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

(2,3)
$\frac{△\mathrm{T}}{\mathrm{T}}=\frac{1}{2}\mathrm{\alpha }\mathrm{t}$
At 30°C, fraction lose of time
$\frac{{\mathrm{T}}_{30}-{\mathrm{T}}_{20}}{{\mathrm{T}}_{20}}=5\mathrm{\alpha }=5\times 19\times {10}^{-4}$
Time lost in 24 hours
$=86400\times 95\times {10}^{-4}=8.25$
On cold day at  10°c
$\frac{{\mathrm{T}}_{10}-{\mathrm{T}}_{20}}{{\mathrm{T}}_{20}}=-5\mathrm{\alpha }$
Time gain in 24 hrs =  $86400\times 95\times {10}^{-4}=8.25$