Answer
a) $45J$
b) $65J$ heat is liberated.
c) Along $ad$ $23J$ of heat is absorbed. Along $db$, $22J$ of heat is absorbed.
Work Step by Step
Let heat ($Q
$) absorbed the system is positive, and liberated out of the system is negative.
Let work ($W$) done by the system is positive, and the work done on the system is negative.
Let $U$ be the change of internal energy.
a) For path $acb$:
$Q=90J \hspace{2mm}$ $W=60J$.
First law of thermodynamics: $Q=U+W$
Thus $U=Q-W=(90-60)J=30J$. $U$ is fixed for initial state $a$ and final state $b$. ( $U_b-U_a=30J$)
Along $adb$:
$W=15J\hspace{3mm}$ $U=30J\hspace{3mm}$ $Q=U+W=(30+15)J=45J$
b) Along the curved path from $b$ to $a$, volume of the system decreases. Thus, work is done on the system.
$W=-35J\hspace{3mm}$
$U=-30J$( internal energy change from $b$ to $a$ is negative of that from $a$ to $b$)
$Q=U+W=(-30-35)J=-65J$
$65J $ of heat is liberated.
c) Process $db$:
$U=U_b-U_d=(U_b-U_a)+(U_a-U_d)=(30-8)J=22J$
($U_a-U_d=-8J$)
Volume does not change. $W=0$
Thus $Q=U+0=U=22J$
$22J$ of heat is absorbed in the process.
Along $adb$, $15J$ of work is done by the system. Along $db$, no work is done.
Thus along $ad$:
$15J$ of work is done by system.
$U$ along $ad$ is $U_d-U_a=8J$
$Q$ along $ad$ is $(8+15)J=23J$
Along $ad$, $23J $ of heat is absorbed.
