A sample of 0.1 g of water at 100°C and normal pressure (1.013×10^5 Nm^-2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

1. 104.3 J

2. 42.2 J

3. 208.7 J

4. 84.5 J

Doubtly Asked 30 Oct 2026 NEET UG 2018 Practice test neet Save WhatsApp Report

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Answer: 3. 208.7 J

First law: $\Delta U=Q-W$, with $W=P\Delta V$.

$Q=54\times4.18\approx225.7$ J.

Volume change: steam 167.1 cc minus water 0.1 cc, $\Delta V=167\ \mathrm{cm^3}=1.67\times10^{-4}\ \mathrm{m^3}$.

$W=1.013\times10^5\times1.671\times10^{-4}\approx16.9$ J.

$$\Delta U\approx225.7-16.9\approx208.7\ \mathrm{J}$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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