Second Law and Entropy Analysis Questions and Answers

Answer: A

Some heat must always be rejected to a colder reservoir, which is why no engine converts all its heat input to work. Heat can certainly move from cold to hot, provided work is supplied, and local entropy can fall as long as the total rises.

Answer: B

Correct answer: heat passing from a colder to a hotter body unaided

Clausius’s statement is that no process can have as its sole result the transfer of heat from a colder to a hotter body. The words “sole result” matter: a refrigerator does exactly that transfer, but not as its only result - work is consumed. Entropy of a *system* may certainly decrease, as it does in the refrigerated space; it is the total that cannot.

Answer: C

A device violating either one can be combined with an ordinary engine or refrigerator to violate the other, which is the classical proof. They are two views of the same restriction. Either one alone is sufficient as a statement of the law.

Answer: D

Correct answer: 50 per cent

\[ \eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0.5 = 50\%. \] The temperatures must be absolute, which is why the result is clean here, and the efficiency can never reach 100 per cent because that would require a sink at absolute zero. Any figure above 100 per cent, as one option offers, would violate the first law before the second.

Answer: A

Both are true and the reason gives the argument: if two reversible engines differed, coupling them would move heat from cold to hot with no work input, which the second law forbids. Working fluid and machine design therefore drop out entirely.

Answer: B

This is the defining relation, and integrating it along a reversible path gives the entropy difference between two states. Because entropy is a property, that difference then holds for any path between the same states.

Answer: C

Both are true and the reason gives the licence: the irreversible path itself gives no usable integral, so a convenient reversible path is invented purely to evaluate the difference. The generated entropy is then found from the balance instead.

Answer: D

Each case is decided by whether the process is reversible and whether heat crosses the boundary. Sorting a process into the right box is the first step in any entropy analysis. Reversibility and heat transfer are the two questions to ask.

Answer: A

It is the measure of irreversibility, and a negative value would mean a process that the second law forbids. Reversible processes are the idealised limiting case in which it vanishes. It is the quantity an exergy analysis ultimately prices.

Answer: B

The first two are correct and the distinction between them matters greatly. The third is false, since a refrigerator moves entropy about by paying work but generates none that is negative. The surroundings gain more entropy than the cold space loses.