Heat pump problem

1. Oct 25, 2014

toothpaste666

1. The problem statement, all variables and given/known data
A)What is the coefficient of performance of an ideal heat pump that extracts heat from 11∘C air outside and deposits heat inside your house at 30∘C?
B) If this heat pump operates on 2000W of electrical power, what is the maximum heat it can deliver into your house each hour?
2. Relevant equations
COP = QH/W
COPideal = TH/(TH-TL)

3. The attempt at a solution
TH = 30 + 270 = 300
TL = 11 + 270 = 281

COPideal = 300/(300-281) = 16
that part came up right
for the second part
W = 2000 J/s

COP = QH/W
QH = COP(W) = 16 (2000 J/s) = 32000 J/s
32000 J/s (3600 s/ h) = 115200000 J/h
i rounded to 2 sig figs because thats what the question wanted but it was wrong. What mistake did I make?

2. Oct 26, 2014

rude man

why are you using oK = oC + 270 when it's oK = oC + 273?
Even though to 2 sig. digits it worked out OK. I would still round off at the end of calculations rather than in the middle somewhere.
I suppose to 2 sig. digits the answer is 1.2e8 J.

Otherwise I see nothing wrong with what you did.

Last edited: Oct 26, 2014
3. Oct 26, 2014

toothpaste666

Thanks! it wouldnt take 1.2e8 so i tried 1.1e8 and it took that for some reason

4. Oct 26, 2014

rude man

1.1e8 is a truncation. 1.2e8 is the correct answer to 2 significant digits. Complain!

5. Oct 26, 2014

toothpaste666

worth a shot. thanks!

6. Oct 26, 2014

rude man

Oops, don't complain yet! Redoing the numbers, it's 1.1e8 after all:
COP = 303/19 = 15.947
W = 15.947 x 2000 x 3600 = 1.148e8 which to 2 sig # is 1.1e8.
(Although I suppose you could argue 1.148 ~ 1.15 which to 2 sig digits is 1.2.)

Again, note that K = oC + 273.15 or 273 for short. Don't use 270 ever.

7. Oct 26, 2014

toothpaste666

ahh yea i gotta get out of the habit of rounding too soon. This one is very close

8. Oct 26, 2014

haruspex

Hmm.. I get ((30+273)/(30-11))*2000*3600=1.148...e8

9. Oct 27, 2014

rude man

Please see my post #6. I believe we have concordance.