Solution to Question 10
A pressure cooker contains superheated water at $120^\circ\text{C}$ (under pressure). When the lid is opened, the pressure drops to atmospheric pressure, where the boiling point of water is $100^\circ\text{C}$.
Since the water is at a temperature higher than the boiling point, it is unstable. It will boil spontaneously (“flash boiling”) using its own internal energy until the temperature drops to $100^\circ\text{C}$.
- Initial Mass, $m_i = 2.6 \text{ kg}$
- Initial Temperature, $T_i = 120^\circ\text{C}$
- Final Temperature, $T_f = 100^\circ\text{C}$
- Specific Heat of Water, $s = 4.2 \text{ kJ/kg}^\circ\text{C} = 4200 \text{ J/kg}^\circ\text{C}$
- Latent Heat of Vaporization, $L = 2.1 \text{ MJ/kg} = 2.1 \times 10^6 \text{ J/kg}$
The heat released by the water as it cools from $120^\circ\text{C}$ to $100^\circ\text{C}$ is used to convert a small portion of the water into steam.
Heat Released ($Q_{released}$) = Heat Absorbed for Vaporization ($Q_{vaporization}$)
$$ Q_{released} = m_i s \Delta T $$
$$ Q_{vaporization} = (\Delta m) L $$
Where:
- $\Delta T = 120^\circ\text{C} – 100^\circ\text{C} = 20^\circ\text{C}$
- $\Delta m$ is the mass of water evaporated.
Calculate the energy released:
$$ Q = 2.6 \text{ kg} \times 4200 \text{ J/kg}^\circ\text{C} \times 20^\circ\text{C} $$
$$ Q = 218,400 \text{ J} $$
Calculate the mass evaporated ($\Delta m$):
$$ \Delta m = \frac{Q}{L} = \frac{218,400}{2.1 \times 10^6} $$
$$ \Delta m \approx 0.104 \text{ kg} $$
The remaining mass of water ($m_f$) is the initial mass minus the evaporated mass.
$$ m_f = m_i – \Delta m $$
$$ m_f = 2.6 \text{ kg} – 0.104 \text{ kg} = 2.496 \text{ kg} $$
Rounding to appropriate significant figures (consistent with the input data precision):
$$ m_f \approx 2.5 \text{ kg} $$