14.5 Conduction

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14.5 Conduction

Resource ID: 4obWZmE9@4
Grade Range: HS - 12

Learning Objectives

Learning Objectives

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

  • Calculate thermal conductivity
  • Observe conduction of heat in collisions
  • Study thermal conductivities of common substances

The information presented in this section supports the following AP® learning objectives and science practices:

  • 1.E.3.1 The student is able to design an experiment and analyze data from it to examine thermal conductivity. (S.P. 4.1, 4.2, 5.1)
  • 5.B.6.1 The student is able to describe the models that represent processes by which energy can be transferred between a system and its environment because of differences in temperature: conduction, convection, and radiation. (S.P. 1.2)
The figure shows an insulated wooden partition in a house. The partition is insulated because it encapsulates a cloth-type material.
Figure 14.13 Insulation is used to limit the conduction of heat from the inside to the outside—in winters—and from the outside to the inside—in summers. (Giles Douglas)

Your feet feel cold as you walk barefoot across the living room carpet in your cold house and then step onto the kitchen tile floor. This result is intriguing, since the carpet and tile floor are both at the same temperature. The different sensation you feel is explained by the different rates of heat transfer: The heat loss during the same time interval is greater for skin in contact with the tiles than with the carpet, so the temperature drop is greater on the tiles.

Some materials conduct thermal energy faster than others. In general, good conductors of electricity—metals like copper, aluminum, gold, and silver—are also good heat conductors, whereas insulators of electricity—wood, plastic, and rubber—are poor heat conductors. Figure 14.14 shows molecules in two bodies at different temperatures. The average kinetic energy of a molecule in the hot body is higher than in the colder body. If two molecules collide, an energy transfer from the molecule with greater kinetic energy to the molecule with less kinetic energy occurs. The cumulative effect from all collisions results in a net flux of heat from the hot body to the colder body. The heat flux thus depends on the temperature difference ΔΤ=Τhot−TcoldΔΤ=Τhot−Tcold size 12{ΔΤ=Τ rSub { size 8{"hot"} } - T rSub { size 8{"cold"} } } {}. Therefore, you will get a more severe burn from boiling water than from hot tap water. Conversely, if the temperatures are the same, the net heat transfer rate falls to zero, and equilibrium is achieved. Owing to the fact that the number of collisions increases with increasing area, heat conduction depends on the cross-sectional area. If you touch a cold wall with your palm, your hand cools faster than if you just touch it with your fingertip.

The figure shows a vertical line labeled 'surface' that divides the figure in two. Just below the line is a horizontal rightward wavy arrow labeled Q, heat conduction. The area left of the surface line is labeled higher temperature and the area right of the surface line is labeled lower temperature. One spherical object, labeled 'high energy before collision' is on the left bottom side, with an arrow from it pointing to the right and up toward the vertical midpoint of the surface line. There is an
Figure 14.14 The molecules in two bodies at different temperatures have different average kinetic energies. Collisions occurring at the contact surface tend to transfer energy from high-temperature regions to low-temperature regions. In this illustration, a molecule in the lower temperature region (right side) has low energy before collision, but its energy increases after colliding with the contact surface. In contrast, a molecule in the higher temperature region (left side) has high energy before collision, but its energy decreases after colliding with the contact surface.

A third factor in the mechanism of conduction is the thickness of the material through which heat transfers. The figure below shows a slab of material with different temperatures on either side. Suppose that T2T2 size 12{T rSub { size 8{2} } } {} is greater than T1,T1, size 12{T rSub { size 8{1} } } {} so that heat is transferred from left to right. Heat transfer from the left side to the right side is accomplished by a series of molecular collisions. The thicker the material, the more time it takes to transfer the same amount of heat. This model explains why thick clothing is warmer than thin clothing in winters, and why Arctic mammals protect themselves with thick blubber.

