Work Done by Water: Calculating/Solving Homework

In summary: Almost done though.In summary, the problem involves a sealed container with 2kg of water that is heated from room temperature to 200 C. The lid on the container is very light, meaning the initial pressure within the container is 1 atm and remains at 1 atm throughout the process. Using the ideal gas law, the final volume can be determined and compared to the initial volume to find the amount of work done by the water molecules. The given answers are in units of kJ.
  • #1
Bgerst103
32
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.
 
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  • #2
The given answers are: A) 234 B) 334 C) 434 D) 534 kJ
 
  • #3
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
 
  • #4
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.
 
  • #5
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
 
  • #6
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.
 
  • #7
Sorry to hear this. Thermo is confusing enough to most people without also trying to master these other subjects (in one day).

Chet
 
  • #8
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.
 

Related to Work Done by Water: Calculating/Solving Homework

1. How is work done by water measured?

The work done by water is measured in joules (J), which is a unit of energy. This measurement takes into account the force applied by the water and the distance over which the force is applied.

2. What is the formula for calculating work done by water?

The formula for calculating work done by water is W = Fd, where W is the work done in joules, F is the force applied in newtons (N), and d is the distance over which the force is applied in meters (m).

3. How do you calculate the force of water?

The force of water can be calculated using the formula F = mg, where F is the force in newtons (N), m is the mass of the water in kilograms (kg), and g is the acceleration due to gravity (9.8 m/s²).

4. Can work done by water be negative?

Yes, work done by water can be negative. This occurs when the water is pushing against an object in the opposite direction of its motion, resulting in a negative value for work done.

5. What factors affect the work done by water?

The work done by water is affected by the force of the water, the distance over which the force is applied, and the angle at which the water is applied. Other factors such as the density and viscosity of the water may also play a role in the calculation.

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