Question

An ideal, or Carnot, heat pump is used to heat a house to a temperature of 294 K (21 oC). How much work must the pump do to deliver 3000 J of heat into the house (a) on a day when the outdoor temperature is 273 K (0 oC) and (b) on another day when the outdoor temperature is 252 K (-21 oC)

Answer 1

Answer:

a) $W_{in} = 214.286\,J$, b) $W_{in} = 428.571\,J$

Explanation:

a) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

Where:

$T_{L}$ - Temperature of surroundings, measured in Kelvin.

$T_{H}$ - Temperature of the house, measured in Kelvin.

Given that $T_{H} = 294\,K$ and $T_{L} = 273\,K$. The Coefficient of Performance is:

$COP_{HP} = \frac{294\,K}{294\,K-273\,K}$

$COP_{HP} = 14$

Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.

$COP_{HP} = \frac{Q_{H}}{W_{in}}$

The input work to deliver a determined amount of heat to the house:

$W_{in} = \frac{Q_{H}}{COP_{HP}}$

If $Q_{H} = 3000\,J$ and $COP_{HP} = 14$, the input work that is needed is:

$W_{in} = \frac{3000\,J}{14}$

$W_{in} = 214.286\,J$

b) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

Where:

$T_{L}$ - Temperature of surroundings, measured in Kelvin.

$T_{H}$ - Temperature of the house, measured in Kelvin.

Given that $T_{H} = 294\,K$ and $T_{L} = 252\,K$. The Coefficient of Performance is:

$COP_{HP} = \frac{294\,K}{294\,K-252\,K}$

$COP_{HP} = 7$

Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.

$COP_{HP} = \frac{Q_{H}}{W_{in}}$

The input work to deliver a determined amount of heat to the house:

$W_{in} = \frac{Q_{H}}{COP_{HP}}$

If $Q_{H} = 3000\,J$ and $COP_{HP} = 7$, the input work that is needed is:

$W_{in} = \frac{3000\,J}{7}$

$W_{in} = 428.571\,J$