# Need help understanding entropy

1. Dec 8, 2007

### maos

Hey. This isn't exactly homework help. I'm studying for a test and i ran into this:

A) Find the equilibrium temperature that results when 1.8 kilograms of liquid water at 380 K is added to 3.0 kilograms of liquid water at 280 K in a perfectly insulated container.

B) When heat is added to or removed from a solid or liquid of mass m and specific heat capacity c, the change in entropy can be shown to be S = mc ln(Tf/Ti), where Ti and Tf are the initial and final Kelvin temperatures. Use this equation to calculate the entropy change for each amount of water. Then combine the two entropy changes algebraically to obtain the total entropy change of the universe. Note that the process is irreversible, so the total entropy change of the universe is greater than zero.

Now here is where the problem starts:

C) Assuming that the coldest reservoir at hand has a temperature of 273 K, determine the amount of energy that becomes unavailable for doing work because of the irreversible process.

I used 273*224.4 because Energy unavailable for doing work = T0∆S. and i get the right answer. But why can i use the change in entropy i found in the previous question?

The entropy found in B) came from completely different temperature changes. Why is it that i can use that entropy in C) where the cold reservoir is now 273K?

I don't know if i'm really making sense, :( this unit's been a mess for me.