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?

Answer 1

`T_C`=118.8 K= 154.2°C

Step-by-step explanation:

COP_max of carnot heat pump=
`(T_(H) )/(T_(H)-T_(C) )`

where T_H and T_C are temperatures of hot and cold reservoirs

Also COP=
`(Q_H)/(W)`

in the question
`COP= (60)/(100) * COP_(max)`

⇒
`(Q_H)/(W) =(60)/(100)*(T_H)/(T_H-T_C)`

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

`Q_H= W`

and T_H= 24° C= 297 K

`1=(60)/(100)* (297)/(297-T_C)`

on calculating the above equation we get

`T_C`=118.8 K

the outdoor temperature for efficient addition of heat to interior of home

`T_C`=118.8 K= 154.2°C