You actually explained why already by yourself and well done for noticing the problem. Dimensional analysis does show that temperature and energy are not the same.
Just to convince ourselves of this, lets look at the ideal gas equation for a single mole, (i.e. number moles n=1 below) of ideal gas:
$$pV=nRT=RT$$
With $nR=8.31\mathrm{m^2kg s^{-2}K^{-1}}$ looking like it has the units we use for energy over those for temperature (i.e. the units are $ \mathrm{\text{ } m^2kg s^{-2}K^{-1}}=JK^{-1}$).
$$\frac{pV}{R}=T$$
What happens to the units now? The units in this expression do not amount tot units of energy. This means $\text{Energy}\neq\text{Temperature}$ and is what we expected.
As others have said, in various ways; temperature is a function of state. The sum of variables the function of state depends on tells us the number of degrees of freedom it has. These are represented by the number of invariables in the function of the state. 28
i.e. general form:
$$g(x_1,x_2,..,x_n)$$
Where subscript n, is a positive integer which gives the degree of freedom. The Empirical temperature is a single valued result found at a particular time (i.e. so we have a result that is independent of time) when we examine our function of state. Look at the function of state variables volume, V and pressure, p which are represented as follows:
$$\Phi(p,V)=T_{empirical}$$
Proof:
Proof for existence of temperature (derivation is from <PERSON> equilibrium thermodynamics and paraphrased a bit here to illustrate the concept) added for anyone who wants to convince themselves of what I just said.
The condition for thermal equilibrium is described by the zeroth law of thermodynamics that is:
If two bodies A and B are in equilibrium with a third, C then they are also in equilibrium with each other.
Consider a simple example where we model a thermodynamic process involving fluid of fixed mass (i.e. independent of mass).
The condition for thermodynamic equilibrium for function, F of two arbitrary states,with body A at state 1 and and body C at state 3. (You may have plotted a p/V indicator diagram so it will help you to think of this now)
$$F_1=(p_1,V_1,p_3,V_3)=0$$
Next the condition for equilibrium for body B at state 2 and C at 3.
$$F_2=(p_2,V_2,p_3,V_3)=0$$
Solving both for $p_3$
$$p_3=f_1(p_1,V_1,V_3)=0$$
$$p_3=f_2(p_2,V_2,V_3)=0$$
and equating the two:
$$p_3=f_1(p_1,V_1,V_3)=f_2(p_2,V_2,V_3)$$
Solve for $p_1$. The reasoning for this change of variables did actually confuse me for a little while, but it is again just a matter of being comfortable manipulating expressions.
$$p_1=g(V_1,p_2,V_2,V_3)$$
$$F_3=(p_1,V_1,p_2,V_2)=0$$
$$p_1=f_3(V_1,p_2,V_2)$$
$$p_3=f_1(p_1,V_1,V_3)=f_2(p_2,V_2,V_3)$$
$$p_3=f_1(p_1,V_1)=f_2(p_2,V_2)$$
$$\Phi_1(p_1,V_1)=\Phi_2(p_2,V_2)$$
Which gives the conditions for thermal equilibrium the existence of temperature as a function of p and V.
Hence the result expressed earlier.
$$\Phi(p,V)=T_{empirical}$$
This should convince you that temperature and energy are not the same thing! 749