the cold water:
[latex]\begin{array}{lll}\hfill {Q}_{\text{cold}}\text{+}{Q}_{\text{hot}}& \text{=}& \text{0,}\\ \hfill {Q}_{\text{cold}}& =& {\text{–Q}}_{\text{hot}},\\ {m}_{W}{c}_{W}\left({T}_{\text{f}}-\text{20.0ºC}\right)& =& {\mathrm{-m}}_{\mathrm{Al}}{c}_{\mathrm{Al}}\left({T}_{\text{f}}-\text{150ºC.}\right)\end{array}[/latex]
- This an equation for the unknown final temperature, [latex]{T}_{\text{f}}[/latex]
- Bring all terms involving [latex]{T}_{\text{f}}[/latex] on the left hand side and all other terms on the right hand side. Solve for [latex]{T}_{\text{f}}[/latex],
[latex]{T}_{\text{f}}=\frac{{m}_{\text{Al}}{c}_{\text{Al}}\left(\text{150ºC}\right)+{m}_{W}{c}_{W}\left(\text{20}\text{.0ºC}\right)}{{m}_{\text{Al}}{c}_{\text{Al}}+{m}_{W}{c}_{W}}\text{,}[/latex]
and insert the numerical values:
[latex]\begin{array}{lll}{T}_{\text{f}}& =& \frac{\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)\left(\text{150ºC}\right)\text{+}\left(\text{0.250 kg}\right)\left(\text{4186 J/kgºC}\right)\left(\text{20.0ºC}\right)}{\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)+\left(\text{0.250 kg}\right)\left(\text{4186 J/kgºC}\right)}\\ & =& \frac{\text{88430 J}}{\text{1496.5 J/ºC}}\\ & =& \text{59}\text{.1ºC.}\end{array}[/latex]
Discussion
This is a typical calorimetry problem—two bodies at different temperatures are brought in contact with each other and exchange heat until a common temperature is reached. Why is the final temperature so much closer to [latex]\text{20.0ºC}[/latex] than [latex]\text{150ºC}[/latex]? The reason is that water has a greater specific heat than most common substances and thus undergoes a small temperature change for a given heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase its temperature appreciably. This explains why the temperature of a lake stays relatively constant during a day even when the temperature change of the air is large. However, the water temperature does change over longer times (e.g., summer to winter).
Take-Home Exploration: How Land and Water Heat Differently
Which heats up faster—land or water? This simple experiment illustrates how different specific heats lead to different temperature responses, and why this matters for climate, weather patterns, and even coastal ecosystems.
To investigate:
- Place equal masses of dry sand (or soil) and water into two small containers. (Since sand is denser than water, use about 50% more water by volume to achieve equal mass.)
- Expose both containers to a heat source—such as a heat lamp or warm sunlight—for the same duration.
- Measure and record the final temperature of each material.
- Next, continue heating both containers until they reach the same final temperature.
- Remove the containers from the heat source and record their temperature every 5 minutes for the next 30 minutes as they cool.
Which material cools faster? This simple exercise demonstrates a real-world effect: land heats and cools faster than water due to its lower specific heat. This difference is responsible for important atmospheric patterns like land breezes and sea breezes, where coastal winds shift directions based on heating and cooling cycles of land versus water.
Check Your Understanding
If 25 kJ of heat is required to raise the temperature of a material from [latex]25^\circ \text{C}[/latex] to [latex]30^\circ \text{C}[/latex], how much heat would be needed to raise it from [latex]45^\circ \text{C}[/latex] to [latex]50^\circ \text{C}[/latex]?
The temperature difference in both cases is [latex]5^\circ \text{C}[/latex], so the amount of heat required is the same—25 kJ. The total heat transfer depends only on the temperature change, not the initial or final temperature.
Section Summary
- The amount of heat [latex]Q[/latex] transferred that results in a temperature change [latex]\Delta T[/latex] in an object with mass [latex]m[/latex] is calculated using the formula:
[latex]Q = mc\Delta T[/latex]
where [latex]c[/latex] is the specific heat of the material. This equation is also used to define specific heat.
Conceptual Questions
- What three factors affect the heat transfer that is necessary to change an object’s temperature?
