We Thought Heat Easily Moves From One Place To Another. We Were Wrong

Are you allowed to introduce a heat engine? Then in principle you should be able to run it for a while off the temperature difference between the $200$ K mass and one $400 K$ mass, and dump the resulting work into the second $400 K$ mass, making it hotter than any mass was to begin with.

The efficiency of the heat engine will start out at $50\%$ and decline as the source temperature differential gets used up. And the first two masses will converge at some temperature a bit below $300 K$ . I'm going to guesstimate that you could get $280+280+440$ . If I get time I'll do an update with the exact calculation.

Update: Anonymous says in comments that it's not allowed to introduce a heat engine, but that would make it a rather brain-dead question, so I'll continue under my original interpretation. My first thought of running a maximally efficient heat engine between two of the masses and dumping the work into the third mass shows why it's an interesting problem, but it's not the optimum answer we were asked for, because turning work straight into heat creates entropy unnecessarily. If we design a custom reversible heat engine for this specific application, we should be able to keep entropy from increasing at all. Therefore the question becomes to find the state that has the same overall entropy and energy, but the maximum temperature of one of the masses.

We know that has to be with the other two masses at the same temperature, because otherwise we could run a heat engine per the original plan and do better. The weighted average temperature is $T_{av} = 1000/3=333.3 K$ , so both the initial and final states are one mass at $T_{av}+\Delta T$ and two at $T_{av}-\Delta T/2$ , where $\Delta T = -400/3=-133.3 K$ initially and later some positive value to be determined.

The entropy of any of the masses can be calculated by integrating $dQ/T=CdT/T$ from some reference temperature. Since we have no other information, let's assume the heat capacity $C$ is constant. That means we can't use $T=0$ as a reference because the integral diverges, but we can use any temperature as a reference so it doesn't matter. (For a real object, $C$ would go to zero at $T=0$ per the 3rd Law.) Then for each object $C\ln(T/T_{av})$ .

That means the total entropy is $C\ln(\frac{T_{av}+\Delta T}{T_{av}})+2C\ln(\frac{T_{av}-\Delta T/2}{T_{av}})$ .

For the initial $\Delta T = -400/3$ , this comes to $[/math]-C\ln(125/108)[/math] = -0.146183 C$ . (It has to be negative because the reference state is thermal equilibrium, so it has to be able to increase to get there.)

The other $\Delta T$ giving the same Q turns out to be $100(17-3\sqrt{17})/3=154.356$ . So the final answer is that plus $1000/3$ or $487.689 K$ .

This question is related to very interesting fact. It can only be done with the help of HEAT PUMP and REFRIGERATOR combination. Two of the temperature as 400K and 200K are used for heat pump which developes WORK and this work is used to run a refrigerator which also has a reservoir having 400K temperature which is third temperature. Thus you make a working system and now calculate it as given by JINAY :P