Afaik. temperature is in relation with the kinetic energy of the individual molecules. In vacuum there are only a few molecules so measuring their kinetic energy is very hard, because vacuum has a very little heat capacity and the thermometer with much higher heat capacity will interfere with measurement. I guess the heat transport will be slow too, because heat conduction and convection will be negligable and only heat radiation will transport energy between the wall of the vacuum container and the thermometer. Is there a better way to measure temperature in vacuum?
Thermodynamics – How to Measure the Temperature of a Vacuum
temperaturethermodynamicsvacuum
Related Solutions
Heat is not a property of a system. Heat is a process function. Temperature is a property of a system because is a state function. For instance, the state of a simple gas is given by temperature, pressure, and composition $(T,p,N)$.
Temperature is defined as the inverse ratio of variation of entropy $S$ to changes in internal energy $U$ $$T \equiv \left( \frac{\partial S}{\partial U} \right)^{-1}$$ This is the thermodynamic concept of temperature, which is more general than the kinetic concept that you are considering. Regarding your question, part of the energy given as heat is used to break the bonds and when are broken if you continue supplying energy this will increase the kinetic energy of the molecules.
Moreover, kinetic temperature is not the average of the kinetic energies of all the molecules of the object. This average of kinetic energies is the average kinetic energy. The kinetic temperature is defined as $2/3$ the average internal energy per number density.
At the other hand, heat $Q$ is defined for a given process as the internal energy interchanged which is neither work nor due to flow of matter $$Q \equiv \Delta U - W - U_{matter}$$ Notice that internal energy is a state function and $\Delta U$ denotes the difference between the initial and final energies, but heat is not a state function and this is why we write $Q$ instead of an incorrect $\Delta Q$.
The concept of process function is most easily understood with the example of a lake. A lake has some amount of water, and this can change by evaporation and raining. You can count the amount of water added to the lake by some raining process, but the lake itself does not have any amount of "raining" or evaporation" only some amount of water. Similarly a thermodynamic system has internal energy but has not heat or work.
For example, rotational kinetic energy of gas molecules stores heat energy in a way that increases heat capacity, since this energy does not contribute to temperature.
This description is misguiding in two ways.
First, the statement that
rotational energy does not contribute to temperature
makes an impression that temperature is a quantity that is closely connected with the translational kinetic energy, but not rotational kinetic energy. But that is false; according to classical theory (applicable when temperatures are high) in thermodynamic equilibrium, all quadratic degrees of freedom, translational and rotational, correspond to kinetic energy $k_BT/2$ on average.
It is only true that rotational energy does not contribute to translational kinetic energy of molecules, since the two energies are exclusive contributions to total kinetic energy.
Second, heat capacity when molecules are allowed to rotate is not higher because rotational energy does not contribute to translational kinetic energy of molecules.
It is higher because for the same temperature, such system has higher energy than system without rotation. This is because there are additional degrees of freedom, to which corresponds additional average kinetic energy.
Equilibrium implies temperature implies average energies of molecules. Value of average kinetic energies of molecules neither implies temperature exists nor implies temperature is only connected to translational kinetic energy.
Best Answer
The temperature of a true vacuum would be a measure for the energy distribution of the photon gas in that vacuum. You can derive the occupation of the electromagnetic modes in a volume with Bose-Einstein statistics, which is essentially what Planck did to describe the emission spectrum of a black body.
However, you don't need to do understand the details of this derivation, because the photon gas will be in thermal equilibrium with the vessel walls. So, stick a thermometer with a high-emissivity surface (e.g., glass) in the vacuum vessel and wait until it has equilibrated with the walls by radiative energy exchange.
This will even work if the walls are very far away, or not there at all, like in outer space. The thermomemeter has temperature $T_t$ and will radiate with a flux $\sigma T_t^4$, where $\sigma$ is the Stefan-Boltzmann constant; at the same time, it will receive a flux $\sigma T_e^4$, where $T_e$ is the temperature of the photon gas in vacuum. The temperature of the thermometer will gradually approach the temperature of the photon gas.