# Entropy change in a reversible carnot cycle with 2 beakers

Consider 2 beakers each containing 1kg of water. One beaker has initial temperature of 25 degrees celsius and the other at 100 degrees celsius. The beaker are now mixed. Assuming no heat exchanged is involved with the surroundings; Calculate the final temperature of the water and the entropy change in the universe.

In working out, I've managed to determine the final temperature of the water to be $334K$. The entropy change in the universe is approximately $57Jk^-1$.

The question continues:

In a completely separate process, a reversible heat engine using the Carnot cycle is to be operated between 2 beakers of water starting in their initial states of 25 degrees celsius and 100 degrees celsius. The final state will have the beakers at the same temperature at which the engine would have stopped. Again, assuming the whole system is isolated so no heat is exchanged with the surrounding;

What is the change in entropy of the Universe from the initial to the final state?

I understand the keyword here is reversible heat engine in a Carnot cycle
but after an hour, I am unable to get started. Really appreciate if anyone could at least get me moving. Thanks in advance.

Edit: The reason for not posting workings in the first question is because of the irrelevance of the working to the second leg of the question. I'm more interested in concepts for which I must recall or being directed to an approach to get me started.

• It sounds like they are asking you to calculate the efficiency of a Carnot engine between the two beakers while it moves dQ of heat. That dQ will lead to a change in both temperatures and the machine will perform a certain amount of work dW (e.g. to raise a weight, but the details are irrelevant). Rinse repeat with infinitesimally small heat steps and integrate over the work. How much work can the machine perform in total? What's the change in entropy? – CuriousOne Jun 7 '15 at 8:56
• To calculate the efficiency, wouldn't I be required to know what the final temperature is? That would enable me to compute the heat energy for each of the beaker. The difference would then allow me to compute the efficiency – Physkid Jun 7 '15 at 9:52
• What you can possibly do is, since the whole system is isolated so the net exchange of heat is zero. Hence, the heat lost by one of the beakers is gained by the other so that finally both the beakers reach a common temperature. Considering this in mind, you can equate the efficiencies in each case with a common final temperature and then solve for that temperature. After you have resolved the common temperature, I assume you would be able to relate the entropy with the Carnot Cycle's efficiency. Let me know, if it resolves your query. – ritvik1512 Jun 7 '15 at 10:41
• @RitvikChoudhary Thank you. Let me work it out later. It's been a long day due to the coming exams. – Physkid Jun 7 '15 at 11:11
• Alright, let me know soon. :) – ritvik1512 Jun 7 '15 at 11:13