Question

10g of water at \(100\degree C \)is converted into steam at the same temperature. If the latent heat of steam is 540cal/g, the change in entropy is

• A. 14.47 cal/K
• B. 144.7 cal/K
• C. 74.47 cal/K
• D. 47.47 cal/K

Hint:

Often, ∆S =mL/T. Watch out for potential factor-of-10 mismatches in the given options or question statements.

Answer 1

The Correct Option is A

Heat absorbed, $Q = m L = 10 \text{ g} \times 540 \text{ cal/g} = 5400 \text{ cal}$.
At $100^\circ \text{C}$, the temperature in Kelvin is $T = 373 \text{ K}$. Change in entropy:
$\Delta S = \frac{Q}{T} = \frac{5400 \text{ cal}}{373 \text{ K}} \approx 14.48 \text{ cal/K}.$
The official “closest” or typical reported answer is $14.47 \text{ cal/K}$. Possibly the question includes a factor of 10 or an alternative unit interpretation