Optimizing Pressure Cooker Steam Release: Weight Calculation for Boiling at 120C

[cd19]

Homework Statement

A pressure cooker has a lid screwed on tight. A small opening with A=5 mm² is covered with a weight which can be lifted to let steam escape. How much mass should the weight have to allow boiling at 120 C with an outside atmosphere at 101.3kPa

Homework Equations

p=f/a therefore f = pa and the mass will be equal to pa/g

saturated water pressure at 120C is .2 Mpa
atmospheric pressure is 101.3 Kpa

m = (200000-101300)(.005)/9.8 = 50.36g

The Attempt at a Solution

 

[cd19]

Right so someone tell me if I'm correct here:

p=f/a therefore f = pa and the mass will be equal to pa/g

saturated water pressure at 120C is .2 Mpa
atmospheric pressure is 101.3 Kpa

m = (200000-101300)(.005)/9.8 = 50.36g

Am I correct?

 

[RTW69]

correct!