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Idealised hot water bottle

  1. Jul 1, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    2)A certain heat source (a heat reservoir or thermal reservoir), is always at temperature ##T_0##. Heat is transferred from it through a slab of thickness ##L## to an object which is initially at a temperature ##T_1 < T_0##. The object, which is otherwise thermally insulated from its surroundings, has a mass m = 0.5 kg and a specific heat capacity ##c = 4 \times 10^{3} J kg^{-1}K^{-1}##. Heat is conducted through the slab at a rate (in J s^{-1}) specified by the
    formula ##KA((T_0 - T)/L),## where K is the thermal conductivity of the slab, A is the area of the slab through which heat is transferred and T is the instantaneous temperature of the object.

    a)Show that provided certain assumptions are made, ##KA((T_0 - T)/L)\Delta t = mc \Delta{T}##
    b)Show that for small t, and hence small ΔT, the above equation can be rearranged and
    integrated to give $$T_2 - T_1 = (T_0 - T_1)\left(1 - \exp\left(-\frac{KA(t_2 - t_1)}{Lmc}\right)\right)$$

    3. The attempt at a solution

    2) a)Assuming all energy gained by the slab is then gained by the object ( and there is then no losses because of insulation), $$\frac{KA(T_o - T)}{L} \Delta t = cm (T + \Delta T - T)$$ and result follows.
    b)Integrate $$\frac{KA}{L} \int_{t_1}^{t_2} dt = cm \int_{T_1}^{T_2} \frac{dT}{T_o - T}$$
    which gives $$\frac{KA}{Lmc}(t_2 - t_1) = \ln\left(\frac{T_o - T_1}{T_o -T_2}\right).$$ It is given that ##T_o > T_1## so the numerator of the log is definitely postive. However, I am not so sure about the denominator. Since there are no heat losses, I think there may exist a temperature ##T_2## where ##T_2 >T_o##.

    Thank you.
     
  2. jcsd
  3. Jul 1, 2013 #2
    I think it is quite obvious that ##T_0## is the highest temperature possible anywhere in the system.
     
  4. Jul 1, 2013 #3
    Your final solution is correct. Just work it into the same mathematical form that they ask for.
     
  5. Jul 2, 2013 #4

    CAF123

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    Hi voko,
    Did you take the system to be reservoir + slab + object? If so, yes, it seems sensible that ##T_o## is the highest temperature since there is no external source of heat.Right?

    However, if you take the object as the system, then there does exist an external source of heat so why is it that the object can never get hotter than the reservoir? (given that we don't know anything about the object)?
     
  6. Jul 2, 2013 #5
    Your equations are already based on the assumption that there is just one source of heat, with a particular temperature. You can't have that both ways.
     
  7. Jul 2, 2013 #6

    CAF123

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    What do you mean by this?
     
  8. Jul 2, 2013 #7
    You cannot model your system as if it had one source with a particular temperature, and at the same time model it as if it had unknown sources.
     
  9. Jul 2, 2013 #8

    CAF123

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    If I take the system to be reservoir + slab + object, then the maximum temperature of this system is ##T_0## since there are no external heat sources. There exists an internal heat source (reservoir) only. Is this reasoning correct for why the temperature of the object can never exceed ##T_o##?
     
  10. Jul 2, 2013 #9
    Since the system is insulated, the heat to the object can only be passed from the slab or from the reservoir via the slab. If the object is already as hot as the reservoir, can any more heat be passed to it by simple conduction? Why?
     
  11. Jul 2, 2013 #10

    CAF123

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    What are you taking as the system here? Just the object?
    I would say no. If we assume the slab is a diathermal wall between the object and the reservoir, then the object and reservoir will settle to the same temperature. Since the reservoir is always at a temperature of ##T_0##, by the zeroth law, the object may not become hotter.
     
  12. Jul 2, 2013 #11
    The entire system, with the reservoir and the conductor. The object alone cannot be analyzed meaningfully, I think.
     
  13. Jul 2, 2013 #12

    CAF123

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    Okay, thank you. Is my analysis correct in my previous post? Also, given that the system is object, reservoir and slab, would it not be correct to state that given no external heat source, it is not possible for the temperature of the system to exceed ##T_0##?
     
  14. Jul 2, 2013 #13
    It is correct in the sense that at equilibrium, the temperature will have to be ##T_0## everywhere. But that does not mean, per se, that the temperature somewhere cannot be higher than ##T_0## while on the way to equilibrium.

    That requires some other proof.

    That is correct, but you have omitted some important details. Such as what law of thermodynamics is at play here.
     
  15. Jul 2, 2013 #14

    CAF123

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    What sort of proof? Can you elaborate?

    The first law of thermodynamics. The system is isolated so no heat exchange to the surroundings.
     
  16. Jul 2, 2013 #15
    Sorry, I cannot elaborate on the proof. You will have to :)

    You have to prove that the temperature in this system cannot exceed ##T_0## even before it reaches equilibrium.

    That is not enough to make ##T_0## the max temperature. Why can the slab not transmit more heat to the object at any given temperature, say ##T_0##, and so make it even hotter?
     
  17. Jul 2, 2013 #16
    This is a standard problem in heat transfer. The system is the mass. Heat is being conducted from the reservoir to the mass through the slab. The formulation of the problem implies that the thermal inertia (ρCp) of the slab can be neglected so that the temperature gradient within the slab is always linear. The formulation also implies that the thermal conductivity of the mass is very high, so that, throughout the process, the temperature within the mass is essentially uniform. The temperature within the mass as well as within the slab can, at no location and no time, be higher than the reservoir temperature. This is implicit in the problem formulation as well as the problem solution.
     
  18. Jul 2, 2013 #17

    CAF123

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    The temperature of the reservoir is always ##T_0##. Assuming that the temperature of the system can exceed ##T_0##, then that means there must exist more heat energy than what we had orginally, (the reservoir is the only source of heat) so contradicting the first law. (Hmm.. is that right?) Or in an isolated system, entropy never decreases. So if the temperature exceeded ##T_0## and then at a later t decreased, this would contradict the second law of thermodynamics.

    I think the question you pose is exactly why I thought that there may exist a ##T > T_0## in the first place. Although with a bit of further reading, is it because in an isolated system, as we have here, equilibrium is at maximum entropy and so the state of the system cannot become more disordered? (which would happen if the object was allowed to become hotter?)

    Sorry for the late response, but I had to read up on the second law before I could answer.
     
  19. Jul 2, 2013 #18
    Nope. Because the reservoir is at a constant temperature no matter how much heat it discharges, it is effectively an infinite energy supply. So conservation of energy won't work here.

    Imagine an anvil and a slab in some isolated space, both at an equal temperature. The anvil is flying at a constant speed towards the slab. When the anvil hits the slab, some parts of the anvil and the slab will get hotter. But then they will get colder. No contradiction.

    Let's put it this way: can heat be transferred from some object to some other object, if the the former object's temperature does not exceed the latter's, purely by conduction, i.e., without any mechanical work? Which law of thermodynamics is involved here?
     
  20. Jul 2, 2013 #19
    I'm having trouble understanding what you are saying. T (=T2) > T0 does not satisfy the equation for the solution (in either form). As long as t2>t1 (which it must, since t1 is the initial time), T < T0.
     
  21. Jul 2, 2013 #20

    CAF123

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    The zeroth law.
     
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