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 = 105 Pa
Specific heat capacity of water, c = 1000 J/Kg °C
Latent heat, L = 2.25 × 106 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 × 106
= 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 × 105 = 1699 J
Using the first law, we get
∆Q = ∆W + ∆U
∆U = ∆Q − ∆W = 29200 − 1699
= 27501 = 2.75 × 104 J
