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

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SUMMARY

The discussion focuses on calculating the necessary mass of a weight to allow steam release from a pressure cooker at a boiling point of 120°C. The saturated water pressure at this temperature is 0.2 MPa, while the atmospheric pressure is 101.3 kPa. The correct calculation for the mass is derived using the formula m = (p_a - p_o) * A / g, resulting in a required mass of 50.36 grams. Participants confirm the accuracy of this calculation, affirming the methodology used.

PREREQUISITES
  • Understanding of pressure concepts, specifically saturated water pressure and atmospheric pressure.
  • Familiarity with basic physics equations, particularly force and mass calculations.
  • Knowledge of units of measurement, including pascals (Pa) and grams (g).
  • Basic principles of thermodynamics related to boiling points and pressure cookers.
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  • Research the relationship between pressure and boiling points in various liquids.
  • Learn about the design and safety mechanisms of pressure cookers.
  • Explore advanced calculations involving steam pressure and heat transfer.
  • Investigate the effects of altitude on boiling points and pressure cooker performance.
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Chemists, culinary professionals, and anyone interested in the physics of cooking and pressure cooking techniques.

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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

 
Last edited:
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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?
 
correct!
 

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