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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.
Heat conduction occurs through any material, represented here by a rectangular bar, whether window glass or walrus blubber. The temperature of the material is T 2 size 12{T rSub { size 8{2} } } {} on the left and T 1 size 12{T rSub { size 8{1} } } {} on the right, where T 2 size 12{T rSub { size 8{2} } } {} is greater than T 1 size 12{T rSub { size 8{1} } } {} . The rate of heat transfer by conduction is directly proportional to the surface area A size 12{A} {} , the temperature difference T 2 T 1 size 12{T rSub { size 8{2} } - T rSub { size 8{1} } } {} , and the substance’s conductivity k size 12{k} {} . The rate of heat transfer is inversely proportional to the thickness d 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 [link] , is given by

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

where Q / t size 12{Q/t} {} is the rate of heat transfer in watts or kilocalories per second, k size 12{k} {} is the thermal conductivity    of the material, A size 12{A} {} and d size 12{d} {} are its surface area and thickness, as shown in [link] , and ( T 2 T 1 ) size 12{ \( T rSub { size 8{2} } - T rSub { size 8{1} } \) } {} is the temperature difference across the slab. [link] gives representative values of thermal conductivity.

Calculating heat transfer through conduction: conduction rate through an ice box

A Styrofoam ice box has a total area of 0 .950  m 2 and walls with an average thickness of 2.50 cm. The box contains ice, water, and canned beverages at 0ºC . 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 35 . 0ºC 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.
    A = 0 . 950  m 2 d = 2 . 50  cm = 0 .0250 m; T 1 = C; T 2 = 35 . C, t = 1 day = 24 hours = 86,400 s.
  2. Identify the unknowns. We need to solve for the mass of the ice, m size 12{m} {} . We will also need to solve for the net heat transferred to melt the ice, Q size 12{Q} {} .
  3. Determine which equations to use. The rate of heat transfer by conduction is given by
    Q t = kA ( T 2 T 1 ) 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 = mL f . size 12{Q= ital "mL" rSub { size 8{f} } } {}
  5. Insert the known values:
    Q t = 0.010 J/s m ⋅º C 0.950  m 2 35. C C 0.0250 m = 13.3 J/s.
  6. Multiply the rate of heat transfer by the time ( 1  day = 86,400  s size 12{1`"day=86,400"`s} {} ):
    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 = mL f size 12{Q= ital "mL" rSub { size 8{f} } } {} . Solve for the mass m size 12{m} {} :
    m = Q L f = 1 . 15 × 10 6  J 334  × 10 3  J/kg = 3 . 44 kg. 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 lbs, 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 [link] shows that Styrofoam is a very poor conductor and thus a good insulator. Other good insulators include fiberglass, wool, and goose-down feathers. Like Styrofoam, these all incorporate many small pockets of air, taking advantage of air’s poor thermal conductivity.

Thermal conductivities of common substances At temperatures near 0ºC.
Substance Thermal conductivity 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
Styrofoam 0.010
Practice Key Terms 3

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Source:  OpenStax, College physics ii. OpenStax CNX. Nov 29, 2012 Download for free at http://legacy.cnx.org/content/col11458/1.2
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