A heat pump delivers 1000 calories to the inside of a house at 300°K (27°C). If
ID: 1479874 • Letter: A
Question
A heat pump delivers 1000 calories to the inside of a house at 300°K (27°C). If the outside air
temperature is 270°K (-3°C), how much mechanical work (in calories) must be supplied to the
pump?
b) Explain (quantitatively, not just words) why, given the efficiency of the heat pump (more calories
of heat delivered than calories of work input to the heat pump), one can't take heat from the hot
reservoir of the heat pump and make a perpetual motion machine by extracting enough energy to
run the heat pump
Please have free body diagram, explain clearly your formulas. Thanks
Explanation / Answer
The coefficient of performance of a heat pump is
K = Th / (Th -Tc)
= 300 / (300 -270)
= 300 /30 = 10
Work done in this process is
W = Qh / k = 1000/10
W = 100 calories
b)
When we closely observe our work above to 100 calrories of work is done to deliver 1000 calories of heat to the hot reservoir.
It is difficult to to built a an engine where heat is trasnferred from cold end to the hot end without the external agent. this is law of thermodynamics which clearly says that without exteranl work input its is highly impossible to dervie heat from the cold reservoir to transfer it to hot reservoir.