## 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**

**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**

**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