SSC JE Mechanical Engineering 2014 Set 1 Paper

According to Thermodynamics First Law, $\mathrm{\Delta }\mathit{U}=\mathrm{\Delta }\mathit{Q}-\mathrm{\Delta }\mathit{W}$, where $\mathrm{\Delta }\mathit{U}$ is Change in Internal Energy (kJ/kg), $\mathrm{\Delta }\mathit{Q}$ is the amount of heat received (in kJ/kg), and $\mathrm{\Delta }\mathit{W}$ is work done (in kJ/kg). The units will be kJ for all three thermodynamic property is specific values (independent of mass) are not being considered. Further, for Isobaric process (the thermodynamic process in which pressure remains constant), the relation for work done is
$\mathrm{\Delta }{\mathit{W}}_{\mathrm{isobaric}}=\mathit{p}\times \left({\mathit{V}}_2-{\mathit{V}}_1\right).$
Here, the change in internal energy is to be obtained in kJ. So,
$\mathit{Q}=10\mathit{k}\mathit{g}\times 5\mathrm{kJ}∕\mathrm{kg}=50$ kJ ; and $\mathit{W}=\mathit{p}\times ∆\mathit{V}=100\mathrm{kPa}\times 0.2$ m${}_{}^3=20$ kJ Thus,
$\mathit{U}=\mathrm{\Delta }\mathit{Q}-∆\mathit{W}=50-20=30$ kJ