With temperature, the rates of substitution reactions and elimination reactions both increase.
For the purpose of answering this question, let us consider $S_N2$ and $E_2$ type reactions.
For $S_N2$ type reactions, the HOMO of the nucleophile has to be oriented with the LUMO of the electrophile. Here, a linear orientation is required. So, the entropic requirement is that the nucleophile should approach the electrophile in a certain way.
For $E2$ type reactions, the base attacks the $\alpha$-proton, which finally leads to the elimination. So, the HOMO of the base interacts with the $\sigma^*$ orbital of the $C-H$ bond, and then the $\sigma$ orbital interacts with the $\sigma^*$ orbital of the leaving group. So here the entropic requirement is the linear orientation of the base, the proton and the carbon it is attached to, and the leaving group should be antiperiplanar with respect to the proton.
With an increase in temperature, the "rate of bond rotation" increases and hence, the probability that a collision with a proper orientation will occur in a given time frame increases. This increase is more prominent in the case of the elimination reaction because it is quite dependent on the orientation of bonds (2 constraints vs. 1 constraint) compared to the substitution reaction.
So, a rise in the temperature will not only affect the thermodynamics of the equation, but it will also directly affect the rate of reaction.