Contents

Temperature Change and Heat Capacity

One of the major effects of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase change and that no work is done on or by the system. Experiments show that the transferred heat depends on three factors—the change in temperature, the mass of the system, and the substance and phase of the substance.

Figure a shows a copper-colored cylinder of mass m and temperature change delta T. The heat Q, shown as a wavy rightward horizontal arrow, is transferred to the cylinder from the left. To the right of this image is a similar image, except that the heat transferred Q prime is twice the heat Q. The temperature change of this second cylinder, which is also labeled m, is two delta T. This cylinder is surrounded by small black wavy lines radiating outward. Figure b shows the same two cylinders as in Figure a. The left cylinder is labeled m and delta T and has a wavy heat arrow pointing at it from the left that is labeled Q. The right cylinder is labeled two m and delta T and has a wavy heat arrow pointing to it from the left labeled Q prime equals two Q. Figure c shows the same copper cylinder of mass m and with temperature change delta T, with heat Q being transferred to it. To the right of this cylinder, Q prime equals ten point eight times Q is being transferred to another cylinder filled with water whose mass and change in temperature are the same as that of the copper cylinder.

Figure 14.4 The heat

Q Q size 12{Q} {}

transferred to cause a temperature change depends on the magnitude of the temperature change, the mass of the system, and the substance and phase involved. (a) The amount of heat transferred is directly proportional to the temperature change. To double the temperature change of a mass

m m size 12{m} {}

, you need to add twice the heat. (b) The amount of heat transferred is also directly proportional to the mass. To cause an equivalent temperature change in a doubled mass, you need to add twice the heat. (c) The amount of heat transferred depends on the substance and its phase. If it takes an amount

Q Q size 12{Q} {}

of heat to cause a temperature change

Δ T Δ T size 12{ΔT} {}

in a given mass of copper, it will take 10.8 times that amount of heat to cause the equivalent temperature change in the same mass of water assuming no phase change in either substance.

The dependence on temperature change and mass are easily understood. Owing to the fact that the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the absolute temperature and the number of atoms or molecules. Owing to the fact that the transferred heat is equal to the change in the internal energy, the heat is proportional to the mass of the substance and the temperature change. The transferred heat also depends on the substance so that, for example, the heat necessary to raise the temperature is less for alcohol than for water. For the same substance, the transferred heat also depends on the phase (gas, liquid, or solid).

Heat Transfer and Temperature Change

The quantitative relationship between heat transfer and temperature change contains all three factors:

Q = mc Δ T, Q = mc Δ T, size 12{Q= ital "mc"ΔT,} {}

14.2

where $Q Q size 12{Q} {}$ is the symbol for heat transfer, $m m size 12{m} {}$ is the mass of the substance, and $Δ T Δ T$ is the change in temperature. The symbol $c c size 12{c} {}$ 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 $1 .00ºC 1 .00ºC$ . The specific heat $c c$ is a property of the substance; its SI unit is $J/ (kg \cdot K) J/ (kg \cdot K)$ or $J/ (kg \cdotºC). J/ (kg \cdotºC).$ Recall that the temperature change $(Δ T) (Δ T)$ is the same in units of kelvin and degrees Celsius. If heat transfer is measured in kilocalories, then the unit of specific heat is $kcal/ (kg \cdotºC). kcal/ (kg \cdotºC).$

Values of specific heat must generally be looked up in tables, because there is no simple way to calculate them. In general, the specific heat also depends on the temperature. Table 14.1 lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak. We see from this table that the specific heat of water is five times that of glass and ten times that of iron, which means that it takes five times as much heat to raise the temperature of water the same amount as for glass and ten times as much heat to raise the temperature of water as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth.

Example 14.1

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 $20.0ºC 20.0ºC$ to $80.0ºC 80.0ºC$ . (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 14.1.

