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`        (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 m2  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.`
4 years ago Navjyot Kalra
654 Points
```							(a) Surface area of the body,A = 1.8 m2Temperature difference, ΔT = (skin surface temperature) – (outer surface temperature of clothing) = 33̊ C – 1̊ C = 32̊ C = (32+273) K = 305 KThickness, Δx = 1.2 cm = (1.2 cm) (1 m/100 cm) = 0.012 mTo 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 WFrom 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)= 15Since 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.
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4 years ago
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