(a) Heat current $H = \frac{KA\mathrm{\Delta}\theta}{l}$
$\therefore$ $\mathrm{\Delta}\theta = \frac{Hl}{KA}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{30 \times 1 \times 10^{- 2}}{0.76 \times 100 \times 10^{- 4}} = 39.47$
$= 40℃(approx.)$