3 Thermodynamics problems [work done by gas, radiation, coefficient of performance]

[clairez93]

PROBLEM 1 (work done by gas)

Homework Statement

1. An ideal monatomic gas originally in state A is taken reversibly to state B along the straight-line path shown in the pressure-volume graph. What ist he work done by the gas?
A) +12 cal
B) +122 cal
C) 0 cal
D) -110 cal
E) +110 cal

Homework Equations

4.186 calories per J

The Attempt at a Solution

I thought the area under the curve would be the work done by the gas, so I split it into a triangle and a rectangle and did this:

$$\frac{1}{2}(1)(101.3*10^{3})(2)(10^{-3}) = 101.3$$ (the area of the triangle)
$$2(101.3*10^{3})(2)(10^{-3}) = 405.2$$ (the area of the rectangle)

I added them up to get 506.5. I then divided by 4.186 to get 120.99, which I supposed was closest to 122.

The answer, however, is +110 cal.

PROBLEM 2 (RADIATION)

Homework Statement

2. Two identical solid spheres have the same temperature. One of the spheres is cut into two identical pieces. These two hemispheres are then separated. The intact sphere radiates an energy Q during a given time interval. During the same interval, the two hemispheres radiate a total energy of Q'. What is the ratio Q'/Q?
A) 1.5
B) 4.0
C) 0.50
D) 0.25
E) 2.9

Homework Equations

$$H = Aert^{4}$$

The Attempt at a Solution

Not a clue how to start this one, as I don't have any values for r or t or A. My guess is those don't matter since H is directly proportional to A, so then I would have to find the surface area, however, I'm not sure how to compute the total surface area for Q' then.

The answer is 1.5, however.

PROBLEM 3 (coefficient of performance)

Homework Statement

3. If the coefficient of performance for a refrigerator is 5.0 and 65 J of work are done on the system, how much heat is rejected to the room?
A) 210 J
B) 260 J
C) 130 J
D) 330 J
E) 390 J

Homework Equations

$$COP = \frac{Q_{h}}{W}$$
$$Q_{h} - Q_{c} = W$$

The Attempt at a Solution

$$COP = \frac{Q_{h}}{W}$$
$$COP = \frac{Q_{h}}{65} = 5$$
$$Q_{h} = 325$$
$$Q_{h} - Q_{c} = W$$
$$325 - Q_{c} = 65$$
$$Q_{c} = 260$$

The answer, however, is 390, so I guess I am supposed to add 325 and 65, though I am not sure why.

 

[clairez93]

Oh, I don't expect help for all of them at once; anyone of them at a time is fine.

 

[clairez93]

Any help would be greatly appreciated, thanks!

 

[clairez93]

Just trying to keep this thread up where it is visible to people who may possibly help. :] Thanks!

 

[xxChrisxx]

For number 3 you add them ( as you stated). You've got to remember sign convention for heat and work.

 

[clairez93]

Sorry, I don't understand? Why would I add them? I can see that adding it would produce the right answer, but I don't know why.

 

[xxChrisxx]

What are Qc and Qh?

Also using the COP is what you wat/what you pay for.

Think about that it is that you want from a refrigirator. Is the useful heat transfer the heat rejected or the heat taken from the space to be refrigirated. (hopwfully this shoudl help)

 

[clairez93]

Qc is heat expelled, Qh is heat absorbed.
I don't understand, the useful heat would be heat absorbed, so that's Qh, and you're solving for heat rejected, which is Qc, right?

 

[xxChrisxx]

Ahhh that's where you are gong wrong. Qc is the heat absorbed from the space to be refrigirated and Qh is the rejected heat into the room, in the case of heat pump and refrigirator. A heat pump is a refrigiration cycle in reverse (kind of).

In the case of the heat pump Qh is the useful heat transfer (to be put in COP equation)
In the case of a refrigirator Qc is the useful heat transfer.

So it would be Qc=COP*W = 325

This goes into the second equation to find the heat rejected at higher temperature (Qh)

http://s260.photobucket.com/albums/ii3/cdcracing/?action=view&current=refrigcycle.jpg