Fermi level position at 310 K

In an n-type semiconductor (non-degenerate, all donors ionised)

$$n = N_c\,e^{-(E_c-E_F)/kT} \;\Rightarrow\; E_c - E_F = kT\ln\frac{N_c}{N_d}$$

In the extrinsic range $n = N_d$ is fixed. If the small change of $N_c$ with temperature is neglected, $\ln(N_c/N_d)$ is constant, so $E_c-E_F \propto T$.

Calculation

$$\begin{aligned} \frac{(E_c-E_F)_{310}}{(E_c-E_F)_{300}} &= \frac{310}{300}\\ (E_c-E_F)_{310} &= 0.5\times\frac{310}{300} \\ &= 0.5167\ \text{eV} \end{aligned}$$

Answer: at 310 K the Fermi level lies about 0.517 eV below the conduction band edge, i.e. it moves down by about 0.017 eV, towards the middle of the gap.

More exact check (including $N_c \propto T^{3/2}$)

$$\begin{aligned} (E_c-E_F)_{310} &= \frac{310}{300}(0.5) + k(310)\tfrac{3}{2}\ln\frac{310}{300}\\ &= 0.5167 + 0.0013 = 0.518\ \text{eV} \end{aligned}$$

The difference is only about 1 meV, so the simple answer is acceptable.

Physical meaning

As temperature rises, more electrons can be thermally excited across the gap and the material moves towards intrinsic behaviour, so the Fermi level drops from near $E_c$ towards $E_{Fi}$ (mid-gap).