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How to apply Thermodynamics First Law to this problem?

  1. Mar 19, 2017 #1

    vcsharp2003

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    1. The problem statement, all variables and given/known data
    A spring (k = 500 N/m) supports a 400 g mass which is immersed in 900 g of water. The specific heat of the mass is 450 J/kg and of water is 4184 J/kg. The spring is now stretched 15 cm and, after thermal equilibrium is reached, the mass is released so it vibrates up and down. By how much has the temperature of the water changed when the vibration has stopped?

    I am ending up with ΔT ( change in temperature of thermodynamic system) and ΔUint (change in internal energy of thermodynamic system) as unknowns and one equation only, so it seems it's not possible to solve for temperature change ΔT.

    2. Relevant equations
    Thermodynamics_First_Law.png
    heat = mass x specific heat X temperature change


    3. The attempt at a solution

    Let's consider the spring + water as a closed system. Also, let's look at the system starting from when the spring is in water and mass hanging from it in an equilibrium position to when mass returns to it's original position and vibration of spring has stopped. Then, for this system the following points will be true.
    • there could be some heat flow across the system's boundary $$\Delta Q_{in} \neq 0$$
    • the system does not do any work on it's surrounds nor is work done on the system $$W_{by} = 0$$
    • the internal energy of the system should increase as it's temperature will rise $$\Delta U_{int} > 0$$
    • the change in KE of the system is zero since it has no translational motion $$\Delta KE = 0 $$
    • the change in PE of the system is change in spring PE + change in gravitational PE ( change in gravitational PE is zero but change in spring PE is not zero since the spring is less stretched at end of interval being considered i.e. spring PE has decreased)
    So, the first law of Thermodynamics becomes
    $$(0 - 0) + (0 - \frac{1}{2} \times k \times x^{2}) + \Delta U_{int} = Q_{in} - 0$$
    $$(0 - 0) + (0 - \frac{1}{2} \times 500 \times 0.15^{2}) + \Delta U_{int} = Q_{in} - 0$$

    Also, by using the second formula under Relevant Equations using specific heat we get another equation. If ΔT = temperature change in K over the time interval being considered, then
    $$Q_{in} = 0.9 \times 4184 \times \Delta T + 0.4 \times 450 \Delta T$$

    If we combine above two equations then we get the following equation.
    $$(0 - 0) + (0 - \frac{1}{2} \times 500 \times 0.15^{2}) + \Delta U_{int} = 0.9 \times 4184 \times \Delta T + 0.4 \times 450 \Delta T - 0$$.

    It's impossible to solve for ΔT since we have two unknowns in above equation. I am not sure if I am applying first law of Thermodynamics correctly.

    NOTE: However, if Qin is assumed to be zero since heat flows are happening within the system due to fluid resistance against vibration, and ΔUint = 0.9 x 4184 x ΔT + 0.4 x 450 ΔT then by substituting into first law of thermodynamics equation, we get only one unknown of ΔT which we can solve. But, I am not sure if this second approach is correct.
     
  2. jcsd
  3. Mar 19, 2017 #2

    Nidum

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    The solution comes out in a few lines using simple sums .

    How much energy did you start with and where does it go ?
     
  4. Mar 19, 2017 #3

    vcsharp2003

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    This problem is given at end of the chapter of First Law of Thermodynamics, so I was applying that law. Can you solve this using first law of Thermodynamics?

    The energy into the system is 0.9×4184×ΔT+0.4×450 x ΔT. I can equate this to spring PE
    .5×500×0.1522 and solve this problem easily.

    But, I was looking for a more rigorous way of doing this using first law of thermodynamics, if it's possible.
     
    Last edited: Mar 19, 2017
  5. Mar 19, 2017 #4
    You were correct that Q is equal to zero, for the reason that you gave. You did a pretty good job of analyzing the rest of the problem. (I'm not so sure that the change in gravitational potential energy is zero in this process, but, apparently, they expected you to make this assumption.)
     
  6. Mar 19, 2017 #5
    The mass starts at rest in the water and is then stretched and released. It returns to its equilibrium position.
    Only the vibrational KE is dissipated in the water, the temperature of the water and the mass rise.
     
  7. Mar 19, 2017 #6

    vcsharp2003

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    I am thinking that the initial equilibrium position of mass should be same as final equilibrium position of mass according to free body diagram below, which will mean initial and final gravitation PE are equal.
    When mass is in initial rest position ( not when its stretched by 15 cm but before that when it's hanging by itself from spring in water) or when it has finally stopped after vibration in water, then the following equation is satisfied.

    Tension in spring = weight + upthrust due to displaced water
    kx = weight + upthrust due to displaced water

    Since k ( spring constant), weight of mass and upthrust due to displaced water are going to be same in both these positions, so x (amount by which spring is stretched) is going to be same in both positions. Therefore, we can say that gravitation PE are same in both these positions.

    Also, then equation from first law of Thermodynamics should be different as below.

    $$\Delta KE = 0 \text {, } \Delta PE = 0 \text { for both gravitational and spring PE, } Q_{int} = 0 \\ \text {but } W_{by} = -\frac{1}{2}kx^{2} \text { since spring was initially stretched by 15 cm and so work was done on the system}$$


    The first law of Thermodynamics would then become as below.
    $$(0) + (0) + \Delta U_{int} = 0 - (-\frac{1}{2}kx^{2})$$

    We can substitute ΔUint = 0.9×4184×ΔT+0.4×450ΔT in above equation.

    Mass_in_Water_attached_to_spring.png
     
    Last edited: Mar 19, 2017
  8. Mar 19, 2017 #7
    If you are going to take as the initial state the equilibrium rest location of the mass, then you need to include the work done in lowering the mass as part of the process, and this will also involve gravitational (buoyancy) effects. If you take as the initial state, the lower mass location with the stretched spring, then there is some mass of water that is at a higher elevation than at equilibrium and the mass itself which is at a lower elevation than at equilibrium.
     
  9. Mar 19, 2017 #8

    vcsharp2003

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    Yes, I realized that after posting my last message and have modified the post accordingly. Thanks for your great help. I think the upthrust is being ignored in this problem else the volume of the mass would be mentioned..
     
  10. Mar 19, 2017 #9
    Yes. Very perceptive.
     
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