Mass of gas, m_{g} = 3 kg

Initial pressure of gas = 2.5 bar

Initial temperature, T_{1}′ = 400 K

Quantity of heat received by gas, Q = 600 kJ

Specific heat of gas, c_{v} = 0.81 kJ/kg K

Surrounding temperature = 290 K

Temperature of infinite source, T_{1} = 1200 K

Heat received by the gas is given by,

Q = m_{g}c_{v} (T_{2}′ – T_{1}′)

600 = 3 × 0.81 (T_{2}′ – 400)

T_{2}′ = \(\cfrac{600}{3\times0.81}\) + 400 = 646.9 K say 647 K

Available energy with the source

= area 1-2-3-4-1

= (1200 – 290) × \(\cfrac{600}{1200}\) = 455 kJ

Change in entropy of the gas

Unavailability of the gas = area 3′- 4′- 5′- 6′- 3′

= 290 × 1.168 = 338.72 kJ

Available energy with the gas

= 600 – 338.72 = 261.28 kJ

∴ Loss in available energy due to heat transfer

= 455 – 261.28 = 193.72 kJ.