Question

During a 2-hour winter afternoon, when the outside temperature was 287K, a house heated by electricity was kept at 320K with the expenditure of 52kW ⋅ hr (kilowatt-hours) of electrical energy. What was the average energy leakage in J/s through the walls of the house to the environment (the outside air and ground)?

Answer 1

To calculate the average energy leakage rate, we convert the energy used to maintain the temperature from kWh to joules and then divide by the total time in seconds, yielding an average leakage rate of 26,000 J/s.

Step-by-step explanation:

The student asked for the average energy leakage in J/s through the walls of a house to the environment when the outside temperature was 287K, the inside temperature was kept at 320K, and the electrical energy expenditure was 52kW·hr over 2 hours. To find the answer, we need to convert the electrical energy provided to the house from kilowatt-hours to joules and then divide by the total time in seconds.

First, we convert 52 kWh to joules (1 kilowatt-hour is equal to 3.6×106 joules):
52 kWh × 3.6×106 joules/kWh = 1.872×108 joules

Next, we find the total time in seconds for 2 hours:
2 hours × 60 minutes/hour × 60 seconds/minute = 7200 seconds

Now we calculate the average energy leakage rate:
1.872×108 joules / 7200 seconds = 26,000 J/s

The average energy leakage rate is therefore 26,000 J/s.