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A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine running backward is called a heat pump if its purpose is to exhaust warm air into the hot reservoir. Heat pumps are widely used for home heating. You can think of a heat pump as a refrigerator that is cooling the already cold outdoors and, with its exhaust heat $Q_H$, warming the indoors. Perhaps this seems a little silly, but consider the following. 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, $15kW$of electric power (generated by doing work at the rate of $15kJ/s$at the power plant) produces heat energy inside the home at a rate of $15kJ/s$. Suppose that the neighbor鈥檚 home has a heat pump with a coefficient of performance of $5.0$, a realistic value. Note that 鈥渨hat you get鈥� with a heat pump is heat delivered, $Q_H$, so a heat pump鈥檚 coefficient of performance is defined as$K=Q_H/W_{in}$.

a. How much electric power $\left(inkW\right)$does the heat pump use to deliver $15kJ/s$of heat energy to the house?

b. An average price for electricity is about $40MJ$per dollar. A furnace or heat pump will run typically $250$ hours per month during the winter. What does one month鈥檚 heating cost in the home with a $15kW$ electric heater and in the home of the neighbor who uses a heat pump?

Step by step solution

01

Step : 1 Given Information 

The coefficient of performance of a heat pump is$5.0$and the heat energy deliver to the house is$57kJ/s$.

02

Step : 2 Calculation of A

Formula to calculate the electric power of the heat pump is,
$K=\frac{Q_{\mathrm{H}}}{W_{\mathrm{HP}}}$

- K is the coefficient of performance.
- $Q_{\mathrm{H}}$ is the heat energy deliver to the house.
- $W_{\mathrm{HP}}$is the electric power input to run the heat pump.

Substitute $15\mathrm{kJ}/\mathrm{s}$for $Q_{\mathrm{H}}$and 5.0 for K to find $W_{in}$

$\begin{array}{c}5.0=\frac{15\mathrm{kJ}/\mathrm{s}}{W_{\mathrm{HP}}}\\ W_{\mathrm{HP}}=\frac{15\mathrm{kJ}/\mathrm{s}}{5.0}\\ =3.0\mathrm{kJ}/\mathrm{s}脳\frac{1\mathrm{W}}{1\mathrm{kJ}/\mathrm{s}}\\ =3.0\mathrm{W}\end{array}$

Conclusion :

Therefore, the heat pump consumes $3.0W$electric power to run the heat pump.

03

Step : 3 Calculation of B

Formula to calculate the heating cost for the one month of the electric heater is,

heating cost

$=\frac{\left(W_{\text{heater}}\right)\left(\text{runninghour}\right)}{\text{(avg.priceperdollar})}$

- $W_{\text{heater}}$is the work input to the heater.

Substitute $15\mathrm{kW}$for $W_{\text{heater}},40\mathrm{MJ}$for avg.price per dollar and 250 hours for running hour g to find heating cost.

$\begin{array}{r}\text{heating cost}=\frac{\left(15\mathrm{kW}脳\frac{{10}^3\mathrm{J}/\mathrm{s}}{1\mathrm{kW}}\right)\left(250\text{hours}脳\frac{3600\mathrm{s}}{1\text{thour}}\right)}{\left(40\mathrm{MJ}/\mathrm{\$}脳\frac{{10}^6\mathrm{J}}{1\mathrm{MJ}}\right)}\\ =\frac{1.35脳{10}^{10}\mathrm{J}}{40脳{10}^6\mathrm{J}/\mathrm{\$}}\\ =\mathrm{\$}337.5\end{array}$

Conclusion :

Therefore, heating cost of the electric heater is $\$337.5$.

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