A sample of 0.1 g of water at 100°C and normal pressure (1.013 105 N m-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
- 104.3 J
- 208.7 J
- 42.2 J
- 84.5 J
Answer
(2) 208.7 J
Solution
Heat supplied: Q = 54 cal = 54 4.18 J = 225.72 J
Volume of 0.1 g of water = 0.1 cc
⇒ V = 167.1 - 0.1 = 167 cc = 167 10-6 m3
Work done by the sample: W = P V
= 1.013 105 167 10-6
= 16.92 J
First law of thermodynamics: U = Q - W
= 225.72 - 16.92
208.8 J 208.7 J
Answer: (2)