$Q=50 \mathrm{cal}$
$\mathrm{W}=20 \mathrm{cal}$
According ot I law of thermodynamics,

$\mathrm{dQ}=\mathrm{dU}+\mathrm{d} \mathrm{W}$
$\text { or } \mathrm{dU}=\mathrm{dQ}-\mathrm{dW}=50-20=30 \mathrm{cal}$
(1) For path iaf
$Q=36\,cal$
${W=?}$
$\mathrm{dU}=30 \mathrm{cal}($ since internal energy depends only on the initial and final positions of the system).
$\therefore W=Q-d U=36-30=6 \mathrm{cal}$
$\begin{aligned} \text { (u) }  \mathrm{W}=-13 \mathrm{cal} \\  \mathrm{d} \mathrm{U}=-30 \mathrm{cal} \\  \mathrm{Q}=? \end{aligned}$
$\therefore \mathrm{Q}=\mathrm{dU}+\mathrm{W}=-43 \mathrm{cal}$
(iii) $\quad \mathrm{E}_{\text {int }, \mathrm{f}}=\mathrm{E}_{\text {int }, i}+\Delta \mathrm{U}=10 \mathrm{cal}+30 \mathrm{cal} .=40$
cal.