- AB - isothermal
- BC - isobaric
- CA - isochoric

_{C}. Step 3 is in contact with the reservoir at 600 K.

a) What is the temperature T

_{C}at C?

b) How much work is done by the engine in one cycle?

c) What is the efficiency of the engine? (Be careful with the heat flows.)

d) What is ΔS of the universe for one cycle of this engine?

e) What is the efficiency of a Carnot engine operating between reservoirs at 600K and T

_{C}? Explain why your answer is either the same or different from your answer to part c.

Question so kindly provided by Joe Kiskis

**Part A**

Since step 2 is isochoric, for the beginning and the end of it, we know that:

V

_{2}⁄V

_{1}= T

_{2}⁄T

_{1}

Solve for T

_{2}, and we get

T

_{C}= 300K**Part B**

To find the total work requires summing the work for the three seperate steps using:

- W
_{1}= Q = nRT ln(V_{2}⁄V_{1}) because it's isothermal - W
_{2}= PΔV because it's isobaric - Note that this is going to be a negative number!

- W
_{3}= 0, because it's isochoric

W

_{total}= 963 J**Part C**

Now that we know the total work produced by the engine, we need to find a Q to compare it to to find the efficiency of the engine.

- Q
_{H total}= Q_{H1}+ Q_{H3} - We calculated Q
_{1}in part B, because for an isothermal W=Q

- Q
_{2}is in contact with T_{C}, so isn't of use here - Q
_{H total}= 3456J + nC_{V}ΔT - Q
_{H total}= 3456J + 1mole ⋅ 3⁄2 ⋅ 8.31J⁄molK ⋅ 300K - Q
_{H total}= 7196J - Now we just plug that into the definition of e
- e = W⁄Q
_{H} - e = 0.134

**Part D**

The important thing to remember here is that the entropy of a cyclic process like an engine doesn't change. But THAT IS NOT WHAT THIS IS ASKING! It wants to know ΔS for the universe. This is rather easy, because the two temperatures of the universe don't change, so we cant just use ∑Q⁄T.

- ΔS
_{total}= ΔS_{C}+ ΔS_{H} - ΔS
_{total}= [Q_{H}−W]⁄T_{C}− 7196J⁄600K - Note the minus sign, because Q into the engine is Q
*out*of the universe. To be honest, I just knew that the two values had opposite signs, and just picked the one that gave me a positive answer. Understand the concept over memorizing the formulas. - ΔS
_{total}= 6233J⁄300K − 7196J⁄600K - ΔS
_{total}= 8.78 J⁄K

**Part E**

- e
_{Carnot}= 1 − T_{C}⁄T_{H} - e
_{Carnot}= 1 − 300K−600K - e
_{Carnot}= 0.50

kenneth, when you get the chance, can you explain how you got parts c,d,and e. at the moment i'm stuck on c. i'm getting a value close to yours. i'm getting .139 for e.

ReplyDeletethank you so much.

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