What is the surface area of the cylinder

Reference no: EM13102675

Suppose you have an electric hot water heater for your house which is an aluminum cylinder which has a 0.6 m radius and is 1.2 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. A.) What is the surface area of the cylinder? B.) 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.) C.)Assume you pay $0.10 per kW-hour for electricity. How much would it cost just to keep the hot water inside the heater for one week? Suppose that you wrap the hot water heater on all sides with a 10 cm thick blanket of fiberglass insulation which has a thermal conductivity of 0.04 W/(m K). D.) Assume the inner surface of the fiberglass insulation is at 90 C and the outer surface is at 27 C, and the total surface area is still what you calculated in part A. E.)Assume you pay $0.10 per kW-hour for electricity. How much would it cost just to keep the hot water inside the fiberglass-wrapped heater for one week?

Reference no: EM13102675

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