A clock is calibrated at a temperature of 20^oC. Assume that the pendulum is a thin brass rod of negligible mass with a heavy bob attached to the end $\left({\mathrm{\alpha }}_{\text{brass }}=19\times {10}^{-6}/\mathrm{K}\right)$

$\begin{array}{l}\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}=2\mathrm{\pi }\sqrt{\frac{{\mathrm{l}}_0+{\mathrm{\alpha l}}_0{\mathrm{\Delta \theta }}_0}{\mathrm{g}}}\\ ={\mathrm{T}}_0\left(1+\frac{1}{2}\mathrm{\alpha \Delta \theta }\right)\end{array}$

At 30^oC, fraction loss of time $=\frac{{\mathrm{T}}_{{30}^{\circ }}-{\mathrm{T}}_{{20}^{\circ }}}{{\mathrm{T}}_{{20}^{\circ }}}$

                                               $=5\mathrm{\alpha }=5\times 19\times {10}^{-6}$

Time lost in 24 hours $=86400\times 95\times {10}^{-6}=8.2\mathrm{s}$

On a cold day at 10^oC, fraction gain of time

$=\frac{{\mathrm{T}}_{{10}^{\circ }}-{\mathrm{T}}_{{20}^{\circ }}}{{\mathrm{T}}_{{20}^{\circ }}}=-5\mathrm{\alpha }$

Time gained in 24 hours = 8.2 s