Question

A 2 kg oxygen execute an internally reversible process whereby heat is received, causing the total entropy to increase by 1.8 kJ/K for an initial temperature of 90oC. Determine the final temperature if the process is constant volume.

Answer 1

Solution;

$\Delta s=mc_pln(\frac{T_2}{T_1})$

For oxygen;

$c_p=918J/kgK$

$T_1=90°c=363K$

$m=2kg$

$1800=2×918×ln(\frac{T_2}{363})$

$0.9803=ln(\frac{T_2}{363})$

$2.6655=\frac{T_2}{363}$

$T_2=967.58K$