13-5. Due to a temperature difference DT, heat is conducted through an aluminum plate that is 0.035 m thick. The plate is then replaced by a stainless plate that has the same temperature difference and cross-sectional area. How thick should the steel plate be, so that the same amount of heat per second is conducted through it?
If the aluminum plate has thickness L1 and the stainless steel plate has thickness L2, then the amounts of heat conducted through the two plates in time t are
Q1 = kAlA(DT)t/L1
and
Q2 = kSSA(DT)t/L2
Setting the two equal gives
kAlA(DT)t/L1 = kSSA(DT)t/L2
Canceling all the common factors
kAl/L1 = kSS/L2
or
L2 = (kSS/kAl)L1 = (14/240)(0.035 m) = 0.0020 m.
13-12. Two identical bars are attached end to end. In a time of 44 minutes, a certain amount of heat is conducted from the left end to the right end. Suppose, instead, that the two bars were placed on top of one another, where the difference in temperature between the ends of the bar is the same as in part a. How long, in minutes, would it take for the same amount of heat to be conducted from left to right along this combination? (See the figure on page 399 of the text book.)
Let each plate have length, in the x-direction, of L, and a cross-sectional area, perpendicular to
the x-direction, of A. When the plates are placed end to end, the total length is 2L, and the heat conducted through the combination in time t1 is
Q = kA(DT)t1/2L
When the plates are placed on top of one another, the total length in the x-direction is just L, and the cross-sectional area of the combination is 2A. The amount of heat conducted through in time t2 is
Q = k(2A)(DT)t2/L
The two amounts of heat are equal if
2t2 = t1/2, or t2 = t1/4 = (44 min)/4 = 11 min.
13-20. Assume that the earth has an average surface temperature of 22C and is a perfect emitter of radiation (e = 1). Find the energy per second per square meter that the earth radiates into space.
Q = esAT4t = (1)(5.67 × 10-8 J/s-m2-K4)(1 m2)(22 + 273.15K)4(1 s) = 430 J
13-26. Suppose the skin temperature of a person is 34C when the person is standing inside a room whose temperature is 25C. The skin area of the individual is 1.5 m2. (a) Assuming the emissivity is 0.80, find the net loss of radiant power from the body. (b) Determine the number of food Calories of energy (1 food Calorie = 4186 J) that is lost in one hour due to the net loss rate obtained in part (a). Metabolic conversion of food into energy replaces this loss.
(a) The person radiates energy at a rate
Q/t = esAT4 = (0.80)(5.67 × 10-8 J/s-m2-K4)(1.5 m2)(34 + 273.15K)4 = 605.6 J/s (1 J/s = 1W)
However, the person is also absorbing heat radiated by the surroundings. If the person and the surroundings were at the same temperature, the two rates would be equal, so the rate of heat absorption must be the same expression, but calculated using the temperature of the surroundings,
Q/t = esAT4 = (0.80)(5.67 × 10-8 J/s-m2-K4)(1.5 m2)(25 + 273.15K)4 = 537.7 J/s
The net heat loss is then (605.6 - 537.7) = 67.9 J/s.
(b) In one hour, the total heat loss is
(67.9 J/s) × (60 s/min) × (60 min/h) = 244,440 J/h = 58.4 Cal/h