Cooling via the quantum vacuum? Suppose you had an isolated cloud of gas at a low temperature in a vacuum with no external sources of radiation (e.g. no CMB). The gas would clearly cool via the emission of low-energy photons. But where do these photons come from? The thermal energy of the gas is simply manifest as elastic collisions between atoms where no energy should be lost. Do we have to appeal to the vacuum state of the electromagnetic field?  But aren’t these virtual photons which could not be transformed into thermal radiation?
 A: The particles of the gas undergo a process called radiative collision. For example, two atoms collide, and you end up with the same two atoms plus one or more photons.
A Feynman diagram for this process would have two atoms (complicated bound states of nucleons and electrons) coming in, and the same two atoms plus some photons coming out. It should be easy to see that such Feynman diagrams are possible. (And just as for any other scattering process, the sum of the amplitudes of all possible diagrams is related to the cross section.)
A: You need to be a little clearer about your assumptions. You haven't said what holds the cloud of gas together. Is it contained in a zero temperature transparent box - e.g., a very deep non-rel. QM deep square well? What is the gas made of - atoms? 
If it's atoms confined by a transparent box that only interacts with them by reflecting them, then the cooling would have to proceed by exciting internal electronic states of the atoms in occasional inelastic high-energy pair collisions, followed by energy loss via electromagnetic decay. No QFT required.
A: I have been pondering this question some.  I would answer yes, if I understand the question in one sense, but if I interpret the question another way I am tempted to say no.  The vacuum is a set of quantum occupation sites $\{|0_k\rangle\}$.  A hot body usually emits a blackbody spectrum of particles which fill these slots in a thermal distribution. So one can say the vacuum cools a body by permitting the emission of radiation   This is pretty standard fair.  
The one thing which I am pondering is whether this question is meant to ask whether the vacuum can absorb thermal energy in some way and remain a vacuum.  In the above example one has a vacuum plus a hot body, and the vacuum transforms into a set of boson states plus vacuum plus a cool body.  So in a box you end up with radiation and not a “pure vacuum.”  A vacuum can change its energy vev.  After all that is what all of this business of Higgs symmetry breaking and inflation and …, is all about.  So I might interpret the question as asking whether the thermal energy of a body can be absorbed by the vacuum so it changes its vev.  However, the vacuum is a zero temperature state.  So it is hard to see how one can cool an object with the vacuum without radiation emitted.  This also appears to be a violation of thermodynamics.  
There appears to be a loop hole, but I think it can be closed.  Suppose we have a hot body with a temperature $T$ and we accelerate it to $a$.  On the frame of the accelerated body there is a temperature $T’~=~2\pi a$ from the horizon, called Unruh radiation, which bounds its Rindler wedge.  On the accelerated frame the pure vacuum observed in an inertial frame becomes a vacuum plus a thermal distribution of boson states.  So it is tempting to think one could accelerate a body with $a$ so that for $T’~<~T$ and the radiation from the body crosses the Rindler horizon with a rate of energy flow that is faster than the energy it emits.  This state of affairs approximates the situation for a body accelerated near the event horizon of a black hole.  So for this hot body suspended above a black hole with a temperature $T~>~2\pi a$ the black hole will absorb more energy than it emits, as observed on that frame.
So what is the problem?  For one the limiting case with $a~\rightarrow~0$ recovers the case above.  So the vacuum does not absorb this energy and change the vacuum vev.  The other problem is that an inertial observer does not detect Unruh radiation.  The body will absorb this radiation, but the inertial observer still detect nothing.  The reason is the time dilation slows down or redshift any radiation the body absorbs so as to cancel out any apparent heating up.  A body then will not be observed to actually heat up, or if it is hotter than the Unruh radiation it is not observed to be influenced by any exterior radiation source.
What has bothered me is the near horizon approximation for a particle near a black hole. For the accelerated observer the life time of the exterior world races by in a flash.  For a stellar mass black hole it requires billions of g-forces to remain a few meters from the horizon, and to get within centimeters requires about a billion billion g's of acceleration.  If by some means an observer could do this the outside world would be racing by, say for a small proper time with $t~=~g^{-1} cosh(gs)$.  So the proper time element is $s~\simeq ~ g^{-1}ln(gt)$ for a time unit $t$ outside.  As a result for t the lifetime of the black hole $\sim~ 10^{67}$year, $g$ in units of distance $\sim~ 1cm ~ 10^{-10} sec~\sim~ 10^{-17}$year the proper time the observer on the accelerated frame observes the BH to evaporate is
$$
s~\sim~ 5\times 10^{18} years.
$$
This is much shorter than the BH life time measured by the exterior world. Assume you get that acceleration up to $10^{33}cm^{-1}$ or $10^{43}sec^{-1}$ or $10^{50}year^{-1}$, then you are hovering practically on the stretched horizon.  The BH evaporates in about $10^{-42}$ seconds, or close to the Planck unit of time!  Bang!; which means all that ingoing and outgoing radiation which interacts with the black hole hits you at once is a colossal thunderclap.  The event horizon appears for larger g close in to be more of a singularity, or a surface region of huge energy density that is radiating and absorbing energy at a ferocious rate.
So I think the result hold for a body suspended near a black hole as well.  The above argument means the same time dilation for the absorption and emission of radiation will hole in the BH case, just as with the accelerated frame case.  This question is interesting for this reason, and frankly this may still bother me for a while.  Yet I feel safe in saying that the vacuum will not cool a body by adjusting its vev in some way and without the emission of radiation.
A: A basic part of the theory is to remember the Boltzmann distribution: N = exp (-E/kT) for each energy level E, where N is the occupation number and T is the Temperature (I am omitting subscripts). So for a given T (=/=0) not everything is at the ground state. These nonground state atoms can decay by spontaneous emission as stated in other Answers.
Spontaneous Emission does have a modern (post Einstein) explanation in terms of the Quantum Vacuum however, which might also interest you.
