Q68PE - College Physics (Urone)

Question

Question: If you want to cook in water at 150°C, you need a pressure cooker that can withstand the necessary pressure. (a) What pressure is required for the boiling point of water to be this high? (b) If the lid of the pressure cooker is a disk 25.0 cm in diameter, what force must it be able to withstand at this pressure?

Step-by-Step Solution

Answer

Answer

1. The pressure at 150°C is  $4.76\times {10}^5 Pa$
2. The force is equal to  $1.836\times {10}^4 N$

Step 1: Concept

The boiling point is the temperature at which vapor becomes equal to the pressure in the surrounding environment.

Step 2: Solution for part (a)

Here, the boiling point is 150°C. The vapor pressure at this boiling point is. $4.76\times {10}^5 Pa$So, the pressure in the surrounding environment is equal to . $4.76\times {10}^5 Pa$.This is the absolute pressure.

Step 3: Solution for part (b)

The calculated absolute pressure is 4.76×10⁵ Pa. It is the sum of atmospheric pressure and gauge pressure. Gauge pressure is the pressure in the cooker relative to the atmospheric pressure. To calculate the force, this gauge pressure has to be calculated. The atmosphere is equal to 1.013×10⁵ Pa. So, the pressure of the cooker is calculated as,

$$\begin{gathered} \text{ Pressure}=Absolute\text{ pressure}-\text{Atmospheric pressure} \\ =4.76\times {10}^5 Pa-1.013\times {10}^5 Pa \\ =3.747\times {10}^5 Pa \end{gathered}$$

Pressure is the force exerted per unit perpendicular area. The area of the circular disc has to be calculated.

$$\begin{gathered} \text{diameter of the disc}=25 cm\left(\frac{1 m}{100 cm}\right) \\ =0.25 m \\ \text{radius of the disc, }r=\frac{0.25}{2} \\ =0.125 m \\ \text{Area of the disc}=\pi r^2 \\ =\pi {\left(0.125 m\right)}^2 \\ =0.049 m^2 \end{gathered}$$

We can calculate the value of force using the pressure and area,

 $\text{pressure}=\frac{force}{area}$

$$\begin{gathered} \text{force}=pressure\times area \\ =3.747\times {10}^5 Pa\times 0.049 m^2 \\ =1.836\times {10}^4 N \end{gathered}$$

Thus, the force required to withstand the pressure is 1.836×10⁴ N.