Chapter 15, Problem 58GP

(a)

To determine

The coefficient of performance of the ideal heat pump.

Answer to Problem 58GP

The coefficient of performance of the ideal heat pumpis 16.5.

Explanation of Solution

Given info:

The outside temperature of air is, $T_l=6°\text{C}=\left(6°\text{C}+273\right)\text{K}=279\text{K}$ .

The inside temperature of house is, $T_h=24°\text{C}=\left(24°\text{C}+273\right)\text{K}=297\text{K}$ .

Formula Used:

The expression to calculate the coefficient of performance of the ideal heat pumpis,

  $COP=\frac{T_h}{T_h-T_l}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}COP=\frac{297\text{K}}{297\text{K}-279\text{K}}\\ =16.5\end{array}$

Conclusion:

Thus, the coefficient of performance of the ideal heat pumpis 16.5.

(b)

To determine

The maximum heat delivered to house.

Answer to Problem 58GP

The maximum heat delivered to house is $7.13\times {10}^7\text{J}$ .

Explanation of Solution

Given info:

The power of the heat pump is, $P=1200\text{W}=1200\text{J}/\text{s}$ .

The time for the heat is, $t=1\text{h}=1\text{h}\times \frac{3600\text{s}}{1\text{h}}=3600\text{s}$ .

Formula Used:

The expression to calculate the maximum heat delivered to house is,

  $Q=P\cdot t\cdot COP$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}Q=\left(1200\text{J}/\text{s}\right)\left(3600\text{s}\right)\left(16.5\right)\\ =7.13\times {10}^7\text{J}\end{array}$

Conclusion:

Thus, the maximum heat delivered to house is $7.13\times {10}^7\text{J}$ .