Calculating Home Heat Pump Efficiency & kW-hrs of Heat Delivered

[Nusc]

In a home heat pump, heat is absorbed from the Earth by a fluid circulating in buried pipes and delivered to the home interior. Suppose that the heat pump is 40% as efficient as a Carnot refrigerator. If the ground surrounding the pipes is at 5`C and the interior temperature of the house is kept to 20`, how many kilowatt-hours of heat would be supplied to the home interior for ever kilowatt-hour of electrical energy needed to operate the refrigerator?

omega = .4
$\ Q_C=20 C$
$\ Q_H=5 C$
$\omega = Q_C/W$

Well we can solve for the work but were looking for kilowatt-hours,

E=Power X time.

Can someone lead me in the right direction and provide the general equation suited for this problem?

Thanks

 

[Astronuc]

What else can one say about the efficiency as it relates to the hot and cold temperatures, which should be in absolute temperature (K) not (C), although when one is looking at temperature difference, either unit is acceptable.

 

[Nusc]

So what formula is associated with this problem?

 

[Andrew Mason]

In a home heat pump, heat is absorbed from the Earth by a fluid circulating in buried pipes and delivered to the home interior. Suppose that the heat pump is 40% as efficient as a Carnot refrigerator. If the ground surrounding the pipes is at 5`C and the interior temperature of the house is kept to 20`, how many kilowatt-hours of heat would be supplied to the home interior for ever kilowatt-hour of electrical energy needed to operate the refrigerator?

omega = .4
$\ Q_C=20 C$
$\ Q_H=5 C$
$\omega = Q_C/W$

As Astronuc says, work out the coefficient of performance of a Carnot cycle between these temperatures.

Since the COP ($\omega$) of this cycle is 40% that of the Carnot cycle, you can determine the input work (W) required using your formula $\omega = Q_C/W$. [Note:$Q_C$ is the heat removed from the cold reservoir. The output heat $Q_H$ is $Q_C + W$]

AM