Question

Calculate the entropy change in the following. Given latent heat of steam is \(540\ \text{cal/g}\), latent heat of ice is \(80\ \text{cal/g}\).
A. \(10g\) of water at \(100^\circ C\) converted to steam at same temperature,
B. \(20g\) of water at \(100^\circ C\) converted to steam at same temperature,
C. \(1g\) of ice at \(0^\circ C\) converted into water at \(0^\circ C\),
D. \(10g\) of ice at \(0^\circ C\) converted into water at \(0^\circ C\).

• A. B \(>\) A \(>\) D \(>\) C
• B. A \(>\) B \(>\) C \(>\) D
• C. A \(<\) B \(<\) C \(<\) D
• D. A \(>\) B \(<\) C \(>\) D

Hint:

For phase change at constant temperature, use \(\Delta S=\frac{mL}{T}\).

Answer 1

The Correct Option is A

Concept:
Entropy change during phase change is: \[ \Delta S=\frac{Q}{T}=\frac{mL}{T} \]

Step 1: Case A. \[ \Delta S_A=\frac{10\times540}{373} \] \[ \Delta S_A\approx14.48 \]

Step 2: Case B. \[ \Delta S_B=\frac{20\times540}{373} \] \[ \Delta S_B\approx28.96 \]

Step 3: Case C. \[ \Delta S_C=\frac{1\times80}{273} \] \[ \Delta S_C\approx0.293 \]

Step 4: Case D. \[ \Delta S_D=\frac{10\times80}{273} \] \[ \Delta S_D\approx2.93 \]

Step 5: Compare values. \[ B>A>D>C \] \[ \therefore \text{Correct Answer is (A)} \]