An inventor claims to have a heat engine that has an efficiency of 40% when it o

Question

An inventor claims to have a heat engine that has an efficiency of 40% when it operates between a high temperature reservoir of 150? C and a low temperature reservoir of 30? C. This engine: A. must violate the zeroth law of thermodynamics B. must violate the first law of thermodynamics C. must violate the second law of thermodynamics D. must violate the third law of thermodynamics E. does not necessarily violate any of the laws of thermodynamics

Explanation / Answer

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) 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.

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