Q1. A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10^5 N m^-2) requires 54 cal of heat energy to convert into steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is:
A 42.2 joule
B 84.5 joule
C 208.7 joule Correct
D 104.3 joule
Explanation
Use the first law: heat supplied equals the rise in internal energy plus external work. The heat supplied is 54 cal, about 54 × 4.2 = 226.8 J. The work done during expansion is PΔV = 1.013 × 10^5 × 167.1 × 10^-6, approximately 16.9 J, as the initial liquid volume is negligible at this scale. The change in internal energy is therefore about 226.8 - 16.9 = 209 J, matching 208.7 joule. The smaller values result from omitting the heat conversion or over-subtracting the work term.