Solution:
Given:
mass of air, m = 0.50 Kg
[tex]T_{1}[/tex] = 25°C = 273+25 = 298 K
[tex]T_{2}[/tex] = 100°C = 273+100 = 373 K
[tex]T_{o}[/tex] = 200°C = 273+100 = 473 K
Solution:
Formulae used:
ΔQ = mCΔT (1)
ΔS = [tex]\frac{\Delta Q}{T_{o}}[/tex] (2)
where,
ΔQ = change in heat transfer
ΔS = chane in entropy
C = specific heat
ΔT = change in system temperature
Using eqn (1)
ΔQ = [tex]0.50\times 1.005\times (373-298)[/tex] = 36.687 kJ
Now, for entropy generation, using eqn (2)
ΔS = [tex]\frac{37.687}{473}[/tex] = 0.0796 kJ
