19.8.1 Conduction

FIGURE 19.22 shows an object sandwiched between a higher temperature $T_{H}$ and a lower temperature $T_{C}$ . The temperature difference causes heat energy to be transferred from the hot side to the cold side in a process known as conduction.

FIGURE 19.22 Full Alternative Text

It is not surprising that more heat is transferred if the temperature difference $Δ T$ is larger. A material with a larger cross section $A$ (a fatter pipe) transfers more heat, while a thicker material, increasing the distance $L$ between the hot and cold sources, decreases the rate of heat transfer.

These observations about heat conduction can be summarized in a single formula. If a small amount of heat dQ is transferred in a small time interval dt, the rate of heat transfer is $d Q / d t$ . For a material of cross-section area $A$ and length L, spanning a temperature difference $Δ T = T_{H} - T_{C}$ , the rate of heat transfer is

\frac{d Q}{d t} = k A \frac{Δ T}{L}

(19.44)

The quantity $k$ , which characterizes whether the material is a good conductor of heat or a poor conductor, is called the thermal conductivity of the material. Because the heat-transfer rate J/s is a power, measured in watts, the units of $k$ are $W / m K$ . Values of $k$ for common materials are given in TABLE 19.5; a material with a larger value of $k$ is a better conductor of heat.

TABLE 19.5 Thermal conductivities

Material	$k$ (W/m K)
Diamond	2000
Silver	430
Copper	400
Aluminum	240
Iron	80
Stainless steel	14
Ice	1.7
Concrete	0.8
Glass	0.8
Styrofoam	0.035
Air $(20 ° C, 1 atm)$	0.023

NOTE Heat conductivity is yet another use of the symbol $k$ . Whenever you see a $k$ , be very alert to the context in which it is used.

The quantity $Δ T / L$ is called the temperature gradient. A large temperature change over a small distance is a large temperature gradient. The heat-conduction equation tells us that a temperature gradient drives an energy flow. We met the idea that gradients drive flows in our discussion of fluid flows; it’s an important concept that we’ll see several more times.

Most good heat conductors are metals, which are also good conductors of electricity. One exception is diamond, in which the strong bonds that make diamond such a hard material lead to a rapid transfer of thermal energy. Air and other gases are poor conductors of heat because there are no bonds between adjacent molecules.

Some of our perceptions of hot and cold have more to do with thermal conductivity than with temperature. For example, a metal chair feels colder to your bare skin than a wooden chair not because it has a lower temperature—both are at room temperature—but because it has a much larger thermal conductivity that conducts heat away from your body at a much higher rate.

EXAMPLE 19.10 Keeping a freezer cold

A 1.8-m-wide by 1.0-m-tall by 0.65-m-deep home freezer is insulated with 5.0-cm-thick Styrofoam insulation. At what rate must the compressor remove heat from the freezer to keep the inside at $- 20 ° C$ in a room where the air temperature is $25 ° C ?$

MODEL Heat is transferred through each of the six sides by conduction. The compressor must remove heat at the same rate it enters to maintain a steady temperature inside. The heat conduction is determined primarily by the thick insulation, so we’ll neglect the thin inner and outer panels.

SOLVE Each of the six sides is a slab of Styrofoam with cross-section area $A_{i}$ and thickness $L = 5.0 cm$ . The total rate of heat transfer is

\frac{d Q}{d t} = ∑_{i = 1}^{6} k A_{i} \frac{Δ T}{L} = \frac{k Δ T}{L} ∑_{i = 1}^{6} A_{i} = \frac{k Δ T}{L} A_{total}

The total surface area is

\begin{matrix} A_{total} = 2 \times (1.8 m \times 1.0 m + 1.8 m \times 0.65 m \\ + 1.0 m \times 0.65 m) = 7.24 m^{2} \end{matrix}

Using $k = 0.035 W / m K$ from Table 19.5, we find

\frac{d Q}{d t} = \frac{k Δ T}{L} A_{total} = \frac{(0.035 W / m K) (45 K) (7.24 m^{2})}{0.050 m} = 230 W

Heat enters the freezer through the walls at the rate 230 J/s; thus the compressor must remove 230 J of heat energy every second to keep the temperature at $- 20 ° C$ .

REVIEW We’ll learn in Chapter 20 how the compressor does this and how much work it must do. A typical freezer uses electric energy at a rate of about 150 W, so our result seems reasonable.