Question

A heat pump has a coefficient of performance that is 60% of the Carnot heat pump coefficient of performance. The heat pump is used to heat a home to 24.0°C during the winter with the low temperature reservoir at the outdoor temperature. At which outdoor temperature would it be more efficient to add the energy directly to the interior of the home than use it to run the heat pump? -154°C -40.0°C -4.00°C -83.4°C -25.2°C

Answer 1

Answer:
`T_L=-154.2^(\circ)`

Step-by-step explanation:

Given

COP= 60 % of carnot heat pump

`COP=(60)/(100)* (T_H)/(T_H-T_L)`

For heat added directly to be as efficient as via heat pump

`Q_s=W`

`COP=(Q_s)/(W)=(60)/(100)* (T_H)/(T_H-T_L)`

`1=(60)/(100)* (T_H)/(T_H-T_L)`

`1=(60)/(100)* (24+273)/(24+273-T_L)`

`T_L=118.8 K`

`T_L=-154.2^(\circ)`