A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10⁵ Nm^–2) requires 60 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 225 cc, the change in internal energy of the sample, is : 

228 J

\(\Delta Q = \Delta U + \Delta W\)

\(\Delta Q = \Delta U + P \Delta V\)

60 x 4.18 = \(\Delta U\) + 1.013 x 10⁵ (225 x 10^-6 - 0)

\(\Delta U\) = 228 J