## Heat Engine and Entropy Change

1. The problem statement, all variables and given/known data
An inventor claims to have invented an engine which operates between constant-temperature reservoirs at 400K and 300K. Data on the engine, per cycle of operation:
Qh = 200J
Qc -175J
W = 40J
Th = 400K
Tc = 300K
Of the first and second laws of thermodynamics, which (if either) does the engine violate?

2. Relevant equations
dE = Q - W (first law of thermodynamics, Q is heat, W is work, dE is internal energy change)
dS = Q/T (entropy change when T is constant, Q is heat, T is temperature)

3. The attempt at a solution
I understand how to do these problems in terms of the first law, using either dE = Q - W or the equivalent (since an engine is a cycle), by setting dE to zero and using W = abs(Qh) - abs(Qc)

The problem stated above thus violates the first law, because
dE = 0 = abs(Qh) - (Qc) = 200 - 175 = 25J /= 40J

but I get confused determining whether or not it violates the second law. Since entropy is a state function, the net entropy change over the course of a full cycle should be zero, correct?

I use the equation for entropy change with a constant T because the hot and cold reservoirs are at constant temperature.

The heat transfer from the hot reservoir to the engine results in an entropy change of -Qh/Th in the environment (negative because the entropy in the environment in this case is decreasing). The heat transfer from the engine to the cold reservoir after the engine does work results in an entropy change of Qc/Tc in the environment (positive because heat is being put back into the environment, thus the entropy increases).

So the net entropy change for the environment should be dS = -Qh/Th + Qc/Tc

This is where I get confused, plugging in numbers to this equation results in
-200/400 + 175/300 = 0.083J/K

Is that all there is to determining whether or not it violates the second law of thermodynamics? Whether or not the resulting dS is >= 0? My question here then is what happened to dS = 0, where did that go? Is the Work done responsible for this difference? The work done on some external thing would raise the entropy of that thing to account for the 0.083J/K increase in entropy? I'm not clear on how the work done factors into the equation for entropy change.

Thanks, I hope my question is somewhat clear.
 Recognitions: Homework Help Science Advisor Carnot's theorem is equivalent to the second law. It states that all heat engines will have an efficiency that is less than or equal to the Carnot efficiency. So what is the Carnot efficiency and what is the efficiency of this proposed engine? AM
 Carnot efficiency is E = 1 - Tc/Th Which means that not all of the heat put into the system produces work. So the efficiency of the proposed engine is 1 - (300K / 400K) = 0.25, which means that 25% of the heat put into the engine (taken from the environment) becomes work. With that efficiency, work should be 0.25 * abs(Qc) = 43.75 which is not the correct amount as provided by the problem. So is calculating the efficiency of the engine and verifying that the ratio of Qc to Qh is correct effectively checking whether or not it violates the second law?

Recognitions:
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## Heat Engine and Entropy Change

If the efficiency does not exceed the Carnot efficiency for an engine operating between those two temperatures, then you do not have a second law violation.

The problem with the entropy calculation as you have done it (and it is done correctly) is that it does not take into account the work that is claimed to be done. If the designer had claimed to produce 60 J of work, by taking in 200 J at 400K and expelling 175 J at 300K, this would violate the Carnot theorem and therefore violate the second law (as well as the first law of course) since the Carnot theorem follows from the second law.

AM
 Makes sense, though I am still not clear on where dS = 0 comes in to play.

Recognitions:
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