Minimum work needed to compensate for a heat leak

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Muthumanimaran
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Homework Statement


A great deal of effort has been expended to find “high temperature superconductors”: materials that are superconductors at temperatures higher than the boiling point of liquid nitrogen (77 K). Most of the older superconductors had to be operated with liquid helium (boiling point 4.2 K) as the cooling fluid. To estimate the savings possible in operating costs through the use of the “high temperature” superconductors, calculate the minimum work needed to compensate for a heat leak of 1 kJ into the superconductor for both “high temperature” superconductors and the older ones. Assume that the ambient temperature is 300 K.

Homework Equations


First law of Thermodynamics
Equation: delta U = Q - W = 0, the first law of thermodynamics with internal energy change set to zero.


Handwritten or typeset equation: Q = C_v ΔT


Handwritten or typeset expression W = C_v ΔT, with W for work, C_v for heat capacity at constant volume and ΔT for temperature change.


The Attempt at a Solution


Minimum work needed to compensate heat leak, so the first law of thermodynamics says

Equation: delta U = Q - W = 0


Equation W = Q, written in plain black text.


Handwritten or typeset equation: Q = C_v ΔT, with Q on the left and C_v ΔT on the right.


so

Handwritten or typeset expression W = C_v ΔT, with W on the left and C_v ΔT on the right.


The specific heat for Liquid Nitrgoen is
Handwritten or typeset value 2.04kJ/kgK, appearing to be a specific heat figure for liquid nitrogen in a homework post.


Equation W = 454.92 kJ


for ordinary superconductor, for liquid helium
Text reading C_v = 3.12 kJ/(kg K), a heat capacity value given as a plain equation.


Handwritten or typeset expression: W = (3.12 × 10^3)(300 − 4.2)


Equation W = 922.89 kJ, the minimum work needed to offset a 1 kJ heat leak into the superconductor.

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That is not the right answer, and if you would have worked with units consistently you would have seen the problem.

The systems you should consider are at the temperatures of liquid nitrogen or helium already. Now 1 kJ of energy is added to this cold reservoir. How much work do you need to remove this 1kJ?
 
mfb said:
That is not the right answer, and if you would have worked with units consistently you would have seen the problem.

The systems you should consider are at the temperatures of liquid nitrogen or helium already. Now 1 kJ of energy is added to this cold reservoir. How much work do you need to remove this 1kJ?
same 1kJ of work