Question

A pure substance $(C_v=0.733kJ/kg.K)$ undergoes a reversible process in which its temperature increases linearity from ${40}^{o}C$ to ${85}^{o}C$ and its specific entropy increases by 600 J/kg.K. The work done by the system is

• A. 160.2 kJ/kg
• B. 164.3 kJ/kg
• C. 168.3 kJ/kg
• D. 172.3 kJ/kg

Answer 1

The correct option is C 168.3 kJ/kg


$Q=\int Tds$

$=\frac{1}{2}\times \Delta s\times [313+358]$

$=\frac{1}{2}\times 0.6kJ/kg.K\times 671K$

-201.3 kJ/kg

$\Delta U=c_v\times dT$

$=0.733\times (85-40)=32.985kJ/kg$

From first law of thermodynamics for closed system

$Q-W=\Delta U$

$W=Q-\Delta U$

= 201.3 - 32.985 = 168.3 kJ/kg