Question

Electricity can be directly used to heat a home by passing an electric current through a heating coil. This is a direct, 100% conversion of work to heat. That is, 10.0 kW of electric power (generated by doing work at the rate 10.0 kJ/s at the power plant) produces heat energy inside the home at a rate of 10.0 kJ/s. Suppose that the neighbor's home has a heat pump with a coefficient of performance of 2.00, a realistic value. NOTE: With a refrigerator, "what you get" is heat removed. But with a heat pump, "what you get" is heat delivered. So the coefficient of performance of a heat pump is K=QH/Win. An average price for electricity is about 40 MJ per dollar. A furnace or heat pump will run typically 300 hours per month during the winter.

Part A) How much electric power (in kW) does the heat pump use to deliver 10.0 kJ/s of heat energy to the house?

Part B) What does one month's heating cost in the home with a 10.0 kW electric heater? Answer in dollar amount

Part C) What does one month's heating cost in the home of a neighbor who uses a heat pump to provide the same amount of heating? Answer in dollar amount

Answer 1

Given, 10.0 kW of electric power produce heat energy at a rate of 10.0 kJ/s i.e.P = 10.0kW givesQH = 10.0kW The neighbor’s home has a heat pumb with a coefficient of performance of 2.00. i.e.K = 2.00

PART A:

Electric power of heat pumb, P = ?

Heat energy delivered, QH = 10.0kW

coefficient of performance K = 2.00

We know the coefficient of performance is given by:

K = QH /Win

Win = P

input work is nothing but the power input p

Therefore,

K = QH /P

P = QH /K

P = 10.0kW 2.00 P = 5.00kW

Hence, we get the input power of the heat pumb as 5kW

PART B

Electricity cost = 40MJ/$ = = 40 × 10⁶J/$

P=10.0 kW

In one month, time,t = 300hr = 300 × 3600seconds

40 × 10⁶W cost 1$ 1W cost 1 /40 × 10⁶ $ per second

we have 10.0 kW

Therefore the cost of electricity per second is,

10kW cost 1/ 40 × 10⁶ × 10 × 10³ $

Therefore in one month, 1/ 40 × 10⁶ × 10 × 10³ × 300 × 3600 $ = 270 $

Thus, we get the cost in the home with a 10.0 kW electric heater $270

PART C

Cost in one month in the home of a neighbor = ?

we know the power delivered by heat pumb in one second is 5 kW.

The cost for one second, 1/ 40 × 10⁶ × 5 × 10³ $ 2

therefore in one month, 1/ 40 × 10⁶ × 5 × 10³ × 300 × 3600 $ = 135$

Thus, we get the cost in the home with a 5.0 kW electric heater $135