Physics - Heat Pump - 2nd Law of Thermodynamics

[JSGhost]

Homework Statement

A heat pump has a coefficient of performance of 3.70 and operates with a power consumption of 6.91 103 W.
(a) How much energy does it deliver into a home during 1 h of continuous operation?
9.2x10^7 J

(b) How much energy does it extract from the outside air?
? J

Homework Equations

COP(coefficient of performance) = |Q_h|/W
P = W/t
W = P*t

e = W_eng/|Q_h| = (|Q_h|-|Q_c|)/|Q_h| = 1 - |Q_c|/|Q_h|

The Attempt at a Solution

(a) I got this part correct.
COP = |Q_h|/W
P = W/t
W = P*t

|Q_h| = COP*W = COP * P*t = 3.70(6.91x10^3 J/s)(3600s) = 9.2x10^7 J

(b) I need help on part (b). I can't get the correct answer.

I tried using e = 1 - |Q_c|/|Q_h|

|Q_c| = |Q_h|(e-1) = 9.2x10^7 J (3.70-1)

but that answer doesn't work.

 

[franklinear]

Use W=H1-H2
arranging equations
H1=H2+W

U have got H2 and W on part a.
Now u need to find H1. Just plus them.

 

[JSGhost]

Thanks for replying. I actually figured it out by looking at a previous problem I did where it was |Q_c| = |Q_h| + W_eng.