Heat Pump Problem: Max Heat Transfer/COP Calculation

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SUMMARY

The coefficient of performance (COP) of an ideal heat pump extracting heat from 11°C air and depositing it at 30°C is calculated using the formula COPideal = TH/(TH-TL), resulting in a COP of 16. When operating at 2000W, the maximum heat delivered per hour is calculated as QH = COP(W) = 16 * 2000 J/s, yielding a total of 115,200,000 J/h. However, the correct conversion from Celsius to Kelvin requires adding 273.15, not 270, which affects the final calculations. The accurate maximum heat output, after correcting for significant figures, is approximately 1.1e8 J/h.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically heat pumps.
  • Familiarity with the coefficient of performance (COP) calculations.
  • Knowledge of unit conversions between Celsius and Kelvin.
  • Basic proficiency in significant figures and rounding rules in calculations.
NEXT STEPS
  • Study the principles of thermodynamics related to heat pumps and their efficiency.
  • Learn about the implications of significant figures in scientific calculations.
  • Explore advanced heat pump designs and their performance metrics.
  • Investigate the impact of ambient temperature variations on heat pump efficiency.
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Students in thermodynamics, engineers working with HVAC systems, and anyone interested in the efficiency of heat pumps and energy transfer calculations.

toothpaste666
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Homework Statement


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?

Homework Equations


COP = QH/W
COPideal = TH/(TH-TL)

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 that's what the question wanted but it was wrong. What mistake did I make?
 
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toothpaste666 said:

Homework Statement


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?

Homework Equations



The Attempt at a Solution


TH = 30 + 270 = 300
TL = 11 + 270 = 281
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:
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Thanks! it wouldn't take 1.2e8 so i tried 1.1e8 and it took that for some reason
 
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toothpaste666 said:
Thanks! it wouldn't take 1.2e8 so i tried 1.1e8 and it took that for some reason
1.1e8 is a truncation. 1.2e8 is the correct answer to 2 significant digits. Complain! :smile:
 
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worth a shot. thanks!
 
toothpaste666 said:
worth a shot. thanks!
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.
 
ahh yea i got to get out of the habit of rounding too soon. This one is very close
 
rude man said:
1.1e8 is a truncation. 1.2e8 is the correct answer to 2 significant digits. Complain! :)
Hmm.. I get ((30+273)/(30-11))*2000*3600=1.148...e8
 
haruspex said:
Hmm.. I get ((30+273)/(30-11))*2000*3600=1.148...e8
Please see my post #6. I believe we have concordance.
 

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