## 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**

**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**

**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