## Heat Pump

Consider a Carnot heat pump operates between 60°F and 120°F. If the energy absorbed from the low temperature space is 300 Btu/hr, what is the net work done on the working substance?

Hint
For a Carnot heat pump cycle, the Coefficient of Performance (COP) is:
$$COP_c=\frac{T_H}{(T_H-T_L)}$$$where $$T_H$$ and $$T_L$$ are the absolute high and low temperatures respectively in either Kelvin or Rankine. Hint 2 Refrigeration cycles are the reverse of heat engine cycles. Heat is moved from low to high temperatures, which requires work, $$W$$ . For heat pumps, the Coefficient of Performance (COP) is: $$COP=\frac{Q_H}{W}$$$
where $$Q_H$$ is the system’s absorbed heat.
For a Carnot heat pump cycle, the Coefficient of Performance (COP) is:
$$COP_c=\frac{T_H}{(T_H-T_L)}$$$where $$T_H$$ and $$T_L$$ are the absolute high and low temperatures respectively in either Kelvin or Rankine. To convert from Fahrenheit to Rankine: $$^{\circ}R\:=\:^{\circ}F+459.69$$$
Thus, the high and low temperatures are:
$$^{\circ}R_H=120^{\circ}F+459.69=579.69^{\circ}R$$$$$^{\circ}R_L=60^{\circ}F+459.69=519.69^{\circ}R$$$
$$COP_c=\frac{579.69^{\circ}R}{(579.69^{\circ}R-519.69^{\circ}R)}=\frac{579.69}{60}=9.66$$$Refrigeration cycles are the reverse of heat engine cycles. Heat is moved from low to high temperatures, which requires work, $$W$$ . For heat pumps, the Coefficient of Performance (COP) is: $$COP=\frac{Q_H}{W}$$$
where $$Q_H$$ is the system’s absorbed heat. Solving for work:
$$W=\frac{Q_H}{COP}=\frac{300Btu/hr}{9.66}=31.1\:Btu/hr$$\$
31.1 Btu/hr