Pressure Cooker Internal Energy Problem

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SUMMARY

The discussion centers on the thermodynamic analysis of a pressure cooker, specifically addressing the energy dynamics involved when 1/2 cup of water is heated. A total of 15,000 J of energy is used to vaporize the water, while 4,000 J leaks out, resulting in a net energy input of 11,000 J. The process does not fit neatly into traditional thermodynamic categories such as isothermal, adiabatic, or isovolumetric due to the energy exchange with the surroundings. The internal energy change is calculated using the equation: change in internal energy = heat added + work done by the system.

PREREQUISITES
  • Understanding of thermodynamic processes (isothermal, adiabatic, isovolumetric, isolated)
  • Familiarity with the first law of thermodynamics
  • Basic knowledge of energy units (Joules)
  • Concept of internal energy in closed systems
NEXT STEPS
  • Study the first law of thermodynamics in detail
  • Learn about different thermodynamic processes and their characteristics
  • Explore practical applications of pressure cookers in thermodynamics
  • Investigate energy conservation principles in closed systems
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Students studying thermodynamics, educators teaching energy concepts, and anyone interested in the practical applications of thermodynamic principles in cooking and engineering.

sharpnova
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Homework Statement


A pressure cooker is a nearly air tight sealed system. When seated on a stove burner and heated, water turns to steam and increases the pressure inside, resulting in higher temperatures so food cooks faster. Suppose 1/2 cup of water is inside the pressure cooker and it is on an active burner on the stove. Suppose that 15000 J of energy vaporize the water and that 4000 J leak out of the system to the atmosphere. What is the name of this kind of process? How much energy is exchanged by heat? Is the energy put in or taken out of the system? How much energy is exchanged by work? Is that work done by or on the system? What is the change in the system's internal energy? (give the sigh of that change as well as the amount)

Homework Equations


change in internal energy = heat added + work done by system

The Attempt at a Solution


For the first question (identifying the type of process) I can't find one that satisfies the situation.

Isovolumetric has the work done by the system as zero. But the system does work on the gas (increasing its pressure)

Isothermal has the change of internal energy as zero. But the system takes in 15000 J and leaks out 4000 J so there is an 11000 J difference.

Adiabatic has the the condition "no heat in or out" but 15000 went in and 4000 went out for a net of 11000 in.

Isolated has no interaction with surroundings but since 4000 J leak out it is interacting with its surroundings.
 
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sharpnova said:
For the first question (identifying the type of process) I can't find one that satisfies the situation.

Isovolumetric has the work done by the system as zero. But the system does work on the gas (increasing its pressure)

Isothermal has the change of internal energy as zero. But the system takes in 15000 J and leaks out 4000 J so there is an 11000 J difference.

Adiabatic has the the condition "no heat in or out" but 15000 went in and 4000 went out for a net of 11000 in.

Isolated has no interaction with surroundings but since 4000 J leak out it is interacting with its surroundings.
It is possible for the pressure of a system to increase without any work being done on the system.
 
Thank you that cleared it up for me
 

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