Kinetic energy seen from two frames of reference The kinetic energy of two particles (moving in free space) depends on the frame in which you look at the particles. So does total momentum.
If I'm standing in the center of mass of the two particles, they have less energy as when looked at from a frame moving along with one of them.
This means that in a the co-moving frame they can heat up something (say a water reservoir) to a higher temperature than a water reservoir in the COM.
So it depends where I put the reservoir. If I put it in the COM, the total kinetic energy will heat up the water less than when I put it in a frame in which one of the masses is at rest.
Now I can somehow feel what's going on but I don't know how to put it in words. Does anyone have them? If I accelerate both to equal but opposite momenta in the COM frame, I will put less energy in them  then accelerating one to their relative velocity in a frame in which one is at rest. But so what? It's like a reservoir falling on the COM between two free particles in a gravity field. A reservoir falling along on one of the particles will be heat up more than one in the COM. Is there potential energy at work?
 A: The fact that they have different KE in different frames doe not mean that the energy trasferred (by collision for example) will be different. This assumption (your third paragraph) does not follow from the second paragraph. In one frame they may transfer all their energy and in a different frame they may transfer just part of it.
If you do the actual calculation you will see that the energy transfer is the same in all reference frames. This question (in different versios) was answered several times on this site. For your case, in the frame in which the "container" is fixed in the COM, after a perfectly innelastic collision the three objects will be at rest so all the KE is converted in heat. However, if you analyse the situation in a different frame, the container and the objects will be moving after the collision so not all the initial KE is transfered to heat.
A: Consider for the sake of the argument two equal masses $m$  moving with velocities $v$ and $-v$ in the center of mass. The kinetic energy in the center of mass is then
$$T_{COM}=2\cdot\frac{m}{2}v^2 = m\cdot v^2 $$
If the center of mass moves itself in the lab frame with velocity $V$, the total kinetic energy is instead
$$T_L=\frac{m}{2}(V+v)^2 +\frac{m}{2}(V-v)^2 =  m\cdot V^2 + m\cdot v^2 $$
So the kinetic energy in the lab frame is the kinetic energy of all masses in the center of mass frame plus the kinetic energy of the sum of all masses moving with the velocity $V$ of the center of mass in the lab frame.
Generally, the kinetic energy is always a minimum in the COM frame. In every other reference frame the kinetic energy is higher. So the OP is correct to conclude that an inelastic absorption of the kinetic energies of the masses will heat up a water bath less if it is at rest in the center of mass compared to being at rest in any other reference frame.
