A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 105 Nm–2) requires 49 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 269 cc, the change in internal energy of the sample, is :

104.3 J 208.7 J 177 J 42.2 J

177 J

\(\Delta Q = \Delta U + \Delta W\) 49 x 4.18 = \(\Delta U\) + 1.013 x 105 (269 x 10-6 - 0) \(\Delta U\) = 177 J