Idealised hot water bottle

  • Thread starter CAF123
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  • #26
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Maybe, I'll tell you my thoughts: If ##T_2## was greater than ##T_0##, then that would mean heat would flow from the object to the reservoir, thus increasing the temperature of the reservoir. We know the temperature of the reservoir is fixed at ##T_0##, so ##T_2## cannot be greater than ##T_0##. Yes?

If the object somehow gets hotter than the reservoir, then the reservoir by the virtue of its constant temperature will eagerly absorb any heat that is given to it.

You need to analyze how the object, in the given system, could possibly get hotter than the reservoir.
 
  • #27
CAF123
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You need to analyze how the object, in the given system, could possibly get hotter than the reservoir.

Since the system is isolated, any heat or matter external influences would not affect the object. In the given system, therefore, the only way for the object to get hotter is if there was an external agent doing work on the object.
 
  • #28
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Since the system is isolated, any heat or matter external influences would not affect the object. In the given system, therefore, the only way for the object to get hotter is if there was an external agent doing work on the object.

If your system is the just the object itself, it cannot be meaningfully analyzed, as I said earlier.

If your system includes the slab and the reservoir, then your conclusion is manifestly wrong.
 
  • #29
CAF123
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If your system is the just the object itself, it cannot be meaningfully analyzed, as I said earlier.
Just out of interest, why would this be the case?

If your system includes the slab and the reservoir, then your conclusion is manifestly wrong.
My aim is to show that ##T_2## cannot be greater than ##T_0##. Since I know (but have yet to understand) why, the object cannot get hotter than ##T_0## by internal influences. So the only way for it to get hotter is if there was some work done on it. I am taking the system as all three components. Why is it incorrect?
 
  • #32
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Just out of interest, why would this be the case?

A single object isolated from anything else, with no internal structure given - what could one possibly analyze about that?

My aim is to show that ##T_2## cannot be greater than ##T_0##. Since I know (but have yet to understand) why, the object cannot get hotter than ##T_0## by internal influences.

Did you not state how the second law of thermodynamics say in which direction heat can flow?

So the only way for it to get hotter is if there was some work done on it. I am taking the system as all three components. Why is it incorrect?

That seems different from the previous statement. It may also be that I get confused because you sometimes say "hotter", and sometimes "hotter than X".
 
  • #33
CAF123
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A single object isolated from anything else, with no internal structure given - what could one possibly analyze about that?
I asked because Chestermiller used the object as his system in post #16

Did you not state how the second law of thermodynamics say in which direction heat can flow?
Since the initial temperature of the object is colder than the reservoir, heat will flow from the reservoir to the object by the Second Law. Considering the slab as a diathermal wall, heat will be transferred until equilibrium is reached, where the two bodies share the same temperature ##T_0##.

If ##T_2 > T_o##on the way to equilibrium then the object will start to transfer heat to the reservoir. I think I am getting close but still not quite there. Thanks for the help by the way.
 
  • #34
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I asked because Chestermiller used the object as his system in post #16

Since the initial temperature of the object is colder than the reservoir, heat will flow from the reservoir to the object by the Second Law. Considering the slab as a diathermal wall, heat will be transferred until equilibrium is reached, where the two bodies share the same temperature ##T_0##.

Yes. This is a correct articulation of what is happening physically.

If the body were hotter than the reservoir, the flow of heat would be in the opposite direction, and the body would cool.
 
  • #35
CAF123
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If the body were hotter than the reservoir, the flow of heat would be in the opposite direction, and the body would cool.
But why is this not allowed?
 
  • #36
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If ##T_2 > T_o##on the way to equilibrium then the object will start to transfer heat to the reservoir.

How is this possible in the given system with the given initial conditions?
 
  • #37
CAF123
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How is this possible in the given system with the given initial conditions?
Do you mean the IC that the object is initially colder than the reservoir? If the temperature of the reservoir stays the same ##T_0## throughout the process, then by taking the hypothetical case that ##T_2 > T_0##, heat would flow from the object to the reservoir thus increasing its temperature. We know the temperature of the reservoir cannot change thus this is a contradiction. Hence ##T_2 \leq T_0## for all t.
 
  • #38
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Do you mean the IC that the object is initially colder than the reservoir? If the temperature of the reservoir stays the same ##T_0## throughout the process, then by taking the hypothetical case that ##T_2 > T_0##, heat would flow from the object to the reservoir thus increasing its temperature. We know the temperature of the reservoir cannot change thus this is a contradiction. Hence ##T_2 \leq T_0## for all t.

You have already tried the approach, and I have said there is no contradiction.

What really surprises me is that you are perfectly willing to accept that ##T_2 > T_0## somehow. But how can the object's temperature become greater than the reservoir's?

