4. 0/2 points | Previous Answers OSColPhys2016 14.5.P.030 My Notes (a) Calculate

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4. 0/2 points | Previous Answers OSColPhys2016 14.5.P.030 My Notes (a) Calculate the rate of heat conduction (in W) through house walls that are 10.5 cm thick and that have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The surface area of the walls is 115 m2 and their inside surface is at 23.0°C, while their outside surface is at 5.00°C (b) How many 1 kW room heaters would be needed to balance the heat transfer due to conduction? (Round your answer to the next whole integer.) 1 kW room heaters Additional Materials Reading

Explanation / Answer

(a) Thermal conductivity of glass wool = 0.04 W/mK

Thermal conductivity of walls k = 0.08 W/mK (2 times glass wool)

Temperature gradient is given by dT/dx = (Tin - Tout)/width = (23-5)/0.105 = 171.4285 K/m

Surface area of walls = 115 m2

Hence heat flow H = kAdT/dx = 0.08 x 115 x 171.4285 W = 1577.14 W

(b) 2 heater can provide 2000 watt for the loss of 1577.14 W, which is required at least.

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