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What kind of dissipative work is done on the sand?

  1. Jan 7, 2008 #1
    1. The problem statement, all variables and given/known data

    A bag of sand of mass m= 50 kg at a temperature T = 300K falls from a height h = 10m onto the ground and
    comes to an abrupt stop. For your considerations, neglect any transfer of heat between the sand and its
    surroundings and assume that the sand changes its shape but not its volume. Answer (a) to (c) by giving a
    general equation, then reduce it to the terms relevant to this problem.
    a. What kind of dissipative work is done on the sand? What is its amount W?
    b. What is the change in the internal energy U of the sand?
    c. What is the entropy increase S of the sand?
    d. Why is it important in these considerations to assume that the sand comes to an abrupt stop?
    (The acceleration due to gravity is g = 9.81ms−1.)


    2. Relevant equations



    3. The attempt at a solution





    The above problem caused me to hit a brick wall when i was looking through some past tutorial example sheets. I really need a kick in the right direction i suppose to start solving.
    I know i can work out potential energy from the given data but thats as far as it goes for me. Any advice is greatly appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 7, 2008 #2

    dynamicsolo

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    What was the sand doing just before it hit the ground (I mean in terms of physically measureable quantities)? What was it doing just after it hit? Why is the work done on the sand described as "dissipative"? How much work must have been done on the sand during the impact?

    What would be the relevant equation to use that connects U and W? How much heat (Q) was introduced from the environment? How much heat altogether must have been delivered to the sand?

    Once you have this, (c) will be straightforward. Consider what assumptions you have to make to find the entropy change: that will be important in answering (d).
     
  4. Jan 8, 2008 #3
    I presume you mean accelerating or something? I have an idea that consists of potential energy being converted strictly into Q. This is because of the volume being constant and therefore there is no work done by the sad.

    U = W + Q shows that if no work was done by the system then U = Q. Is this correct?
     
  5. Jan 8, 2008 #4

    Andrew Mason

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    There are two aspect to "work" here. Work is being done on the sand by gravity and that work is then dissipated in the form of heat flow to the sand. In this case [itex]W = \Delta Q[/itex]. But there is also the thermodynamic aspect to work done by the sand in changing shape. You are correct that no work is done there since volume does not change: the heat increases the internal energy of the sand. So [itex]\Delta Q = \Delta U[/itex]

    The essential point is that the total energy does not change. In other words, the internal, kinetic and potential energy of the sand before being released must be equal to the internal, kinetic and potential energy immediately before impact which must be equal to the internal, kinetic and potential energy after the impact.

    You are correct that the change in potential energy must equal the increase in internal energy of the sand (heat flow into the sand due to dissipative work done on sand by the impact).

    The only sticky point here is the change in potential energy. Since the shape changes, presumably the centre of mass does as well. But I think you are supposed to ignore that and assume that the centre of mass drops 10 m.

    To determine the change in entropy of the sand, you would have to know the heat capacity of the sand. But I think you can assume that the temperature of the sand will not change much.

    I am not sure why you would have to assume that the sand comes to an abrupt stop. It doesn't make a difference to the calculation that I can see. Besides, it doesn't come to an abrupt stop because it is not a rigid body.

    AM
     
    Last edited: Jan 8, 2008
  6. Jan 8, 2008 #5
    So the dissipative work is the potential energy being converted to heat energy on the sand? so therefore the answer to (a) is just (mass) x (accel. due to gravity) x (height).

    If not then im not sure about the two types of work.
     
  7. Jan 8, 2008 #6

    Andrew Mason

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    Correct.

    The work done on the sand by gravity gives it kinetic energy. The sand then performs dissipative work on itself by converting that kinetic energy to thermal energy (by increasing the temperature of the sand).

    AM
     
  8. Jan 8, 2008 #7
    Ok thank for the help!
    If Q = W for part (a) then for part (b) using the equation U = W + Q does this mean that the change in internal energy of the sand is zero?
     
  9. Jan 8, 2008 #8

    Andrew Mason

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    No. There are two aspects to work: The work done by gravity on the sand (W1) and the non-dissipative work (ie the work that does not result in heat flow) done by the sand during the collision (W2).

    Prior to the collision, [itex]W1 = \Delta{PE} = mgh = \Delta{KE} = \frac{1}{2}mv^2[/itex]. After the fall, [itex]\Delta Q = W1 = \frac{1}{2}mv^2[/itex].

    If you want to use the first law of thermodynamics in the collision, [itex]\Delta U = \Delta Q - W[/itex] where W is W2, the non-dissipative work done by the sand in the collision (= 0), so [itex]\Delta Q = \Delta U[/itex]. All of the kinetic energy before the collision is changed into heat without doing any thermodynamic work.

    AM
     
  10. Jan 8, 2008 #9
    sorry about this im struggling a bit.
    for part (a) all i need to work it out is the mgh equation right?

    but, in relation to part (b), then you mentioned [itex]\Delta Q = W1 = \frac{1}{2}mv^2[/itex] and also that this was equal to [itex]\Delta Q = \Delta U[/itex] so arent they the same value?
    sorry im not following all that well :S
     
  11. Jan 8, 2008 #10

    Andrew Mason

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    Right.

    Right.

    [tex]\Delta Q = mgh = \Delta U[/tex]

    This is only true, as you point out, because the energy is all dissipated as heat in the sand on impact, increasing the temperature of the sand.

    It need not be dissipated as heat. If, for example, instead of a bag of sand you had a golf ball, the energy compressing of the ball on impact would not all be dissipated as heat but some would be stored as potential (spring) energy in the ball. Then you would have to determine how much energy was stored as spring energy in the ball and subtract that from the total energy (mgh) to determine how much energy is disspated as heat.

    AM
     
    Last edited: Jan 8, 2008
  12. Jan 9, 2008 #11

    dynamicsolo

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    I believe you need to make this assumption to keep the entropy calculation simple. If the sand stops "abruptly", you can get away with assuming that there is no significant change in its temperature and can just use delta_S = Q/T , with the given value of T. Otherwise, you'd need to deal with some sort of integration with T changing, which means you'd need a heat conduction model for how the temperature changes during impact, heat capacity and thermal conductivity information for the sand, etc. It is easier (though only an estimate) to just treat the sand as if it heats up and deforms "instantaneously."
     
  13. Jan 9, 2008 #12

    Andrew Mason

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    Well, you can see that the temperature of the sand won't change much because the energy (mgh) is only 50*9.8*10 = 4900 J. = 1172 cal. That would raise the temperature of 50 kg of water about one 50th of a degree.

    It would heat up sand a bit more but not enough to significantly affect the entropy calculation.

    AM
     
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