A passive solar house that is losing heat to the outdoors at an average rate of
ID: 1534496 • Letter: A
Question
A passive solar house that is losing heat to the outdoors at an average rate of 50,000 kJ/h is maintained at 22 degree C at all times during a winter night for 10 h. The house is to be heated by 50 glass containers each containing 20 L of water that is heated to 80 degree C during the day by absorbing solar energy. A thermostat-controlled 14-kW back-up electric resistance heater turns on whenever necessary to keep the house at 22 degree C. The density and specific heat of water at room temperature are rho = 1 kg/L and c = 4.18 kJ/kg degree C. How long would the electric heater run that night if the house incorporated no solar heating ? (Round the final answer to two decimal places.) If the house incorporated no solar heating, the heater would run for 9.26 hours.Explanation / Answer
method one:
The energy lost through the walls etc is (50 MJ/hr)(10 hr) = 500 MJ.
In each hour, the 14kW electric heater produces (14kJ/s) of energy, so the required duration of its operation would be 500 x10^6 J/( 14x10^3 J/s) = 35714 s = 9 hrs 55 minutes approx
method 2:
Let's start with how much heat will be delivered by the 1000 L of water that has been heated to 80C. As it cools down to 22C, this water releases (4.18 kJ/kgK)(1000 kg)(58 K) = 242.4 MJ of energy.
The energy lost through the walls etc is (50 MJ/hr)(10 hr) = 500 MJ. So the electric resistance heater is required to provide 257.6 MJ of energy during the night. In each hour, the 14kW electric heater produces (14kJ/s)(3600s) = 50.4 MJ of energy, so the required duration of its operation would be 257.6 MJ/(50.4 MJ/hr) = 5.11 hours = 5 hrs 7 minutes
different ans in two methods......
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