A building is maintained at a temperature T_h by means of an ideal heat pump. This operates like a refrigerator, but the objective is to deliver heat (IQ_h|) to the high tem- perature reservoir (the building) by drawing energy from a low temperature reservoir (a nearby river) at T_c. So, the COP is |Qh|/|W|], where |W| is the electrical energy used by the heat pump.
(a) Derive an expression for the COP of this ideal heat pump, which is based on the first and second laws of thermodynamics. Your expression should involve only the temperatures of the reservoirs.

The expression is

C.O.P. (ideal heat pump) =

1 + (T_c)÷((T_h-T_c))

Explanation:

At start point, the machine cylinder is filled with air which is expanded adiabatically to pressure p, with corresponding temperature drop from T_h to T_c. Contact is then made with the river which is at temperature T_c. The air is now expanded isothermally to n, extracting heat from the river. The cylinder then undergoes adiabatic compression to temperature T_h. At this point, the cylinder is in contact with the building which is also at T_h. The air is then compressed isothermally, rejecting heat to the building in the process

For the heat pump

C.O.P. (heat pump)

= (Heat rejected to the building/cycle)÷(Work done per cycle)

= (T_h×l•m)÷((T_h-T_c)×p•n)

Where l•m = p•n for an ideal cycle, we have

C.O.P. (heat pump) = (T_h×p•n)÷((T_h-T_c)×p•n)

= (T_h)÷((T_h-T_c))

= 1 + (T_c)÷((T_h-T_c))

answer:

200k

explanation: