Calculating Maximum Work from Cooling Liquid Reversibly

[phizzle86]

Homework Statement

One mole of a liquid with a constant molar heat capacity of 132 j/mol K is initially at a temp of 80 C. The heat capacity is independent of tempearture. Calculate the maximum work that could have been done onthe surroundings while cooling the liquid reversibly. The surroundings are at a temp of 25 C.

Homework Equations

Given: q= -7260 J , change in entropy= 22.3J/K for the cooling

Change entropy = (Cvdt)/T

The Attempt at a Solution

I guess my problem here is not understanding what maximum work is, i would think that the maximum work here is in a reversible process would be the total amount of heat that is released that has the potential to do work, so wouldnt' that b 7260 J? The answer provided is 601 J

thanks guys

 

[RTW69]

I'm not sure this is helpful because I don't see how the answer of 601 J was arrived at, but from first law ΔU=Q-W where ΔU=n∫Cv*dt. For a liquid Cv≈Cp but Cp is constant therefore ΔU=n*Cp*ΔT. Maximum work would occur if Q=0 (insulated container). Therefore Wmax=ΔU

Maybe I am missing something