d
(d) $\frac{{dQ}}{{dt}} = \frac{{KA}}{l}d\theta $ $ = \frac{{0.01 \times 1}}{{0.05}} \times 30$ $= 6J/sec$
Heat transferred in on day ($86400 sec$)
$\theta = 6 \times 86400 = 518400\,J$
Now $Q = mL$ $⇒$ $m = \frac{Q}{L}$$ = \frac{{518400}}{{334 \times {{10}^3}}}$
$= 1.552 kg = 1552g.$