Mechanical Energy into Thermal

[velvetymoogle]

Homework Statement

Students are designing an experiment to demonstrate the conversion of mechanical energy into thermal energy. They have designed the apparatus shown in the figure above. (To see the image, visit http://www.cs.utexas.edu/users/mfkb/Halo/CPL/Physics1998APexam.pdf and check short answer 3). Small lead beads of total mass M and specific heat c fill the lower hollow sphere. The valves between the spheres and the
hollow tube can be opened or closed to control the flow of the lead beads. Initially both valves are open.
1. The lower valve is closed and a student turns the apparatus 180 degrees about a horizontal axis, so that the filled sphere is now on top. This elevates the center of mass of the lead beads by a vertical distance h. What minimum amount of work must the student do to accomplish this?
2. The valve is now opened and the lead heads tumble down the hollow tube into the other hollow sphere. If all of the gravitational potential energy is converted into thermal energy in the lead beads, what is the temperature increase of the lead?
3. The values of M, h. and c for the students' apparatus are M = 3.0 kg, h = 2.00 m. and c = 128 J/(kg * K). The students measure the initial temperature of the lead beads and then conduct 100 repetitions of the "elevate-and-drain" process. Again, assume that all of the gravitational potential energy is converted into thermal energy in the lead beads. Calculate the theoretical
cumulative temperature increase after the 100 repetitions.
4. Suppose that the experiment were conducted using smaller reservoirs, so that M was one-tenth as large (but h was unchanged). Would your answers to parts (b) and (c) be changed? If so, in what way, and why? If not, why not?
5. When the experiment is actually done, the temperature increase is less than calculated in part (c). Identify a physical effect that might account for this discrepancy and explain why it lowers the temperature.

Homework Equations

Q = mcT
PE = mgh

The Attempt at a Solution

1) Work is an energy, and the energy to move the mass to the top is just a change in potential so isn't it Work = Mgh?

2) Since Q = mc(delta)T, just resolve for (delta)T to get Q/cM, right?

3) Q is the heat added, an energy, which is equal to work. I plugged Mgh in for Q and so my new equation is (delta)T = Mgh/cM. The M's drop out and then I just plug in my values and I got 15.625 degrees.

4) For part 2, the value would be different, a higher change in Temp, because you have a smaller denominator, right?
For part 3, the value would be the same because the M's cross out.

5) I said that a physical effect might be experimental error. To do the math, you have to multiply the height by 100 because of the rotations. However, after each rotating, you have to stop the system so that the beads can drop. There will be a drop in temperature during this time.




Are my answers on the whole correct? Can you spot any errors? This is my first time doing it.

 

[Mentz114]

You seem to have it sorted. This is not bad

http://en.wikipedia.org/wiki/Mechanical_equivalent_of_heat

 

[m2003]

I would like to provide some feedback and clarifications on your responses.

1) Your answer for the minimum amount of work is correct. The work done to elevate the center of mass of the lead beads is equal to the change in potential energy, which is Mgh.

2) Your approach is correct, but there is one small error in your equation. It should be (delta)T = Q/Mc, since Q is the energy and not the heat capacity. Otherwise, your answer is correct.

3) Your equation and approach are correct, and your answer of 15.625 degrees is also correct. However, I would suggest using the actual formula for temperature change in this case, which is (delta)T = Q/mc. This takes into account the fact that the mass of the lead beads is changing after each repetition, as they are being drained out and refilled.

4) For part 4, you are correct in saying that the value for the temperature change would be higher for a smaller mass of lead beads (M). However, the value for the cumulative temperature increase after 100 repetitions (part 3) would not be the same, as it depends on the total mass (M) and not just the mass of lead beads in the lower sphere. So, for part 4, the value for the cumulative temperature increase would be lower.

5) Your answer for part 5 is partially correct. Yes, experimental error could be a factor, but there is also another physical effect that could explain the discrepancy. When the lead beads are tumbling down the hollow tube, some of the mechanical energy is lost due to friction and sound. This would result in a lower temperature increase than calculated.