Entropy Questions and Answers

Answer: D

The total entropy change is \(2Mc_{P}\ln\left(\frac{T_{1}+T_{2}}{2\sqrt{T_{1}T_{2}}}\right)\).

Equal masses of the same liquid reach the mean temperature \(T_{f} = (T_{1}+T_{2})/2\). Add the entropy changes of the two portions:

\[ \begin{aligned} \Delta S &= Mc_{P}\ln\frac{T_{f}}{T_{1}} + Mc_{P}\ln\frac{T_{f}}{T_{2}} \\ &= Mc_{P}\ln\frac{T_{f}^{2}}{T_{1}T_{2}} \\ &= 2Mc_{P}\ln\frac{T_{1}+T_{2}}{2\sqrt{T_{1}T_{2}}} \end{aligned} \]

The arithmetic mean always exceeds the geometric mean, so the argument exceeds 1 and \(\Delta S \gt 0\), as an irreversible mixing must give. The option with factor M instead of 2M counts only one portion, and the options without the 2 in the denominator use the wrong final temperature.

Answer: A

The answer given is that it remains constant.

A reversible process generates no entropy. The total entropy of the system and its surroundings therefore stays constant, \(\Delta S_{universe} = 0\), and in a reversible adiabatic process the entropy of the system itself is constant, which is why such a process is called isentropic.

An increase in total entropy is the mark of an irreversible process, and a decrease is impossible for an isolated system. 'Zero' confuses the change in entropy with the entropy itself, which is not zero.

Answer: A

The entropy of an isolated system or the universe increases.

This is the principle of increase of entropy, a statement of the second law: for an isolated system \(\Delta S \geq 0\), with equality only for an ideal reversible process. Every real process is irreversible, so the entropy of the universe always increases.

A decrease would violate the second law. Remaining constant holds only in the reversible limit, which real processes never reach, and the entropy is not zero.

Answer: D

The net entropy generation is closest to 8.51 J/K.

The reservoir at 400°C, that is 673 K, gives up 1 kJ, so its entropy falls. The system's entropy rises by 10 J/K.

\[ \begin{aligned} \Delta S_{res} &= -\frac{1000}{673} = -1.49\ \text{J/K} \\ S_{gen} &= 10 - 1.49 = 8.51\ \text{J/K} \end{aligned} \]

The temperature must be in kelvin: using 400 directly would give \(10 - 2.5 = 7.50\) J/K, which is the trap among the options. 10 J/K counts only the system and ignores the reservoir's loss of entropy.

Answer: D

The key's answer is its entropy change will be zero.

A reversible adiabatic process is isentropic. Entropy changes by \(dS = \frac{\delta Q_{rev}}{T}\); with no heat transfer and no irreversibility (no friction, no unrestrained expansion), the gas's entropy stays exactly the same.

Entropy would increase only in an irreversible adiabatic process, and it can never decrease in an adiabatic process, since that would violate the second law. The statement about the surroundings is also true in itself, because no heat reaches them, but it describes the surroundings rather than the gas; the defining property of the gas's own process is zero entropy change, which is what the key accepts.

Answer: B

The change in entropy is \( C \ln\left(\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}}\right) \).

Identical blocks reach the mean temperature \( T_{f} = \frac{T_{1} + T_{2}}{2} \). Each block's entropy changes by \( C\ln(T_{f}/T_{i}) \):

\[ \begin{aligned} \Delta S &= C \ln\frac{T_{f}}{T_{1}} + C \ln\frac{T_{f}}{T_{2}} \\ &= C \ln\frac{T_{f}^{2}}{T_{1}T_{2}} \\ &= C \ln\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}} \end{aligned} \]

Since the arithmetic mean exceeds the geometric mean, the argument is greater than 1 and \( \Delta S \gt 0 \), as an irreversible process requires. The option with the inverted fraction is negative; the ones with 1 or 2 in place of 4 forget that \( T_{f} \) is a half-sum, squared.

Answer: C

The entropy increase is C ln\(\left[(T_{1} + T_{2})^{2}/4T_{1}T_{2}\right]\).

Identical blocks with equal heat capacity reach the final temperature \(T_{f} = (T_{1} + T_{2})/2\). Each block's entropy change is \(C\ln(T_{f}/T_{i})\), so:

\[ \begin{aligned} \Delta S &= C\ln\frac{T_{f}}{T_{1}} + C\ln\frac{T_{f}}{T_{2}} = C\ln\frac{T_{f}^{2}}{T_{1}T_{2}} \\ &= C\ln\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}} \end{aligned} \]

Because the arithmetic mean exceeds the geometric mean, the argument is greater than 1 and \(\Delta S \gt 0\), as required for an irreversible process. Option D is the inverse ratio, which would give a decrease.

Answer: A

The entropy change is \(a\Big[2\Big(\frac{T_{1}^{2}+T_{2}^{2}}{2}\Big)^{1/2}-T_{1}-T_{2}\Big]\).

Energy is conserved, so the heat lost by one block equals the heat gained by the other. With \(C = aT\):

\[ \begin{aligned} \int_{T_{1}}^{T_{f}} aT\,dT + \int_{T_{2}}^{T_{f}} aT\,dT &= 0 \\ T_{f} &= \left(\frac{T_{1}^{2} + T_{2}^{2}}{2}\right)^{1/2} \end{aligned} \]

Then \(dS = C\,dT/T = a\,dT\), so each block changes entropy by \(a(T_{f} - T_{i})\):

\[ \Delta S = a(2T_{f} - T_{1} - T_{2}) \]

The ln forms belong to a constant heat capacity, where \(T_{f}\) is the arithmetic mean; they do not apply when C grows with T.

Answer: C

The entropy will remain unchanged.

In a reversible adiabatic process, no heat is exchanged, \(dQ = 0\), so \(dS = \frac{dQ_{rev}}{T} = 0\). The process is isentropic: the temperature changes from T₁ to T₂ because work is done, but the entropy stays the same.

Entropy rises only in an irreversible adiabatic process, such as free expansion, which is not the standard meaning of 'adiabatic expansion' here. It never decreases in an adiabatic process, and the ratio of temperatures does not decide the sign.

Answer: B

The entropy change is \(C_{A}\left[\ln\frac{T_{B}}{T_{A}} + \frac{T_{A}}{T_{B}} - 1\right]\).

Because \(C_{B} \gg C_{A}\), the bath stays at \(T_{B}\) and the solid ends at \(T_{B}\). The solid absorbs heat \(Q = C_{A}(T_{B} - T_{A})\), which the bath gives up at constant temperature:

\[ \begin{aligned} \Delta S_{A} &= \int_{T_{A}}^{T_{B}}\frac{C_{A}dT}{T} \\ &= C_{A}\ln\frac{T_{B}}{T_{A}} \\ \Delta S_{B} &= -\frac{C_{A}(T_{B} - T_{A})}{T_{B}} \\ &= C_{A}\left(\frac{T_{A}}{T_{B}} - 1\right) \end{aligned} \]

Adding them gives the keyed result, which is never negative (since \(\ln x \ge 1 - 1/x\)), as the second law requires.

The options with \(C_{B}\ln\) terms treat the bath as if its temperature changed. The option with \(T_{B}/T_{A}\) inverts the bath's ratio and does not follow from \(Q/T_{B}\).