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Temperature change of Water after dropping a ball into it

  1. Apr 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose we drop a lead ball of mass M into water of mass m from a height h and allow everything to settle down. What is the temperature change of the water? Assume that the container is well insulated.

    2. Relevant equations
    Potential Energy = mgh
    dU = mcdT

    where dU is the change in thermal energy, m is mass, g is gravitational constant, h is height, c is specific heat, dT is temperature change.

    3. The attempt at a solution
    In this case, I actually fully solved the problem. Just apply conservation of energy, and you'll get a neat formula. The only thing I'm confused about is that I had to make a sneaky assumption that the change in temperature of the water is the same as the change in temperature of the lead ball. I do not see that this is justified. Could someone explain why this assumption is valid?
     
  2. jcsd
  3. Apr 16, 2016 #2

    haruspex

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    It says "and allow everything to settle down". Presumably that extends to any temperature distribution.
     
  4. Apr 16, 2016 #3
    I don't understand your comment. For example, we worked a similar problem where a lead ball crashes into the ground, and we assumed the ground doesn't have any temperature change at all. Here, we are assuming the lead ball and water have the same temperature change, and I don't see why.
     
  5. Apr 16, 2016 #4

    haruspex

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    It's a question of timescale.
    Did the ball-into-ground problem say"and allow everything to settle down"? If so, the answer should be that the lead ball comes to the pre-existing temperature of the ground.
    Over shorter timescales, it depends where the heat is generated and how insulating the bodies are. A lead ball onto concrete will generate nearly all the internal energy within the ball, which will relatively slowly transfer to the ground. A lead ball into soft earth will generate nearly all the internal energy within the ground, which then seeps into the ball as well as further afield.
    In the current question, the collision will generate nearly all the internal energy within the water. How long that takes to even out within the water, compared to within water+lead, is a complicated question. Lead is the better conductor, but water enjoys convection.
     
  6. Apr 16, 2016 #5
    We assume that the ball and the water have the same initial temperature, right?
     
  7. Apr 16, 2016 #6
    No, the problem does not indicate that. I already thought about that before.

    I just redid the problem where I assume that the "Iron" ball does not undergo any temperature change. I almost get the exact same answer. I think we're supposed to assume that the "Iron" ball does not undergo any temperature change. Almost all of the temperature change is in the water.

    I put "Iron" in "" because I had written earlier that it was lead. It's actually Iron.
     
  8. Apr 16, 2016 #7

    haruspex

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    Then either interpretation will do.
     
  9. Apr 16, 2016 #8
    If the Iron ball and water have different initial temperatures (and the book certainly didn't say they have the same initial temperature) then delta temperature iron = T_f - T_i(Iron) which is not equal to T_f - T_i(water). They would certainly share the same final temperature after an equilibrium is reached by definition. I think the only way to read the question is that the Iron ball doesn't experience much of any temperature change. Thanks for the insights. I think book questions can sometimes be a bit ambiguous about the assumptions they're making even if they are very reasonable.
     
  10. Apr 17, 2016 #9
    When they say "everything settles down," what them mean is that everything reaches thermodynamic equilibrium. At thermodynamic equilibrium, the ball and the water will have the same temperature. This is why they tell you that the container is well-insulated. They forgot to mention that the ball and the water start off at the same temperature.

    Let me guess. You're studying the first law of thermodynamics.

    Chet
     
  11. Apr 17, 2016 #10

    haruspex

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    That was my response in post #2, but did you read post #3? It would be interesting to see the exact text of the other question.
     
  12. Apr 17, 2016 #11
    Yes, I agree with you. Obviously, if one waits long enough in the other problem, all the heat goes into the surroundings and virtually none stays with the ball.
     
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