Heat Transfer and Temperature Change

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 $\text{20 ºC}$ to $\text{80 ºC}$. (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.

1. Calculate the temperature difference.
   14.4 $$\text{Δ}T=T_{\text{f}}-T_{\text{i}}=\text{60 ºC}$$
2. Calculate the mass of water. Because the density of water is $\text{1,000}{\text{kg/m}}^3\text{,}$ one liter of water has a mass of 1 kg, and the mass of 0.250 liters of water is $m_{\mathrm{w}}=0\text{.}\text{250}\text{kg}$.
3. Calculate the heat transferred to the water. Use the specific heat of water in Table 14.1.
   14.5 $$Q_{\mathrm{w}}=m_{\mathrm{w}}{c}_{\mathrm{w}}\text{Δ}T=\left(0\text{.}\text{250}\text{kg}\right)\left(\text{4186}\text{J/kg}\text{ºC}\right)\left(\text{60 ºC}\right)=\text{62}\mathrm{.8\; kJ}$$
4. Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 14.1:
   14.6 $$Q_{\text{Al}}=m_{\text{Al}}{c}_{\text{Al}}\text{Δ}T=\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)\left(\text{60 ºC}\right){\text{= 27 × 10}}^4\text{J = 27 kJ}$$
5. Compare the percentage of heat going into the pan versus that going into the water. First, find the total transferred heat.
   14.7 $$Q_{\text{Total}}=Q_{\text{W}}+Q_{\text{Al}}=\text{62}\text{.}8\text{kJ}+\text{27}\text{.}0\text{kJ = 89}\text{.}8\text{kJ}$$

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 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 $\text{800 J/kg}\cdot \text{ºC}$ if the material retains 10 percent of the energy from a 10,000-kg truck descending 75 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 $(\text{Mgh})$ that the entire truck loses in its descent and then find the temperature increase produced in the brake material alone.

Solution

1. Calculate the change in gravitational potential energy as the truck goes downhill.
   14.10 $$\text{Mgh}=\left(\text{10,000 kg}\right)\left(9\text{.}\text{80 m}{\text{/s}}^2\right)\left(\text{75 m}\right)=\text{7.}\text{35}\times {\text{10}}^6\text{J}$$
2. Calculate the temperature from the heat transferred using $Q\text{=}\text{Mgh}$ and
   14.11 $$\text{Δ}T=\frac{Q}{\text{mc}}\text{,}$$

   where $m$ is the mass of the brake material. Insert the values $m=\text{100 kg}$ and $c=\text{800 J/kg}\cdot \text{ºC}$ to find

   14.12 $$\text{Δ}T=\frac{\left(7\text{.35}\times {\text{10}}^6\text{J}\right)}{\left(\text{100 kg}\right)\left(\text{800 J/kgºC}\right)}=\text{92 ºC.}$$

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, the battery.

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 $\text{20}\text{ºC}$ water—about a cup—into a 0.500-kg aluminum pan off the stove with a temperature of $\text{150 ºC}$. 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 $\mid Q_{\text{hot}}\mid =Q_{\text{cold}}\text{.}$

Solution

1. Use the equation for heat transfer $Q=\text{mc}\text{Δ}T$ 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.
   14.13 $$Q_{\text{hot}}=m_{\text{Al}}{c}_{\text{Al}}\left(T_{\text{f}}-\text{150 ºC}\right)$$
2. 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.
   14.14 $$Q_{\text{cold}}=m_{\mathrm{W}}{c}_{\mathrm{W}}\left(T_{\text{f}}-\text{20.0 ºC}\right)$$
3. Note that $Q_{\text{hot}}<mn>0</mn>$ and $Q_{\text{cold}}>0$ 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.
   14.15 $$\begin{array}{lll}\hfill Q_{\text{cold}}\text{+}{Q}_{\text{hot}}& \text{=}& \text{0,}\\ \hfill Q_{\text{cold}}& =& {\text{–Q}}_{\text{hot}},\\ m_{\mathrm{W}}{c}_{\mathrm{W}}\left(T_{\text{f}}-\text{20.0 ºC}\right)& =& {\mathrm{-m}}_{\mathrm{Al}}{c}_{\mathrm{Al}}\left(T_{\text{f}}-\text{150 ºC}\right)\end{array}$$
4. This an equation for the unknown final temperature, $T_{\text{f}}\text{.}$
5. Bring all terms involving $T_{\text{f}}$ on the left hand side and all other terms on the right hand side. Solve for $T_{\text{f}}$
   14.16 $$T_{\text{f}}=\frac{m_{\text{Al}}{c}_{\text{Al}}\left(\text{150 ºC}\right)+m_{\mathrm{W}}{c}_{\mathrm{W}}\left(\text{20}\text{.0 ºC}\right)}{m_{\text{Al}}{c}_{\text{Al}}+m_{\mathrm{W}}{c}_{\mathrm{W}}}\text{,}$$

   and insert the numerical values

   14.17 $$\begin{array}{lll}{T}_{\text{f}}& =& \frac{\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)\left(\text{150 ºC}\right)\text{+}\left(\text{0.250 kg}\right)\left(\text{4,186 J/kgºC}\right)\left(\text{20.0 ºC}\right)}{\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)+\left(\text{0.250 kg}\right)\left(\text{4,186 J/kgºC}\right)}\\ & =& \frac{\text{88,430 J}}{\text{1496.5 J/ºC}}\\ & =& \text{59}\text{.1 ºC.}\end{array}$$

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 $\text{20 ºC}$ than $\text{150 ºC}$? 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, for example, summer to winter.

Take-Home Experiment: Temperature Change of Land and Water

Footnotes

• 1 The values for solids and liquids are at constant volume and at $\text{25 ºC}$, except as noted.
• 2 These values are identical in units of $\text{cal/g}\cdot ºC$.
• 3 $c_{\text{v}}$ at constant volume and at $\text{20}\mathrm{ºC}$, except as noted, and at 1.00 atm average pressure. Values in parentheses are $c_{\text{p}}$ at a constant pressure of one atm.