Calculating Max Work from Reversible Cooling of Liquid

[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

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

 

[I like Serena]

Welcome to PF, phizzle86!

Did you consider that you cannot extract the maximum of energy as work?
The best you can do in any heat related process is the efficiency of a Carnot engine and in this case you can't even build one.
In other words, you need to respect the second law of thermodynamics.