# Thermodynamics (Carnot Engines)

1. The amount of heat that an engine rejects to the cold reservoir is 3 times the amount of work that it does. What is the efficiency of the engine?

e = W/ Qh = 1 - Qc/Qh

For this problem, I said e= w/3w. Thus e would equal 1/3 = 0.333. This however is not correct. What did I do wrong?

2. A Carnot engine operates between the temperatures 390 K and 300 K. How much heat is discarded for the engine to produce 1900 J of work?

e = 1 - Tc/Th
e = 1 - 300K/390K = 0.2307
e = W / Qh
thus 0.2307 = 1900J/ Qh
I solved for Qh and i got the answer to be 8233.3 J

This isn't correct either, but I'm not sure what I'm doing wrong.

3. During each cycle a reversible engine absorbs 1550 J of heat from a high-temperature reservoir and performs 1200 J of work. How much heat is exhausted to the low-temperature reservoir during each cycle?

Qh = W + Qc
1550 = 1200 + Qc and then I solved for Qc. This isn't correct either.

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Mapes
Homework Helper
Gold Member
You're rushing through these and making algebra errors. How did you go from e = 1 - Qc/Qh to e = W/3W? What is the relationship between W, Qc, and Qh, according to conservation of energy?

Shooting Star
Homework Helper
1. The amount of heat that an engine rejects to the cold reservoir is 3 times the amount of work that it does. What is the efficiency of the engine?

e = W/ Qh = 1 - Qc/Qh

For this problem, I said e= w/3w. Thus e would equal 1/3 = 0.333. This however is not correct. What did I do wrong?
Who said e=W/Qc?

It's given that Qc = 3W => e = 1 - 3W/Qh. Also, e = W/Qh.

Eliminate Qh from these two eqns and see how much is e.

Same for the other eqns.