Answer
$9:59:56:8AM$
Work Step by Step
The standard time period is given as $t_{\circ}=2\pi\sqrt{\frac{L}{g}}$ and the shortened time period is given as $t_s=2\pi \sqrt{\frac{L(1+\alpha \Delta T)}{g}}$
Now $\frac{t_s}{t_{\circ}}=\frac{2\pi\sqrt{\frac{L(1+\alpha \Delta T)}{g}}}{2\pi\sqrt{\frac{L}{g}}}$
$\implies t_s=t_{\circ}\sqrt{1+\alpha \Delta T}$
We plug in the known values to obtain:
$t_s=(1.00s)\sqrt{1+(19\times 10^{-6}K^{-1})(-7.9K)}$
$t_s=0.999925s$
Now the elapsed time of the pendulum can be determined as
$t=t_pt_s$
We plug in the known values to obtain:
$t=(43200s)(0.999925s)$
$\implies t=(43200s)(\frac{1hr}{3600s})$
$\implies t=11 hr 59 min 56.8s=9:59:56:8AM$
