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Mechanical Energy into Thermal

  1. Mar 28, 2007 #1
    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
    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.

    2. Relevant equations
    Q = mcT
    PE = mgh

    3. 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.
  2. jcsd
  3. Mar 28, 2007 #2


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