Initial temperature of fluid, T_{1} = 20 + 273 = 293 K

Final temperature of fluid, T_{2} = 80 + 273 = 353 K

Temperature of the furnace, T_{f} = 1500 + 273 = 1773 K

Atmospheric temperature, T_{0} = 15 + 273 = 288 K

Specific heat of liquid, c_{pl} = 6.35 kJ/kg K

Increase of availability of the liquid

= b_{2} – b_{1} = (h_{2} – h_{1}) – T_{0}(s_{2} – s_{1})

i.e., b_{2} – b_{1} = c_{pl} (T_{2} – T_{1}) – T_{0} × cpl log_{e} \(\cfrac{T_2}{T_1}\)

= 6.3 (353 – 293) – 288 × 6.3 × loge \(\left(\cfrac{353}{293}\right)\) = 39.98 kJ/kg

Now, the heat rejected by the furnace = Heat supplied to the liquid, (h_{2} – h_{1}).

If this quantity of heat were supplied to a heat engine operating on the Carnot cycle its thermal efficiency would be,

Work which could be obtained from a heat engine

= Heat supplied × Thermal efficiency

Possible work of heat engine = (h_{2} – h_{1}) × 0.837

The possible work from a heat engine is a measure of the loss of availability of the furnace.

∴ Loss of availability of surroundings

= (h_{2} – h_{1}) × 0.837 = c_{pl} (T_{2} – T_{1}) × 0.837

= 6.3 ( 353 – 293) × 0.837 = 316.38 kJ/kg

Then, effectiveness of the heating process

∈ = \(\cfrac{Increase \,of \,availability of\, the \,liquid}{ Loss \,of \,availability \,of \,surroundings}\)

= \(\cfrac{39.98}{316.38}\) = 0.1263 or 12.63%.