Work Done by Water: Calculating/Solving Homework

Click For Summary

Homework Help Overview

The problem involves calculating the work done by water in a sealed container as it is heated from room temperature to a higher temperature. The context includes considerations of pressure and volume changes in relation to the properties of water and the setup of the container.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of the container's light lid on initial and final pressures. There are questions about the relevant equations to use, such as the ideal gas law and work done calculations.

Discussion Status

Some participants have provided insights into the pressure conditions throughout the process, suggesting that the pressure remains constant at 1 atm. Others express confusion about thermodynamic concepts and seek clarification on where to begin with the calculations.

Contextual Notes

Participants mention the challenge of balancing multiple physics topics, indicating a potential overload of concepts that may affect their understanding of thermodynamics specifically.

Bgerst103
Messages
32
Reaction score
0

Homework Statement



You have a sealed container with 2kg of water (molar mass: 18g/mol, density: 1000kg/m^3). The lid on the container is very light, and the container side walls are very tall, like a tall graduated cylinder. You start with the water at room temperature 25 C and boil all the water until you get to 200 C. What is the work done by he water molecules.

Homework Equations





3. The Attempt at a Solution [/b

I'm not really sure how to go about starting this. If someone could give me a set of equations or explanation it would be much appreciated.
 
Physics news on Phys.org
The given answers are: A) 234 B) 334 C) 434 D) 534 kJ
 
The key to this problem are the words "the lid on the container is very light." What does this mean to you about what the initial pressure is? What does this mean to you about what the final pressure is? If you know the final pressure, then, from the ideal gas law, you know the final volume. What is it? What is the initial volume? What is the equation for the amount of work done?

Chet
 
Chestermiller said:
The key to this problem are the words "the lid on the container is very light." What does this mean to you about what the initial pressure is? What does this mean to you about what the final pressure is? If you know the final pressure, then, from the ideal gas law, you know the final volume. What is it? What is the initial volume? What is the equation for the amount of work done?

Chet

Well, work done is W=Fd. However the lid on the container is very light doesn't really help me understand how to find the initial pressure. Are we working with PV=nRT or P1V1/T2=P2V2/T2 or both? Either way pressure and volume are both unknown. I need a place to start and then I can probably work my way through the rest of the problem. I just don't know where to start.
 
Bgerst103 said:
Well, work done is W=Fd. However the lid on the container is very light doesn't really help me understand how to find the initial pressure. Are we working with PV=nRT or P1V1/T2=P2V2/T2 or both? Either way pressure and volume are both unknown. I need a place to start and then I can probably work my way through the rest of the problem. I just don't know where to start.
If the lid to on the container is very light (and the system is initially in thermodynamic and mechanical equilibrium), then the initial pressure in the container must be 1 atm. When the system reaches its final equilibrium state, the pressure within the container must still be 1 atm. In fact, during the entire process, the pressure within the container remains at 1 atm. (to match the pressure on the other side of the lid). Does this make sense?

Chet
 
Chestermiller said:
If the lid to on the container is very light (and the system is initially in thermodynamic and mechanical equilibrium), then the initial pressure in the container must be 1 atm. When the system reaches its final equilibrium state, the pressure within the container must still be 1 atm. In fact, during the entire process, the pressure within the container remains at 1 atm. (to match the pressure on the other side of the lid). Does this make sense?

Chet

Yes, that makes sense. Sorry, I've had to learn momentum, thermodynamics, collision properties, and fluid properties all over the course of a day so my head is cluttered with equations and concepts. Thermodynamics is still very confusing to me.
 
Sorry to hear this. Thermo is confusing enough to most people without also trying to master these other subjects (in one day).

Chet
 
Chestermiller said:
Sorry to hear this. Thermo is confusing enough to most people without also trying to master these other subjects (in one day).

Chet

Trust me I found out the hard way. Had to do 125 problems, have 5 left, all thermodynamics, this being one of them.
 

Similar threads

Replies
37
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 18 ·
Replies
18
Views
4K
  • · Replies 44 ·
2
Replies
44
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 6 ·
Replies
6
Views
7K