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Calculate the unavailable energy in 60 kg of water at 60°C with respect to the surroundings at 6°C, the pressure of water being 1 atmosphere.

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Mass of water, m = 60 kg 

Temperature of water, T1 = 60 + 273 = 333 K 

Temperature of surroundings, T0 = 6 + 273 = 279 K 

Pressure of water, p = 1 atm. 

If the water is cooled at a constant pressure of 1 atm. from 60°C to 6°C the heat given up may be used as a source for a series of Carnot engines each using the surroundings as a sink. It is assumed that the amount of energy received by any engine is small relative to that in the source and temperature of the source does not change while heat is being exchanged with the engine.

Consider that the source has fallen to temperature T, at which level there operates a Carnot engine which takes in heat at this temperature and rejects heat at T0 = 279 K. If δs is the entropy change of water, the work obtained is 

δW = – m(T – T0) δs

where δs is negative

∴ δW = – 60 (T – T0\(\cfrac{c_pδT}T\) = – 60 c\(\left(1-\cfrac{T_0}T\right)\) δT

With a very great number of engines in the series, the total work (maximum) obtainable when the water is cooled from 333 K to 279 K would be 

Wmax = Available energy

∴ Unavailable energy = Q1 – Wmax 

= 13565.9 – 1165.7 = 12400.2 kJ.

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