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Due to a temperature difference T, heat is conducted through an aluminum plate that is 0.0460...

Due to a temperature difference T, heat is conducted through an aluminum plate that is 0.0460 m thick. The plate is then replaced by a stainless steel 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? Thermal conductivities of steel and aluminum are 14 J/(s·m·Co) and 240 J/(s·m·Co), respectively.

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Answer #1

tal = thickness of aluminum plate = 0.046 m

ts = thickness of steel plate

kal = thermal conductivity of aluminum plate = 240

ks = thermal conductivity of steel plate = 14

A = area of cross-section

T = Temperature difference

Heat conducted through aluminum = Heat conducted through steel

kal A T/tal = ks A T/ts

kal /tal = ks/ts

240/0.046 = 14/ts

ts = 0.0027 m

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