Is temperature something that can just randomly take on any value?
 
  • #39
CAF123
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.
What really surprises me is that you are perfectly willing to accept that ##T_2 > T_0## somehow.
No, I am not accepting that because I want to prove the exact opposite statement. What I was trying to do was answer your question given in #26.

Is temperature something that can just randomly take on any value?

Certainly not, but is the reasoning quantum mechanical?
 
  • #40
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No, I am not accepting that because I want to prove the exact opposite statement. What I was trying to do was answer your question given in #26.

The question was "how it could get hotter". I do not think it can be answered by postulating something and then showing a contradiction. That can never explain "how".

Certainly not, but is the reasoning quantum mechanical?

It does not have to be, no.
 
  • #41
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But why is this not allowed?

This is allowed, provided the initial temperature of the mass is higher than than the (constant) temperature of the reservoir. Under these circumstances, the mass would cool monotonically to the temperature of the reservoir (heat flowing from hot to cold). But, if the initial temperature of the mass is lower than that of the (constant) temperature of the reservoir (as in your problem), the mass would heat up monotonically to the temperature of the reservoir (heat flowing from hot to cold). In both cases, heat is flowing from hot to cold.
 
  • #42
CAF123
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Hi voko,
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?

If the object and reservoir are at the same temperature ##T_0##, then there no longer exists further heat transfer since neither body is hotter than the other.

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?
I am just a little bit confused with the wording here. You say at any given temperature, but if the temperature of the object is less than the temperature of the reservoir then heat exchange via reservoir → object can still take place. For ##T_2 = T_0##, same reasoning as above: neither body is hotter than the other so no heat exchange can take place.

Is it correct?
 
Last edited:
  • #43
CAF123
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Your final solution is correct. Just work it into the same mathematical form that they ask for.

I am having a little difficulty here, because the show that has two instances of ##T_1## where I have two instances of ##T_0##.

From the OP, I have got to $$T_0 \left( 1 - \exp \left( - \frac{KA}{Lmc}(t_2 - t_1) \right) \right) = T_2 - T_1 \exp \left(-\frac{KA}{Lmc}(t_2 - t_1) \right)$$
 
  • #44
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I am just a little bit confused with the wording here. You say at any given temperature, but if the temperature of the object is less than the temperature of the reservoir then heat exchange via reservoir → object can still take place. For ##T_2 = T_0##, same reasoning as above: neither body is hotter than the other so no heat exchange can take place.

Can you see how that prevents the object from ever getting hotter than the reservoir?
 
  • #45
CAF123
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Can you see how that prevents the object from ever getting hotter than the reservoir?
Heat flows from the reservoir to the object until ##T_2 = T_0##. The object may not get hotter than the reservoir since by the zeroth law, the two bodies will remain in an equilibrium at the common temperature ##T_0##. Or maybe more simply if ##T_2 = T_0##, then no heat exchange can take place further, so this must be the equilibrium.
 
  • #46
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Heat flows from the reservoir to the object until ##T_2 = T_0##. The object may not get hotter than the reservoir since by the zeroth law, the two bodies will remain in an equilibrium at the common temperature ##T_0##. Or maybe more simply if ##T_2 = T_0##, then no heat exchange can take place further, so this must be the equilibrium.

Good!
 
  • #47
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I am having a little difficulty here, because the show that has two instances of ##T_1## where I have two instances of ##T_0##.

From the OP, I have got to $$T_0 \left( 1 - \exp \left( - \frac{KA}{Lmc}(t_2 - t_1) \right) \right) = T_2 - T_1 \exp \left(-\frac{KA}{Lmc}(t_2 - t_1) \right)$$

Here is your original solution:

$$\frac{KA}{Lmc}(t_2 - t_1) = \ln\left(\frac{T_o - T_1}{T_o -T_2}\right).$$

This is the same as:
$$ \ln\left(\frac{T_o - T_2}{T_o -T_1}\right)=-\frac{KA}{Lmc}(t_2 - t_1).$$
So, [tex]\left(\frac{T_o - T_2}{T_o -T_1}\right)=\exp\left(-\frac{KA}{Lmc}(t_2 - t_1)\right)[/tex]
Now, [tex]T_o - T_2=(T_o - T_1)+(T_1 - T_2)[/tex]
So,[tex]\left(1+\frac{T_1 - T_2}{T_o -T_1}\right)=\exp\left(-\frac{KA}{Lmc}(t_2 - t_1)\right)[/tex]
I leave the rest of the algebra up to you.
 
  • #48
CAF123
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Thanks voko and Chestermiller. I feel I still do not completely understand why my attempt at describing why ##T_2 \leq T_0## is incorrect using the entropy argument, so I may come back to this thread when I study that in full later on.
 

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