Question

In a cyclic process, heat transfers are $+14.7kJ$ , $-25.2kJ$, $-3.56kJ$ and $+31.5kJ$. What is the net work done in the net cyclic process?

• A. $+10.56kJ$
• B. $-10.56kJ$
• C. $-17.44kJ$
• D. $+17.44kJ$

Answer 1

The correct option is C $-17.44kJ$

The first law of thermodynamics states that energy cannot be created or destroyed in an isolated system but can be transferred form one form to another.

According to first law of thermodynamics

$\delta Q=\delta W+\Delta U$

$WhereUistheinternalenergy.$

$Forthecyclicprocesses\Delta U=0$

In the cyclic process, $△U_{cyc}=0$
$but,W+q=△U$
Therefore, $W_{cyc}+q_{cyc}=0...\left(1\right)$

$q_{cyc}=[14.7-25.2-3.56+31.5]kJ...\left(2\right)$

Putting (2) in (1) we get,

$W_{cyc}=-17.44kJ$