Suppose you have an electric hot water heater for your housewhich is an aluminum cylinder which has a 0.56 m radius and is 2 mhigh. The walls are 1.0 cm thick. The thermal conductivity ofAluminum is 217 W/(m K). Assume that the temperature of the hotwater inside the hot water heater is kept at a constant 90 C, andthe external temperature is 27 C.
Part A:
What is the surface area of the cylinder?
Part B:
How much energy is lost through the walls of the hot waterheater in one week? (Assume thinner surface of the heater is 90 Cand the outer surface is 27 C.)
Part C:
Assume you pay $0.10 per kW-hour for electricity. How much wouldit cost just to keep the hot water inside the heater for oneweek?
Part D:
Suppose that you wrap the hot water heater on all sides with a10 cm thick blanket of fiberglass insulation which has a thermalconductivity of 0.04 W/(m K). Assume the inner surface of thefiberglass insulation is at 90 C and the outer surface is at 27 C,and the total surface area is still what you calculated in partA.
Part E:
Assume you pay $0.10 per kW-hour for electricity. How much wouldit cost just to keep the hot water inside the fiberglass-wrappedheater for one week?