Two rectangular blocks are shown with the right one labeled T one and the left one labeled T two. The blocks are placed on a surface at a distance d from each other, so that their largest face faces the opposite block. The block T one is cold and the block T two is hot. The blocks are connected to each other with a conducting rectangular block of thermal conductivity k and cross-sectional area A. A wavy line labeled Q is inside the conducting block and points from the hot block to the cold block.
Figure 14.15 Heat conduction occurs through any material, represented here by a rectangular bar, whether window glass or walrus blubber. The temperature of the material is T2T2 size 12{T rSub { size 8{2} } } {} on the left and T1T1 size 12{T rSub { size 8{1} } } {} on the right, where T2T2 size 12{T rSub { size 8{2} } } {} is greater than T1.T1. size 12{T rSub { size 8{1} } } {} The rate of heat transfer by conduction is directly proportional to the surface area A,A, size 12{A} {} the temperature difference T2−T1,T2−T1, size 12{T rSub { size 8{2} } - T rSub { size 8{1} } } {} and the substance’s conductivity k.k. size 12{k} {} The rate of heat transfer is inversely proportional to the thickness dd size 12{d} {}.

Lastly, the heat transfer rate depends on the material properties described by the coefficient of thermal conductivity. All four factors are included in a simple equation that was deduced from and is confirmed by experiments. The rate of conductive heat transfer through a slab of material, such as the one in Figure 14.15, is given by

14.27 Qt=kA(T2−T1)d,Qt=kA(T2−T1)d, size 12{ { {Q} over {t} } = { { ital "kA" \( T rSub { size 8{2} } - T rSub { size 8{1} } \) } over {d} } } {}

where Q/tQ/t size 12{Q/t} {} is the rate of heat transfer in watts or kilocalories per second, kk size 12{k} {} is the thermal conductivity of the material, AA size 12{A} {} and dd size 12{d} {} are its surface area and thickness, as shown in Figure 14.15, and (T2−T1)(T2−T1) size 12{ \( T rSub { size 8{2} } - T rSub { size 8{1} } \) } {} is the temperature difference across the slab. Table 14.3 gives representative values of thermal conductivity.

Example 14.5 Calculating Heat Transfer Through Conduction: Conduction Rate Through an Ice Box

A polystyrene foam ice box has a total area of 0.950 m20.950 m2 and walls with an average thickness of 2.50 cm. The box contains ice, water, and canned beverages at ºC ºC. The inside of the box is kept cold by melting ice. How much ice melts in one day if the ice box is kept in the trunk of a car at 35ºC35ºC size 12{"35" "." "0°C"} {}?

Strategy

This question involves both heat for a phase change (melting of ice) and the transfer of heat by conduction. To find the amount of ice melted, we must find the net heat transferred. This value can be obtained by calculating the rate of heat transfer by conduction and multiplying by time.

Solution

  1. Identify the knowns.
    14.28 A=0.950 m2; d=2.50 cm=0.0250 m;T1=0 ºC;T2=35ºC,t=1 day=24 hours=86,400 s.A=0.950 m2; d=2.50 cm=0.0250 m;T1=0 ºC;T2=35ºC,t=1 day=24 hours=86,400 s.
  2. Identify the unknowns. We need to solve for the mass of the ice, m.m. size 12{m} {} We will also need to solve for the net heat transferred to melt the ice, Q.Q. size 12{Q} {}
  3. Determine which equations to use. The rate of heat transfer by conduction is given by
    14.29 Qt=kA(T2−T1)d.Qt=kA(T2−T1)d. size 12{ { {Q} over {t} } = { { ital "kA" \( T rSub { size 8{2} } - T rSub { size 8{1} } \) } over {d} } } {}
  4. The heat is used to melt the ice, Q=mLf.Q=mLf. size 12{Q= ital "mL" rSub { size 8{f} } } {}
  5. Insert the known values.
    14.30 Qt = 0.010 J/s⋅m ⋅ºC 0.950 m2 35.0 ºC−0 ºC 0.0250 m =13.3 J/s Qt = 0.010 J/s⋅m ⋅ºC 0.950 m2 35.0 ºC−0 ºC 0.0250 m =13.3 J/s
  6. Multiply the rate of heat transfer by the time (1 day = 86,400 s1 day = 86,400 s size 12{1`"day=86,400"`s} {}).
    14.31 Q = Q / t t = 13 . 3  J/s 86 , 400  s = 1 . 15 × 10 6  J Q = Q / t t = 13 . 3  J/s 86 , 400  s = 1 . 15 × 10 6  J size 12{Q= left ( {Q} slash {t} right )t= left ("13" "." 3`"J/s" right ) left ("86","400"`s right )=1 "." "15" times "10" rSup { size 8{6} } `J} {}
  7. Set this equal to the heat transferred to melt the ice: Q=mLf.Q=mLf. size 12{Q= ital "mL" rSub { size 8{f} } } {} Solve for the mass mm size 12{m} {}.
    14.32 m=QLf=1.15×106 J334 ×103 J/kg=3.44kgm=QLf=1.15×106 J334 ×103 J/kg=3.44kg size 12{m= { {Q} over {L rSub { size 8{f} } } } = { {1 "." "15" times "10" rSup { size 8{6} } `J} over {"334" times "10" rSup { size 8{3} } `"J/kg"} } =3 "." "44"`"kg"} {}

