# Naive question about thermodynamics for use in a riddle

I am a mathematician with very little awareness of the physical world, begging for wisdom. I would like to have verification that my idealized physical model makes sense, for use in a riddle.

I believe an idealized model of the physical world has it that when the two objects which are identical except for mass and temperature make contact both objects will settle to an equilibrium temperature which is the mass-weighted average of their two tempratures. Consider two objects $$A$$ and $$B,$$ having masses $$m_A$$ and $$m_B$$ and temperatures $$t_A$$ and $$t_B$$. They are both made of the same material and otherwise identical. Then $$t_{\text{equilibrium}} = \frac{t_A\cdot m_A + t_B\cdot m_B}{m_A + m_B}.$$ This of course assumes that no energy escapes the system I described, which I am happy to assume!

Suppose, however, that if $$A$$ and $$B$$ have equal masses, i.e. $$m_A=m_B$$, and $$t_A>t_B$$, I have a clever way of creating a temperature gradient that transfers more than half of the total system energy $$(t_A\cdot m_A+t_B\cdot m_B)$$ to $$B$$. For concreteness, suppose that in this case I can transfer a fraction $$f=3/5$$ of the total temperature difference to $$B$$, leaving $$A$$ with $$2/5$$ of the total energy.

My question is, how does the fact that I can create this gradient for equal masses generalize to the case where $$m_A\neq m_B$$? I want to know how much energy I can transfer from $$A$$ to $$B$$ given that I have attained a certain "transfer rate" for the equal mass case.

Assuming for simplicity that $$t_B=0$$ and $$t_A=1$$, I think that $$B$$ should get $$\frac{m_A}{m_A + w\cdot m_B},$$ where $$w=\frac{1-f}{f}$$. This would allow that in the case where $$f=1/2$$, when we have are creating no unusual temperature gradient, we recover the weighted average we expect from physics. But I lack confidence with the physics of this and I want to check with physicists whether this makes sense.

Edit: I have edited the question to emphasize that the two objects are identical except for their different masses and initial temperatures.

Edit 2: To clarify a question in the comments, I do not want to allow for phase changes and the like. I am only interested in a very idealized set-up that doesn't contradict fundamental principles of physics.

• The final temperature depends on the specific heats of both materials, not just their masses and initial temperatures. Nov 29, 2018 at 19:49
• Thanks for pointing this out. I am assuming that both objects are otherwise similar except their different temperatures and masses. I will add this in. Nov 29, 2018 at 19:52
• If I understand correctly, you want the masses and temperatures to be in such a way that a given fraction $f$ of the total thermal energy remains with A and the rest of the energy goes to B. Is that right? Nov 29, 2018 at 19:59
• Yes, but the temperature needs to be proportional to the masses. If someone tells me that they can transfer a fixed percentage of the total energy given the equal mass case, I want to infer what they can do in the case of unequal masses. Does this make sense? Nov 29, 2018 at 20:04
• This riddle may offer some context: puzzling.stackexchange.com/questions/56259/… Nov 29, 2018 at 20:05

Suppose, however, that if A and B have equal masses, i.e. $$m_A=m_B$$, and $$t_A>t_B$$, I have a clever way of creating a temperature gradient that transfers more than half of the total system energy $$(t_A \cdot m_A+t_B \cdot m_B)$$ to B. For concreteness, suppose that in this case I can transfer a fraction $$f=3/5$$ of the total temperature difference to B, leaving A with 2/5 of the total energy.
The Second Law of Thermodynamics actually prohibits what you describe. Given that $$A$$ starts out hotter than $$B$$, there is no way of making $$B$$ hotter than $$A$$ without changing the state of a third system*. Doing so would involve taking a system from an equilibrium state ($$A$$ and $$B$$ at the same temperature, which must occur somewhere in the middle of the process you describe) and transforming it into a non-equilibrium state ($$B$$ hotter than $$A$$). If this were possible then a pot of room temperature water could split into ice cubes and boiling water all by itself.
Although you could have a third system which changes its state to facilitate the the process you describe, this essentially ruins the riddle because the process is no longer simple to describe. You can't say that $$A$$ and $$B$$ don't interact with anything else - that offends the physicists - but you can't say that they interact with a third system in such a way that everything works out - that confuses the non-physicists (and perhaps also the physicists). I would recommend that you not create a riddle based on this idea.
*Technically you could scratch the labels off of the systems and rename $$A$$ as $$B$$ and $$B$$ as $$A$$, thereby making $$B$$ hotter than $$A$$, but if that is the solution then this becomes more of a logic riddle/trick question than a math riddle.