Simple reversed heat engine problem

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SUMMARY

The discussion centers on solving a reversed heat engine problem involving a work input of 85 kJ and a heat transfer of 260 kJ from the low temperature region. The heat transfer to the high temperature region is calculated to be 345 kJ. The coefficient of performance (COP) as a refrigerator is determined to be 3.05, calculated using the formula COP = Q(cold) / W, where Q(cold) is 260 kJ and W is the work input of 85 kJ.

PREREQUISITES
  • Understanding of thermodynamic principles related to heat engines
  • Familiarity with the concept of coefficient of performance (COP)
  • Knowledge of the first law of thermodynamics
  • Ability to manipulate equations involving heat transfer and work
NEXT STEPS
  • Study the first law of thermodynamics in detail
  • Learn about different types of heat engines and their efficiencies
  • Explore advanced concepts in thermodynamics such as Carnot cycles
  • Investigate real-world applications of reversed heat engines in refrigeration
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Students studying thermodynamics, engineers working with heat engines, and anyone interested in the principles of refrigeration and energy transfer.

Joon
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Homework Statement


In a reversed heat engine, the work done on the engine is 85 kJ and the heat transfer to the engine from the low temperature region is 260 kJ. Determine:
1. the heat transfer to the high temperature region
2. the coefficient of performance as a refrigerator

Homework Equations


Q(hot) + Q(cold) + W = 0

The Attempt at a Solution


260 + 85 = 345 kJ for question 1 if I am correct.
COP = W / Q(input) = 260 / (345-260) =3.05
 
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Your results are correct (even though the symbols of your last equation are wrong - it should read ##COP=\frac{Q_0}{W}##).
 
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Thanks!
 

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