Carnot Engine problem (including entropy?)

[yakattack]

1.
Homework Statement
1) A 200 litre container of boiling water and a 200 litre of ice cold water are used as heat source and sink for a carnot engine. Calculate the maximum amount of useful work that can be obtained from the system and the final temperature of the two containers of water.

2) If the containers were connected by a conductiong bar so that they came into hermal equilibrium calculate the final temperature and change in entropy of the system.

Homework Equations

\Deltas = \int(dQ/T)dt
w = \etaQ1

The Attempt at a Solution

1) Entropy is is conserved, so cln373+cln273 = 2clnT(final)
hence T(final) = 319K
not sure if this is correct.

\eta = (1 - 273/373) = 0.268
w = 0.268*Q1
is this the correct way to calculate the work? If so how do i know Q1?

2) not sure how to do this part.
If we assume a reversible process, then s is constant but how do i know the final temperature?

 

[fantispug]

For part 1:
Your final temperature sounds roughly correct, and your efficiency (eta) is correct initially - but, as you have asserted by calculating a final temperature, the temperatures of the two containers will change over time and hence so will their efficiency.
My suggestion is to look at how much heat the two containers gain/lose in order to reach thermal equilibrium and consider how the heat changes are related to the work done (hint: conservation of __).

For part 2: You should know how to do this given that you've already done it.