- #36
voko
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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?
CAF123 said:If ##T_2 > T_o##on the way to equilibrium then the object will start to transfer heat to the reservoir.
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.voko said:How is this possible in the given system with the given initial conditions?
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.
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.voko said:.
What really surprises me is that you are perfectly willing to accept that ##T_2 > T_0## somehow.
Is temperature something that can just randomly take on any value?
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.
Certainly not, but is the reasoning quantum mechanical?
CAF123 said:But why is this not allowed?
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?
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.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?
Chestermiller said:Your final solution is correct. Just work it into the same mathematical form that they ask for.
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.
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.voko said:Can you see how that prevents the object from ever getting hotter than the reservoir?
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.
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 $$T_0 \left( 1 - \exp \left( - \frac{KA}{Lmc}(t_2 - t_1) \right) \right) = T_2 - T_1 \exp \left(-\frac{KA}{Lmc}(t_2 - t_1) \right)$$