Question

0.50 kg of air is heated at constant pressure from 25°C to 100°C. The source of the heat is at 200°C. What is the entropy generation for the process?

Answer 1

Given:

mass of air, m = 0.50 Kg

`T_(1)` = 25°C = 273+25 = 298 K

`T_(2)` = 100°C = 273+100 = 373 K

`T_(o)` = 200°C = 273+100 = 473 K

Solution:

Formulae used:

ΔQ = mCΔT (1)

ΔS =
`(\Delta Q)/(T_(o))` (2)

where,

ΔQ = change in heat transfer

ΔS = chane in entropy

C = specific heat

ΔT = change in system temperature

Using eqn (1)

ΔQ =
`0.50* 1.005* (373-298)` = 36.687 kJ

Now, for entropy generation, using eqn (2)

ΔS =
`(37.687)/(473)` = 0.0796 kJ