The precise thermodynamic definition of temperature is T = (∂U/∂S)v,n,..., where U and S are internal energy and entropy, respectively. Subscripted variables are all independent variables upon which the internal energy depends, volume, no. of particles, etc. The equations of thermodynamics are differential equations, so it best handles changes in its critical variable values. In that spirit, suppose we create a volume of space devoid of matter/energy such as the CMB, perhaps separated from its environment by an adiabatic barrier that keeps out radiation. This volume contains no energy and no entropy (remember, it's only a hypothetical case, so S=0.). Now, remove all barriers between this 'perfect' vacuum and the rest of the universe, thus adding CMB, dark matter, vacuum energy, whatever to our volume. Clearly, internal energy increases. What about entropy? Does it increase? I would argue that it does, because the distribution of the CMB and any other mass/energy that may be present is not uniformly distributed. I base this on observations of the CMB across large volumes of space, which is not perfectly uniform. Also it would be a very peculiar situation if the volume should have photons of CMB moving with identical directions and exact wavelength; and any other candidate for constituent of this vacuum should be particulate as well. Since the energy is not uniformly distributed, the volume can exist in many states consistent with the energy in the volume we have carved out of space. Pick the usual symbol, W, for the number of ways this system's energy can be distributed.Since S=K Ln(W), the volume now contains entropy. We could also base the existence of nonzero entropy on the third law which specifies this fact, but I like the Boltzmann definition better for this purpose. What we have just described is the transition from a true vacuum containing no energy and no entropy to one that contains both. Therefore, T = (∂U/∂S)v,n,... > 0 and the region of space has nonzero positive temperature.