Calculating Heat Expelled from a Heat Pump

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Homework Help Overview

The discussion revolves around calculating the heat expelled by a heat pump, given its coefficient of performance and the heat absorbed from the cold environment. The subject area is thermodynamics, specifically focusing on energy conservation and the operation of heat pumps.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the coefficient of performance to the heat absorbed and expelled but expresses uncertainty about the equations needed. Some participants question the conservation of energy in the context of the heat pump's operation. Others discuss the work done by the pump and how it relates to the heat transferred.

Discussion Status

Participants are exploring different aspects of the problem, including the relationship between work, heat absorbed, and heat expelled. There is a mix of attempts to clarify concepts and calculations, with some guidance provided regarding the conservation of energy principle. However, there is no explicit consensus on the final calculations or interpretations.

Contextual Notes

Some participants note the lack of complete information in the original question, which may affect the discussion and calculations. The focus remains on understanding the relationships between the variables involved in the heat pump's operation.

jdog6
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A heat pump has a coefficient of performance of 7.05. If the heat pump absorbs 20 cal of heat from the cold outdoors in each cycle, find the heat expelled to the warm indoors. Answer in units of cal.
I believe COP (heating mode) : 7.05 and COP = Qh/W
Qc = 20 cal
so I have to find Qh= ? cal
I don't know an equation to put all this together?
Please help, thank you.
 
Last edited:
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Is Energy conserved?

If the coeff. of performance is 7.05,
how much Work is done in one cycle?
 
the question does not say all i have is what i posted
 
The entire subject of Thermodynamics
is FOUNDED on Conservation of Energy
(Energy can be moved around, but not created/destroyed).

So, YES, Energy is conserved.

Where do you think the Energy comes from,
that is dumped into the (warm) room?
 
The energy comes from rods outside.
 
Not all of it ... it takes *Work* to operate the pump!
Qh/W = 7.05 , so you have to PAY for the Energy
(electricity, probably) for the W = Qh/7 of the heat.

So, what fraction do you get "for free", from outside?
 
Qh=W+Qc so W=(W+20)/7.05
7.05W-W=20
W(7.05-1)=20
6.05W=20
W=20/6.05
W=3.306
Qh=3.306+20
Qh=23.306 cal/cycle
and to check it 23.306/3.306=7.05
 

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