Pressure Cooker Internal Energy Problem

AI Thread Summary
The discussion revolves around the energy dynamics in a pressure cooker, which is a nearly airtight system. It involves 15,000 J of energy used to vaporize water, with 4,000 J leaking to the atmosphere, resulting in a net energy gain of 11,000 J. The participants explore different thermodynamic processes, noting that isovolumetric, isothermal, adiabatic, and isolated processes do not accurately describe the situation due to energy exchange with the surroundings. The conclusion highlights that the pressure can increase without work being done on the system. Understanding these energy exchanges is crucial for analyzing the internal energy changes in such systems.
sharpnova
Messages
43
Reaction score
1

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.
 
Physics news on Phys.org
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
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top