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Find specific heat of liquid

  1. Apr 27, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    A point source s emits thermal heat energy H every second uniformly in all directions. A cylindrical container with radius R, which contains a liquid of mass m, is placed at a distance h above the source, look at the picture. After t seconds, the temperature of the liquid increases from T1 to T2. If the heat lost from the container in one second is H', determine the specific heat of the liquid. Neglect the heat capacity of the container.

    http://www.luiseduardo.com.br/thermology/heat/heatproblems_arquivos/image029.gif [Broken]

    3. The attempt at a solution

    Let's assume a hypothetical spherical surface enclosing the point source S whose radius changes with time at a constant rate(similar to a wavefront in case of sound wave). Let the rate of change of radius be v. Then time taken when radius of the surface becomes h is equal to t_0=h/v and time when its radius becomes equal to √(R^2+h^2) is t'=√(R^2+h^2)/v. The liquid present in the container starts gaining heat as soon as radius becomes h. Simultaneously it loses heat at a rate H'. Let after a time t, the radius changes to R+dR in time dt, where t_0<t<t'.

    Heat gained = Hdt
    Heat lost = H'dt
    Hdt-H'dt = (dm) s d(ΔT)

    But from here, I'm lost. I don't have any idea how to proceed ahead? I'm also not sure what I'm doing here is correct or even makes sense. Forgive me for that. :redface:
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 28, 2014 #2

    BvU

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    Heat waves travel very fast in the surroundings of the container. We can assume air or vacuum, no information is provided in the formulation, so we don't have to worry. A certain fraction f of H goes into the liquid. The fraction is determined by the stereo angle (dependent on h and R).
    If the formulation mentions only one temperature of the liquid, I am inclined to see this as a stirred tank: same T everywhere in the liquid. And your eqn simplifies quite a bit:

    (fH - H') dt = m cp dT​
     
  4. Apr 28, 2014 #3

    utkarshakash

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    Stereo Angle?? What's that?
     
  5. Apr 28, 2014 #4
    Solid angle I guess.
     
  6. Apr 28, 2014 #5

    BvU

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    Yep. Solid angle.
    And heat waves don't necessarily travel fast. In this exercise, they do.
     
  7. Apr 30, 2014 #6

    utkarshakash

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    The answer is

    [itex] \dfrac{Ht \left( 1- \dfrac{h}{\sqrt{h^2+R^2}} \right) -2H't}{2m(T_2 - T_1)} [/itex]
     
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