(a) Calculate the rate at which body heat flows out through the clothing of a skier, given the following data: the body surface area is 1.8 m 2 and the clothing is 1.2 cm thick; skin surface temperature is 33ºC, whereas the outer surface of the clothing is at 1.0ºC; the thermal conductivity of the clothing is 0.040 W/m .K. (b) How would the answer change if, after a fall, the skier’s clothes become soaked with water? Assume that the thermal conductivity of water is 0.60 W/m .K.

							(a) Surface area of the body,
A = 1.8 m2
Temperature difference, ΔT = (skin surface temperature) – (outer surface temperature of clothing)
 = 33̊ C – 1̊ C
 = 32̊ C
 = (32+273) K
 = 305 K
Thickness, Δx = 1.2 cm
 = (1.2 cm) (1 m/100 cm)
 = 0.012 m
To find out the rate H at which heat flows out through the clothing of a skier, substitute 1.8 m2 for the area A, 305 K for temperature difference ΔT and 0.012 m for thickness Δx and 0.040 W/m. K for the thermal conductivity k in the equation H = kA ΔT/ Δx,
H = kA ΔT/ Δx
 = (0.040 W/m. K) (1.8 m2 ) (305 K) / (0.012 m)
 = 1830 W
From the above observation we conclude that, the rate H at which heat flows out through the clothing of a skier would be 1830 W.
(b) From the equation H = kA ΔT/ Δx, we observed that rate H at which heat flows out through the clothing of a skier is directly proportional to the thermal conductivity k of clothing.
Now the thermal conductivity of water is changed to 0.60 W/m. K.
Thus thermal conductivity k is increased by a factor of,
= (0.60 W/m. K) / (0.04 W/m. K)
= 15
Since that rate H at which heat flows out through the clothing of a skier is directly proportional to the thermal conductivity k of clothing, therefore the rate H would increase by a factor of 15.