Pressure Cooker Problem Solution

We are given the following details:

Volume of the pressure cooker: 3 liters
Pressure inside the cooker: 10 atmospheres
Cross-sectional area of the orifice: 4 mm²
Wattage of heat supplied: 300 watts
Time after which the cooker whistles: 3 minutes

We are asked to calculate:

The weight of the stopper
The mass flow rate of steam

Step 1: Convert Units

First, we need to convert the pressure from atmospheres to Pascals and the cross-sectional area from mm² to m².

\( P = 10 \, \text{atm} = 10 \times 1.013 \times 10^5 \, \text{Pa} = 1.013 \times 10^6 \, \text{Pa} \)

\( A = 4 \, \text{mm}^2 = 4 \times 10^{-6} \, \text{m}^2 \)

Step 2: Calculate Force on the Stopper

The force exerted by the pressure on the stopper is given by:

\( F = P \times A = 1.013 \times 10^6 \, \text{Pa} \times 4 \times 10^{-6} \, \text{m}^2 = 4.052 \, \text{N} \)

Step 3: Calculate the Weight of the Stopper

The weight of the stopper is equal to the force due to pressure. The mass of the stopper can be calculated using \( W = m \times g \), where \( g = 9.81 \, \text{m/s}^2 \):

\( m = \frac{W}{g} = \frac{4.052}{9.81} \, \text{kg} \approx 0.413 \, \text{kg} \)

Weight of the stopper: 0.413 kg

Step 4: Calculate Energy Supplied

The total energy supplied to the pressure cooker over 3 minutes is:

\( \text{Energy} = \text{Power} \times \text{Time} = 300 \, \text{W} \times 180 \, \text{s} = 54000 \, \text{J} \)

Step 5: Calculate the Mass of Steam Released

The amount of steam generated can be found using the latent heat of vaporization \( L = 2.26 \times 10^6 \, \text{J/kg} \):

\( m_{\text{steam}} = \frac{\text{Energy}}{L} = \frac{54000}{2.26 \times 10^6} \, \text{kg} \approx 0.0239 \, \text{kg} \)

Step 6: Calculate the Mass Flow Rate

The mass flow rate of the steam is:

\( \dot{m} = \frac{0.0239 \, \text{kg}}{180 \, \text{s}} \approx 1.33 \times 10^{-4} \, \text{kg/s} \)

Mass flow rate of steam: 1.33 × 10^-4 kg/s