Suppose you have an electric hot water heater for your house which is an aluminu

Question

Suppose you have an electric hot water heater for your house which is an aluminum cylinder which has a 0.56 m radius and is 1.8 m high. The walls are 1.0 cm thick. The thermal conductivity of Aluminum is 217 W/(m K). Assume that the temperature of the hot water inside the hot water heater is kept at a constant 90 C, and the external temperature is 27 C.How much energy is lost through the walls of the hot water heater in one week? (Assume thinner surface of the heater is 90 C and the outer surface is 27 C.)

Explanation / Answer

energy lost =Q=K*A*(T2-T1)/L =217*pi*0.28^2*(90-27)/0.01 =3.367*10^5 W

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