Discussion

The result of 3.44 kg, or about 7.6 lb., seems about right, based on experience. You might expect to use about a 4 kg (7–10 lb.) bag of ice per day. A little extra ice is required if you add any warm food or beverages.

Inspecting the conductivities in Table 14.3 shows that polystyrene foam is a very poor conductor and thus a good insulator. Other good insulators include fiberglass, wool, and goose-down feathers. Like polystyrene foam, these all incorporate many small pockets of air, taking advantage of air’s poor thermal conductivity.

Substance Thermal Conductivity k (J/s⋅m⋅ºC) k (J/s⋅m⋅ºC)
Silver 420
Copper 390
Gold 318
Aluminum 220
Steel iron 80
Steel (stainless) 14
Ice 2.2
Glass (average) 0.84
Concrete brick 0.84
Water 0.6
Fatty tissue (without blood) 0.2
Asbestos 0.16
Plasterboard 0.16
Wood 0.08–0.16
Snow (dry) 0.10
Cork 0.042
Glass wool 0.042
Wool 0.04
Down feathers 0.025
Air 0.023
Polystyrene foam 0.010
Table 14.3 Thermal Conductivities of Common Substances7

A combination of material and thickness is often manipulated to develop good insulators—the smaller the conductivity kk size 12{k} {} and the larger the thickness dd size 12{d} {}, the better. The ratio of d/kd/k size 12{d/k} {} will thus be large for a good insulator. The ratio d/kd/k size 12{d/k} {} is called the RR size 12{R} {} factor. The rate of conductive heat transfer is inversely proportional to R.R. size 12{R} {} The larger the value of R,R, size 12{R} {} the better the insulation. RR size 12{R} {} factors are most commonly quoted for household insulation, refrigerators, and the like—unfortunately, it is still in non metric units of ft.2·°F·h/Btu, although the unit usually goes unstated—1 British thermal unit [Btu] is the amount of energy needed to change the temperature of 1.0 lb. of water by 1.0 °F. A couple of representative values are an RR size 12{R} {} factor of 11 for 3.5-in.-thick fiberglass batts—pieces—of insulation and an RR size 12{R} {} factor of 19 for 6.5-in.-thick fiberglass batts. Walls are usually insulated with 3.5-in. batts, while ceilings are usually insulated with 6.5-in. batts. In cold climates, thicker batts may be used in ceilings and walls.

The figure shows two thick rectangular pieces of fiberglass batt lying one upon the other.
Figure 14.16 The fiberglass batt is used for insulation of walls and ceilings to prevent heat transfer between the inside of the building and the outside environment.

Note that in Table 14.3, the best thermal conductors—silver, copper, gold, and aluminum—are also the best electrical conductors, again related to the density of free electrons in them. Cooking utensils are typically made from good conductors.

