Answer:
[tex]3.32\ \text{kW}[/tex]
Explanation:
[tex]T_c[/tex] = Outside temperature = [tex]-5^{\circ}\text{C}[/tex]
[tex]T_h[/tex] = Temperature of room = [tex]21^{\circ}\text{C}[/tex]
[tex]Q_h[/tex] = Heat loss = 135000 kJ/h = [tex]\dfrac{135000}{3600}=37.5\ \text{kW}[/tex]
Coefficient of performance of heat pump
[tex]\text{COP}=\dfrac{1}{1-\dfrac{T_c}{T_h}}\\\Rightarrow \text{COP}=\dfrac{1}{1-\dfrac{273.15-5}{273.15+21}}\\\Rightarrow \text{COP}=11.3[/tex]
Input power
[tex]W_i=\dfrac{Q_h}{\text{COP}}\\\Rightarrow W_i=\dfrac{37.5}{11.3}\\\Rightarrow W_i=3.32\ \text{kW}[/tex]
The minimum power required to drive this heat pump is [tex]3.32\ \text{kW}[/tex].