- The brakes in a car increase in temperature by [latex]\text{Δ}T[/latex] when bringing the car to rest from a speed [latex]v[/latex]. How much greater would [latex]\text{Δ}T[/latex] be if the car initially had twice the speed? You may assume the car to stop sufficiently fast so that no heat transfers out of the brakes.
Problems & Exercises
- On a hot day, the temperature of an 80,000-L swimming pool increases by [latex]1\text{.}\text{50ºC}[/latex]. What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation.
- Show that [latex]1\phantom{\rule{0.25em}{0ex}}\text{cal/g}\cdot \text{ºC}=1\phantom{\rule{0.25em}{0ex}}\text{kcal/kg}\cdot \text{ºC}[/latex].
- To sterilize a 50.0-g glass baby bottle, we must raise its temperature from [latex]\text{22}\text{.}0\text{ºC}[/latex] to [latex]\text{95}\text{.}0º\text{C}[/latex]. How much heat transfer is required?
- The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at [latex]\text{20}\text{.}0º\text{C}[/latex]: (a) water; (b) concrete; (c) steel; and (d) mercury.
- Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm per rub, and with an average frictional force of 40.0 N, what is the temperature increase? The mass of tissues warmed is only 0.100 kg, mostly in the palms and fingers.
- A 0.250-kg block of a pure material is heated from [latex]\text{20}\text{.}0º\text{C}[/latex] to [latex]\text{65}\text{.}0º\text{C}[/latex] by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed.
- Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water?
- (a) The number of kilocalories in food is determined by calorimetry techniques in which the food is burned and the amount of heat transfer is measured. How many kilocalories per gram are there in a 5.00-g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a [latex]\text{54}\text{.}9º\text{C}[/latex] temperature increase? (b) Compare your answer to labeling information found on a package of peanuts and comment on whether the values are consistent.
- Following vigorous exercise, the body temperature of an 80.0-kg person is [latex]\text{40}\text{.}0º\text{C}[/latex]. At what rate in watts must the person transfer thermal energy to reduce the the body temperature to [latex]\text{37}\text{.}0º\text{C}[/latex] in 30.0 min, assuming the body continues to produce energy at the rate of 150 W? [latex]\left(\text{1 watt = 1 joule/second or 1 W = 1 J/s}\phantom{\rule{0.25em}{0ex}}\right)[/latex].
- Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails [latex]\left(\text{1 watt = 1 joule/second or 1 W = 1 J/s and 1 MW = 1 megawatt}\right)[/latex]. (a) Calculate the rate of temperature increase in degrees Celsius per second
- [latex]\left(\text{ºC/s}\right)[/latex] if the mass of the reactor core is [latex]1\text{.}\text{60}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex] and it has an average specific heat of [latex]\text{0.3349 kJ/kgº}\cdot \text{C}[/latex]. (b) How long would it take to obtain a temperature increase of [latex]\text{2000º}\text{C}[/latex], which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the [latex]5×{\text{10}}^{5}\text{-kg}[/latex] steel containment vessel would also begin to heat up.)
Footnotes
1 The values for solids and liquids are at constant volume and at [latex]\text{25ºC}[/latex], except as noted.
2 These values are identical in units of [latex]\text{cal/g}\cdot ºC[/latex].
3[latex]{c}_{\text{v}}[/latex] at constant volume and at [latex]\text{20}\text{.}0ºC[/latex], except as noted, and at 1.00 atm average pressure. Values in parentheses are [latex]{c}_{\text{p}}[/latex] at a constant pressure of 1.00 atm.
Glossary
- specific heat
- the amount of heat necessary to change the temperature of 1.00 kg of a substance by 1.00 ºC
the spontaneous transfer of energy due to a temperature difference
the transfer of energy by a force that causes an object to be displaced; the product of the component of the force in the direction of the displacement and the magnitude of the displacement
the quantity of matter in a substance; measured in kilograms
the energy an object has by reason of its motion, equal to [latex]\frac{1}{2}{\text{mv}}^{2}[/latex] for the translational (i.e., non-rotational) motion of an object of mass [latex]m[/latex] moving at speed [latex]v[/latex]
the amount of heat necessary to change the temperature of 1.00 kg of a substance by 1.00 ºC
