Calculate the increase in the internal energy of 10 g of water when it is heated from 0°C to 100°C and converted into steam at 100 kPa. The density of steam = 0.6 kg m^{−3}. Specific heat capacity of water = 4200 J kg^{−1} °C^{−1} and the latent heat of vaporization of water = 2.25 × 10 6J kg^{−1}.

#### Solution

Given:-

Mass of water, m = 10 g = 0.01 kg

Pressure, P = 10^{5} Pa

Specific heat capacity of water, c = 1000 J/Kg °C

Latent heat, L = 2.25 × 10^{6} J/Kg

∆ t = Change in temperature of the system = 100°C = 373 K

∆Q = Heat absorbed to raise the temperature of water from 0°C to 100°C + Latent heat for conversion of water to steam

∆Q = \[mc ∆ t + mL\]

= 0.01 × 4200 × 100 + 0.01 × 2.5 × 10^{6}

= 4200 + 25000 = 29200 J

∆W = P∆V

∆V \[= \text{mass}\left( \frac{1}{\text{final density}} - \frac{1}{\text{initial density}} \right)\]

\[∆ V = \left( \frac{0 . 01}{0 . 6} \right) - \left( \frac{0 . 01}{1000} \right) = 0 . 01699\]

∆W = P∆V = 0.01699 × 10^{5} = 1699 J

Using the first law, we get

∆Q = ∆W + ∆U

∆U = ∆Q − ∆W = 29200 − 1699

= 27501 = 2.75 × 10^{4} J