What is the minimum work needed to cool an object using a carnot refrigerator?

AI Thread Summary
To determine the minimum work required to cool an object using a Carnot refrigerator, 433J of heat is extracted from a cold reservoir at 0°C and rejected to a hot reservoir at 19°C. The coefficient of performance (COP) is calculated using the formula COP = Tc/(Th-Tc), resulting in a COP of approximately 14.4. The work done (W) can then be found using W = Qc/COP, yielding a minimum work of about 30.1J. The calculations demonstrate the efficiency of the Carnot cycle in refrigeration. This analysis highlights the relationship between heat transfer and work in thermodynamic systems.
Runaway
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Homework Statement



433J of heat is extracted from a massive object at 0\circC while rejecting heat to a hot reservoir at 19\circC.
What minimum amount of work will accomplish this? Answer in units of J.

Homework Equations


COP= Qc/(Qh-Qc) = Tc/(Th-Tc)


The Attempt at a Solution


433J * 273.15K/(292.15K-273.15K)=6224.944 joules
 
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Runaway said:

Homework Statement



433J of heat is extracted from a massive object at 0\circC while rejecting heat to a hot reservoir at 19\circC.
What minimum amount of work will accomplish this? Answer in units of J.

Homework Equations


COP= Qc/(Qh-Qc) = Tc/(Th-Tc)


The Attempt at a Solution


433J * 273.15K/(292.15K-273.15K)=6224.944 joules
COP = Qc/W. So express W in terms of COP and Qc and solve.

AM
 


W= 433j/(273.15k/(292.15k-273.15k)) = 30.119j?
 


Runaway said:
W= 433j/(273.15k/(292.15k-273.15k)) = 30.119j?
Yes. But you should explain your reasoning.

For a Carnot refrigerator:

COP = Qc/W = Tc/(Th-Tc) = 273/19 = 14.4

W = Qc/COP = 433/14.4 = 30.1 J

AM
 
Last edited:


Thanks for your help
 

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