Solution

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

Calculate the temperature difference:
$Δ T = T_{f} - T_{i} = 60.0ºC. Δ T = T_{f} - T_{i} = 60.0ºC.$
14.3
Calculate the mass of water. Because the density of water is $10001000$ , one liter of water has a mass of 1 kg, and the mass of 0.250 liters of water is $m_{w} = 0 . 250 m_{w} = 0 . 250 size 12{m rSub { size 8{w} } =0 "." "25"`"kg"} {}$ .
Calculate the heat transferred to the water. Use the specific heat of water in Table 14.1:
$Q_{w} = m_{w} c_{w} Δ T = (0.250) (4186 J/kg ºC) (60.0 ºC) = 62 .8 kJ. Q_{w} = m_{w} c_{w} Δ T = (0.250) (4186) (60.0 ºC) = 62 .8 kJ.$
14.4
Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 14.1:
$Q_{Al} = m_{Al} c_{Al} Δ T = (0.500 kg) (900 J/kgºC) (60.0ºC) {= 27.0 × 10}^{4} J = 27.0 kJ. Q_{Al} = m_{Al} c_{Al} Δ T = (0.500 kg) (900 J/kgºC) (60.0ºC) {= 27.0 × 10}^{4} J = 27.0 kJ.$
14.5
Compare the percentage of heat going into the pan versus that going into the water. First, find the total transferred heat:
$Q_{Total} = Q_{W} + Q_{Al} = 62 . 8 .0 Q_{Total} = Q_{W} + Q_{Al} = 62 . 8 .0 size 12{Q rSub { size 8{"Total"} } =Q rSub { size 8{W} } +Q rSub { size 8{"Al"} } ="62" "." 8`"kJ "+" 89" "." 5`"kJ = 152" "." 3`"kJ"} {}$
14.6

Thus, the amount of heat going into heating the pan is

\frac{27.0}{89.8} \times 100% =30.1%, \frac{27.0}{89.8} \times 100% =30.1%, size 12{ {  {"62" "." 8`"kJ"}  over  {"152" "." 3`"kJ"} }  times "100"%="41"%} {}

14.7

and the amount going into heating the water is

\frac{62.8}{89.8} \times 100% =69.9% . \frac{62.8}{89.8} \times 100% =69.9% . size 12{ {  {"62" "." 8`"kJ"}  over  {"89" "." 8`"kJ"} }  times "100"%="69.9"% "." } {}

14.8

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.

The figure shows a truck coming from the left and moving on a road which is sloping downhill to the right. Smoke is coming from the area of the wheels of the truck.

Figure 14.5 The smoking brakes on this truck are a visible evidence of the mechanical equivalent of heat.

Example 14.2

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 $800 J/kg \cdot ºC 800 J/kg \cdot ºC$ 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 $(Mgh) (Mgh) size 12{ $ ital "Mgh" $ } {}$ 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
$Mgh = (10,000 kg) (9 . 80 m {/s}^{2}) (75.0 m) = 7. 35 \times 10^{6} J. Mgh = (10,000 kg) (9 . 80 m {/s}^{2}) (75.0 m) = 7. 35 \times 10^{6}$
14.9
Calculate the temperature from the heat transferred using $Q = Mgh Q = Mgh size 12{Q"= " ital "Mgh"} {}$ and
$Δ T = \frac{Q}{mc}, Δ T = \frac{Q}{mc}, size 12{ΔT= { {Q} over { ital "mc"} } } {}$
14.10

where $m m$ is the mass of the brake material. Insert the values $m = 100 kg m = 100 kg$ and $c = 800 J/kg \cdot ºC c = 800 J/kg \cdot ºC$ to find
$Δ T = \frac{(7.35 \times 10^{6} J)}{(100 kg) (800 J/kgºC)} = 92ºC. Δ T = \frac{(7.35 \times 10^{6})}{(100 kg) (800 J/kgºC)} = 92ºC.$
14.11

Discussion

This same idea underlies the recent hybrid technology of cars, where mechanical energy (gravitational potential energy) is converted by the brakes into electrical energy (battery).

Substances	Specific heat (c)
Solids	J/kg⋅ºC	kcal/kg⋅ºC1 1 These values are identical in units of $cal/g \cdotºC cal/g \cdotºC$ .
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
Gases 2 2 $c_{v} c_{v}$ at constant volume and at $20 . 0ºC 20 . 0ºC$ , except as noted, and at 1.00 atm average pressure. Values in parentheses are $c_{p} c_{p}$ at a constant pressure of 1.00 atm.
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)

Table 14.1 Specific Heats3 of Various Substances

The values for solids and liquids are at constant volume and at

25ºC 25ºC

, except as noted.

Note that Example 14.2 is an illustration of the mechanical equivalent of heat. Alternatively, the temperature increase could be produced by a blow torch instead of mechanically.

