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

[Runaway]

Homework Statement

433J of heat is extracted from a massive object at 0^$$\circ$$C while rejecting heat to a hot reservoir at 19^$$\circ$$C.
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

 

[Andrew Mason]

Homework Statement

433J of heat is extracted from a massive object at 0^$$\circ$$C while rejecting heat to a hot reservoir at 19^$$\circ$$C.
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

 

[Runaway]

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

 

[Andrew Mason]

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

 

[Runaway]

Thanks for your help