Example 14.6 Calculating the Temperature Difference Maintained by a Heat Transfer: Conduction Through an Aluminum Pan

Water is boiling in an aluminum pan placed on an electrical element on a stovetop. The sauce pan has a bottom that is 0.800 cm thick and 14 cm in diameter. The boiling water is evaporating at the rate of 1 g/s. What is the temperature difference across—through—the bottom of the pan?

Strategy

Conduction through the aluminum is the primary method of heat transfer here, and so we use the equation for the rate of heat transfer and solve for the temperature difference.

14.33 T2−T1=QtdkAT2−T1=QtdkA size 12{T rSub { size 8{2} } - T rSub { size 8{1} } = { {Q} over {t} } left ( { {d} over { ital "kA"} } right )} {}

Solution

  1. Identify the knowns and convert them to the SI units.

    The thickness of the pan, d=0.800 cm=8.0×10−3 m,d=0.800 cm=8.0×10−3 m, the area of the pan A=π(0.14/2)2 m2=1.54×10−2 m2A=π(0.14/2)2 m2=1.54×10−2 m2, and the thermal conductivity, k=220 J/s⋅m⋅°C.k=220 J/s⋅m⋅°C.

  2. Calculate the necessary heat of vaporization of 1 g of water.
    14.34 Q=mLv=1.00×10−3 kg2,256×103 J/kg=2,256 JQ=mLv=1.00×10−3 kg2,256×103 J/kg=2,256 J size 12{Q= ital "mL" rSub { size 8{v} } = left (1 "." 0 times "10" rSup { size 8{ - 3} } `"kg" right ) left ("2256" times "10" rSup { size 8{6} } `"J/kg" right )="2256"`J} {}
  3. Calculate the rate of heat transfer given that 1 g of water melts in one second.
    14.35 Q/t=2,256 J/s or 2.26 kWQ/t=2,256 J/s or 2.26 kW size 12{Q/t="2256"`"J/s"} {}
  4. Insert the knowns into the equation and solve for the temperature difference.
    14.36 T2−T1=QtdkA=2,256 J/s8.00 × 10−3m220 J/s⋅m⋅ºC1.54×10−2 m2=5.33 ºCT2−T1=QtdkA=2,256 J/s8.00 × 10−3m220 J/s⋅m⋅ºC1.54×10−2 m2=5.33 ºC

Discussion

The value for the heat transfer Q/t = 2.26kW or 2,256 J/s Q/t = 2.26kW or 2,256 J/s size 12{Q/t"=2" "." "26"`"kW"`"or"`"2256"`"J/s "} {} is typical for an electric stove. This value gives a remarkably small temperature difference between the stove and the pan. Consider that the stove burner is red hot while the inside of the pan is nearly 100 ºC100 ºC size 12{"100°C"} {} because of its contact with boiling water. This contact effectively cools the bottom of the pan in spite of its proximity to the very hot stove burner. Aluminum is such a good conductor that it only takes this small temperature difference to produce a heat transfer of 2.26 kW into the pan.

Conduction is caused by the random motion of atoms and molecules. As such, it is an ineffective mechanism for heat transport over macroscopic distances and short time distances. Take, for example, the temperature on Earth, which would be unbearably cold during the night and extremely hot during the day if heat transport in the atmosphere was to be only through conduction. In another example, car engines would overheat unless there was a more efficient way to remove excess heat from the pistons.

Check Your Understanding

How does the rate of heat transfer by conduction change when all spatial dimensions are doubled?

