## Carnot Cycle

Consider a Carnot refrigerator operates between -40°F and 100°F. If the energy absorbed from the low temperature space is 250 Btu/hr, what is the net work done on the working substance?

Hint
For a Carnot refrigeration cycle, the Coefficient of Performance (COP) is:
$$COP_c=\frac{T_L}{(T_H-T_L)}$$$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 refrigerators and air conditioners, the Coefficient of Performance (COP) is: $$COP=\frac{Q_L}{W}$$$
where $$Q_L$$ is the system’s rejected heat.
For a Carnot refrigeration cycle, the Coefficient of Performance (COP) is:
$$COP_c=\frac{T_L}{(T_H-T_L)}$$$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:
$$^{\circ}R_H=100^{\circ}F+459.69=559.69^{\circ}R$$$$$^{\circ}R_L=-40^{\circ}F+459.69=419.69^{\circ}R$$$
$$COP_c=\frac{419.69^{\circ}R}{(559.69^{\circ}R-419.69^{\circ}R)}=\frac{419.69}{140}=2.998$$$Refrigeration cycles are the reverse of heat engine cycles. Heat is moved from low to high temperatures, which requires work, $$W$$ . For refrigerators and air conditioners, the Coefficient of Performance (COP) is: $$COP=\frac{Q_L}{W}$$$
where $$Q_L$$ is the system’s rejected heat. Solving for work:
$$W=\frac{Q_L}{COP}=\frac{250Btu/hr}{2.998}=83.4\:Btu/hr$$\$
83.4 Btu/hr