Thermal Conductivity and Heat Transfer in a Hot Water Bottle

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In summary, a heat source at temperature ##T_0## transfers heat through a slab to an initially colder object with a specific heat capacity ##c## and mass ##m##. This is represented by the formula ##KA((T_0 - T)/L)## where ##K## is the thermal conductivity of the slab, ##A## is the area of the slab, and ##T## is the instantaneous temperature of the object. By making certain assumptions, it can be shown that ##KA((T_0 - T)/L)\Delta t = mc \Delta{T}## and for small time intervals, this equation can be rearranged and integrated to give the final temperature of the object as ##T_2 = T_1
  • #36
CAF123 said:
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?
 
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  • #37
voko said:
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
CAF123 said:
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
voko said:
.
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
CAF123 said:
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
CAF123 said:
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
Hi voko,
voko said:
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.

voko said:
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?
 
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  • #43
Chestermiller said:
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
CAF123 said:
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
voko said:
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
CAF123 said:
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!
 
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  • #47
CAF123 said:
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
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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