Heat Pump Q: Cost of Winter Heat w/ 75% CoP

[ewnair]

Homework Statement

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?

Homework Equations

I understand that K = Th/(Th-Tc)

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?

 

[Redbelly98]

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.

... 75% of that is 5.33. Equated this to Qh/Win and assumed that Qh is the same for both actual and ideal pump.

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

 

[ewnair]

I got $1330 which isn't 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.

 

[Redbelly98]

K is not 7.11, it is 5.33.

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 ...

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

 

[honkon]

K is not 7.11, it is 5.33.
Yes, 5.33 should be equated with Qh/Win. Then solve the equation for Win.

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)