Heat given to the system during path 1-4-3 is 70 cal.

Internal energy change from state 1 to 3 is 60 cal. Heat given to the system in path 1-
4-3 is 60 cal + 10 cal = 70 cal

$∆\mathrm{U} + ∆\mathrm{W} = \mathrm{Q}$

Work done during path 1-4-3 = 10 cal.
Area of rectangle = 40-10 = 30 cal
Area of half rectangle = 15 cal
Work done 3 – 1 = 10+15 = 25 cal
Now, according to thermodynamics first law,

$\mathrm{Q} = ∆\mathrm{W} + ∆\mathrm{U}$

For path 3 – 1; $\mathit{∆}U\mathit{ }\mathit{=}\mathit{ }\mathit{-}\mathit{60}\mathit{ }cal$

$∆\mathrm{W} = -25 \mathrm{cal}$

Hence, - 85 cal Heat is ejected through path 3 – 1.