# Heat Pump Question

1. Oct 4, 2008

### ewnair

1. The problem statement, all variables and given/known data

Assume that you heat your home with a heat pump whose heat exchanger is at Tc = 2degrees celcius and which maintains the baseboard radiators at Th = 47 degrees celcuiys. If it would cost $1000 to heat the house for one winter with ideal electric heaters (which have a coefficient of performance of 1), how much would it cost if the actual coefficient of performance of the heat pump were 75% of that allowed by thermodynamics? 2. Relevant equations I understand that K = Th/(Th-Tc) 3. The attempt at a solution I converted the temperature to kelvin and solved for the ideal coefficient which was 7.11. 75% of that is 5.33. Equated this to Qh/Win and assumed that Qh is the same for both actual and ideal pump. I multiplied 1000 with 7.11/5.33 and got an answer of$1330. However the answer is incorrect. Can anyone help please?

2. Oct 4, 2008

### Redbelly98

Staff Emeritus
Once you have figured out that the heat pump's actual COP is 5.33, the 7.11 figure should not enter into the calculation again.

Yes, absolutely. So what is Win, using this equation?

3. Oct 4, 2008

### ewnair

I got $1330 which isnt correct. i equated the 2 equations and plugged in Win = 1000 and K = 7.11 to find out Qh. Then i used this value of Qh and K= 5.33 to get Win which is$1330 but the answer is wrong.

4. Oct 5, 2008

### Redbelly98

Staff Emeritus
K is not 7.11, it is 5.33.

Yes, 5.33 should be equated with Qh/Win. Then solve the equation for Win.

5. Apr 23, 2009

### honkon

The Problem states that K=1 (which is kind of misleading). So if we have a actual engine with K=5.33, our work should be less.

So, 1=Qh/(1000) and
5.33=Qh/(dollars we pay)

you will get \$187.5 as your answer (you might think that is weird because an "actual" engine is more efficient than a carnot engine)