Example 14.3

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 $20 .0ºC 20 .0ºC$ water (about a cup) into a 0.500-kg aluminum pan off the stove with a temperature of $150ºC 150ºC$ . 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 $∣ Q_{hot} ∣ = Q_{cold} ∣ Q_{hot} ∣ = Q_{cold} size 12{ \lline Q rSub { size 8{"hot"} } \lline =Q rSub { size 8{"cold"} } } {}$ .

Solution

Use the equation for heat transfer $Q = mc Δ T Q = mc Δ T size 12{Q= ital "mc"ΔT} {}$ 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:
$Q_{hot} = m_{Al} c_{Al} (T_{f} - 150ºC) . Q_{hot} = m_{Al} c_{Al} (T_{f} - 150ºC) .$
14.12
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:
$Q_{cold} = m_{W} c_{W} (T_{f} - 20.0ºC) . Q_{cold} = m_{W} c_{W} (T_{f} - 20.0ºC) .$
14.13
Note that $Q_{hot} < 0 Q_{hot} < 0 size 12{Q rSub { size 8{"hot"} } <0} {}$ and $Q_{cold} > 0 Q_{cold} > 0 size 12{Q rSub { size 8{"cold"} } >0} {}$ 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:
$\begin{array}{l} Q_{cold} + Q_{hot} & = & 0, \\ Q_{cold} & = & {–Q}_{hot}, \\ m_{W} c_{W} (T_{f} - 20.0ºC) & = & {−m}_{Al} c_{Al} (T_{f} - 150ºC.) \end{array} \begin{array}{l} Q_{cold} + Q_{hot} & = & 0, \\ Q_{cold} & = & {–Q}_{hot}, \\ m_{W} c_{W} (T_{f} - 20.0ºC) & = & {−m}_{Al} c_{Al} (T_{f} - 150ºC.) \end{array}$
14.14
Bring all terms involving $T_{f} T_{f} size 12{T rSub { size 8{f} } } {}$ on the left hand side and all other terms on the right hand side. Solve for $T_{f} T_{f} size 12{T rSub { size 8{f} } } {}$ ,
$T_{f} = \frac{m_{Al} c_{Al} (150ºC) + m_{W} c_{W} (20 .0ºC)}{m_{Al} c_{Al} + m_{W} c_{W}}, T_{f} = \frac{m_{Al} c_{Al} (150ºC) + m_{W} c_{W} (20 .0ºC)}{m_{Al} c_{Al} + m_{W} c_{W}},$
14.15

and insert the numerical values:

$\begin{array}{l} T_{f} & = & \frac{(0.500 kg) (900 J/kgºC) (150ºC) + (0.250 kg) (4186 J/kgºC) (20.0ºC)}{(0.500 kg) (900 J/kgºC) + (0.250 kg) (4186 J/kgºC)} \\ = & \frac{88430 J}{1496.5 J/ºC} \\ = & 59 .1ºC. \end{array} \begin{array}{l} T_{f} & = & \frac{(0.500 kg) (900 J/kgºC) (150ºC) + (0.250 kg) (4186 J/kgºC) (20.0ºC)}{(0.500 kg) (900 J/kgºC) + (0.250 kg) (4186 J/kgºC)} \\ = & \frac{88430 J}{1496.5 J/ºC} \\ = & 59 .1ºC. \end{array}$
14.16

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 $20.0ºC 20.0ºC$ than $150ºC 150ºC$ ? 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 Experiment: Temperature Change of Land and Water

What heats faster, land or water?

To study differences in heat capacity:

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 achieve approximately equal masses by using $50% 50% size 12{"50% "} {}$ more water by volume.)
Heat both (using an oven or a heat lamp) for the same amount of time.
Record the final temperature 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.

Which sample cools off the fastest? This activity replicates the phenomena responsible for land breezes and sea breezes.

Check Your Understanding

If 25 kJ is necessary to raise the temperature of a block from $25ºC 25ºC$ to $30ºC 30ºC$ , how much heat is necessary to heat the block from $45ºC 45ºC$ to $50ºC 50ºC$ ?

Show solution

Solution

The heat transfer depends only on the temperature difference. Since the temperature differences are the same in both cases, the same 25 kJ is necessary in the second case.