Question

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.

Answer 1

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$\times {10}^6\mathrm{J}/\mathrm{Kg}$
$∆t=\mathrm{Change}\mathrm{in}\mathrm{temperature}\mathrm{of}\mathrm{the}\mathrm{system}=100°\mathrm{C}=373\mathrm{K}$

∆Q = Heat absorbed to raise the temperature of water from 0$°\mathrm{C}\mathrm{to}100°\mathrm{C}$ + 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$=\mathrm{mass}\left(\frac{1}{\mathrm{final}\mathrm{density}}-\frac{1}{\mathrm{initial}\mathrm{density}}\right)$
$∆\mathrm{V}=\left(\frac{0.01}{0.6}\right)-\left(\frac{0.01}{1000}\right)=0.01699$

∆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