3 identical exotic rocks each at temperature T1=2400K, T2=1200K, T3=600K are giv

Question

3 identical exotic rocks each at temperature T1=2400K, T2=1200K, T3=600K are given to you. They can be stored in a special chamber so that they will not lose energy until you’re ready to use them. They also have the same heat capacity = 1.00GJ/K. You have a handful of Carnot engines and would like to extract as much work as possible out of the system. Each Carnot engine has a hot port and a cold port. To run the engine, you can put a hot rock in the hot port (a.k.a. hot bath) and connect the cold port (cold bath) to the ambient surrounding or to another hot rock at a lower temperature. A) Come up with a strategy to extract the maximum amount of work from the system. Ambient temperature is 300K, and can be assumed to be constant. What is the overall maximum efficiency? Note: You’re allowed to mix two rocks together so that they can come to thermal equilibrium as well, say the 1200K and 600K together gives two rocks at 900K. However, is this wise to do? You decide. B) How does your answer change if you cannot use the ambient surroundings as the cold bath? (This means you can only do work by attaching your Carnot engine between the rocks.) What are the final temperatures of the rocks? C) You should have 3 rocks with identical temperatures from B) Now, continue to extract the heat using the ambient as the cold bath. What is the total work done, including the work extracted in B)? D) If you are allowed to run the engines in reverse and pump heat into the rocks, and you are allowed as many super efficient batteries to store the work done by the engine. Would your answers in A) to C) change? Why?

HINT: To extract maximum work, we must keep _____ constant. HINT: Find out how much heat is accessible to us. HINT: Find out how much heat must be dumped into the cold bath. HINT: Think about what are the final temperatures of the system.

Explanation / Answer

at any time, the efficiency of a Crnot engine cannot exceed 1-Tcold/THot (absolute T's)

a. t1 vs amb gives eff of 1- 300/2400 = 87.50%
b. t2 vs amb= 1- 300/1200 = 75.00%
c. t3 vs amb = 1 - 300/600 = 50.00%
d. t1 t2 vs amb = 1 - 300/1800 = 83.33%
e. t1 t3 vs amb = 1 - 300/1500 = 80.00%
f. t2 t3 vs amb = 1 - 300/900 = 66.67%
g. t1 t2 t3 vs amb = 1 - 300/1400=78.57%


adding case a,b,c together = 70.83% total for all rocks
adding case a and f = 78.61% (a + 2 x f)/3
adding case b and e = 78.33%
adding case c and d = 72.22%
case g provides 78.57% which is the best

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