Solution

Because area is the product of two spatial dimensions, it increases by a factor of four when each dimension is doubled Afinal=(2d)2=4d2=4AinitialAfinal=(2d)2=4d2=4Ainitial size 12{A rSub { size 8{"final"} } = \( 2d \) rSup { size 8{2} } =4d rSup { size 8{2} } =4A rSub { size 8{i"nitial"} } } {}. The distance, however, simply doubles. Because the temperature difference and the coefficient of thermal conductivity are independent of the spatial dimensions, the rate of heat transfer by conduction increases by a factor of four divided by two, or two:

14.37 Qtfinal=kAfinalT2−T1dfinal=k4AinitialT2−T12dinitial=2kAinitialT2−T1dinitial=2Qtinitial.Qtfinal=kAfinalT2−T1dfinal=k4AinitialT2−T12dinitial=2kAinitialT2−T1dinitial=2Qtinitial. size 12{ left ( { {Q} over {t} } right ) rSub { size 8{"final"} } = { { ital "kA" rSub { size 8{"final"} } left (T rSub { size 8{2} } - T rSub { size 8{1} } right )} over {d rSub { size 8{"final"} } } } = { {k left (4A rSub { size 8{"initial"} } right ) left (T rSub { size 8{2} } - T rSub { size 8{1} } right )} over {2d rSub { size 8{"initial"} } } } =2 { { ital "kA" rSub { size 8{"initial"} } left (T rSub { size 8{2} } - T rSub { size 8{1} } right )} over {d rSub { size 8{"initial"} } } } =2 left ( { {Q} over {t} } right ) rSub { size 8{"initial"} } } {}

Applying the Science Practices: Estimating Thermal Conductivity

The following equipment and materials are available to you for a thermal conductivity experiment:

  • 1 high-density polyethylene cylindrical container
  • 1 steel cylindrical container
  • 1 glass cylindrical container
  • 3 cork stoppers
  • 3 glass thermometers
  • 1 small incubator
  • 1 cork base (2 cm thick)
  • crushed ice
  • 1 digital timer
  • 1 metric balance
  • 1 meter stick or ruler
  • 1 Vernier caliper
  • 1 micrometer

Notes—The three cylindrical containers have equal volumes and are tested in sequence. All cork stoppers fit snugly into the open tops of the containers and have small holes through which a thermometer can be placed securely. There is enough ice to fill each of the containers. Each container with thermometer fits inside the incubator on the cork base. The incubator has been uniformly pre-heated to a temperature of 40 °C. The thermometers can be observed through the incubator window

Exercise 1

Describe an experimental procedure to estimate the thermal conductivity (k) for each of the container materials. Point out what properties need to be measured, and how the available equipment can be used to make all of the necessary measurements. Identify sources of error in the measurements. Explain the purpose of the cork stoppers and base, the reason for using the incubator, and when the timer should be started and stopped. Draw a labeled diagram of your setup to help in your description. Include enough detail so that another student could carry out your procedure. For assistance, review the information and analysis in Example 14.5: Calculating Heat Transfer through Conduction.

The dimensions of each container are measured, so that the side surface area (A) and the thickness of the sides (d) are determined. Weigh an empty container, fill it with ice, weigh it again, insert the cork stopper, and insert the thermometer through the stopper so that the bulb is near the bottom of the container. Place the container in the incubator on the cork base. The incubator provides a uniform high temperature, which evenly surrounds the container and will melt the ice within 20 minutes. Because the cork is an effective insulator, most of the heat transfer will occur through the sides of the containers. By using the incubator temperature of 40 °C (T2) and the temperature of the ice (T1 = 0 °C), the temperature difference is uniform until all of the ice melts, and the temperature of the water in the container rises. During the time (t) the container is placed in the incubator and the ice completely melts, the amount of heat transferred into the container is almost all of the heat needed to melt the ice, which equals the mass of the ice (m) multiplied by the latent heat of ice (Lf). By using all the measured quantities in the rearranged equation (14.26) for thermal conductivity, k= d(m L f ) tA( T 2 − T 1 ) , k= d(m L f ) tA( T 2 − T 1 ) , the thermal conductivity (k) can be estimated for each of the container materials. Sources of error include the measurements of the length and radius of the container, which are affected by the precision of the calipers and meter stick, and the measurement of the container thickness, which is made with the micrometer. The mass of the ice, the measured temperatures, and the time interval are also subject to precision limits of the balance, thermometers, and timer, respectively.

Footnotes

  • 7 At temperatures near 0 ºC.

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