Question

A pressure cooker contains 0.5m^3 water and water vapour mixture at 300 Celsius. Calculate the mass of each if their volumes are equal.

Answer 1

From the steam table at $300 ^\circ C$ (https://www.tlv.com/global/TI/calculator/steam-table-temperature.html):

$$v_f = 0.00140422 \frac{m^3}{kg}$$

$$v_g = 0.0216631 \frac{m^3}{kg}$$

(i) Calculate the mass of water:

$V_L = \frac{0.500 m^3}{2} = 0.250 m^3$

$v_f = \frac{V_L}{m_L}= 0.00140422 \frac{m^3}{kg}$

$\therefore m_L = \frac{V_L}{v_f} = \frac{0.250 m^3}{0.00140422 \frac{m^3}{kg}} = 178 kg$

(ii) Calculate the mass of water vapour mixture:

$V_v = \frac{0.500 m^3}{2} = 0.250 m^3$

$v_g = \frac {V_v}{m_v} = 0.0216631 \frac{m^3}{kg}$

$\therefore m_v = \frac{0.250 m^3}{0.0216631 \frac{m^3}{kg}}= 11.54 kg$

Answer: (i) $m_L = 178 kg$

(ii) $m_v = 11.54 kg$