Second Law and Entropy Analysis Questions and Answers

Answer: A

Correct answer: 5

\[ \mathrm{COP} = \frac{T_C}{T_H - T_C} = \frac{250}{300-250} = \frac{250}{50} = 5. \] A coefficient of performance above one is normal and not a violation of anything, since heat is being moved rather than created. Note that the temperature *difference* is the denominator, so the performance falls sharply as the lift increases.

Answer: B

Correct answer: 2 - 4 - 1 - 3

Entropy increases with both temperature and disorder, and here the two effects point the same way. The cold solid is lowest, the solid just below melting has been warmed and is higher, the liquid just above melting adds the entropy of fusion, and the vapour at its boiling point adds the much larger entropy of vaporisation. The two jumps at the phase changes are what dominate the ordering.

Answer: C

Correct answer: True

That is the Nernst statement of the third law. Its practical importance is that it provides an absolute zero for the entropy scale, so entropies can be tabulated as absolute values rather than as differences - unlike energy and enthalpy, which are always referred to an arbitrary datum. It also implies that absolute zero cannot be reached in any finite number of steps.

Answer: D

Correct answer: an error in the calculation or an impossible case

Entropy generation cannot be negative - that is the second law in its most usable form, \( S_{gen} \ge 0 \). A negative result therefore signals a mistake: a wrong sign on a heat term, a temperature in Celsius where absolute was needed, or a set of specified inlet and outlet conditions that no real device could achieve. It is one of the most useful checks available on a process calculation.

Answer: A

Correct answer: an adiabatic process can still generate entropy

Adiabatic means no heat crosses the boundary; isentropic means the entropy does not change. They coincide only when the process is also reversible, since \[ \Delta S = \int\frac{\delta Q}{T} + S_{gen}, \] and with \( Q=0 \) the entropy still rises by whatever is generated internally. A real turbine is very nearly adiabatic and distinctly not isentropic, which is precisely what its isentropic efficiency measures.

Answer: B

The first two are correct and follow from irreversibility. The third is inverted: less work extracted means more energy left in the stream, so the real exit temperature is higher. That extra enthalpy is the work that was never extracted.

Answer: C

Entropy accompanies heat at the temperature of the surface it crosses, so a large difference between system and surroundings shows up as generation. Choosing the boundary carefully decides where that generation is booked.

Answer: D

Both are true and the reason gives the deficiency: energy is conserved everywhere, so an energy balance cannot point to a loss, whereas exergy is destroyed exactly where the irreversibility is. It puts a work value on each loss in the flowsheet.

Answer: A

The same pressure drop taken through a turbine would deliver work, so the valve discards an opportunity. Enthalpy is unchanged across an ideal throttle, which is precisely why the loss is invisible in an energy balance.