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## Q. 4.4

Given that the latent heat of vaporization of water at 100°C is $2257 J ⋅ g^{−1}$, estimate the latent heat at 300°C.

## Verified Solution

Let      $Δ H _{1}$ = latent heat at 100° $C = 2257 J⋅ g^{ −1}$

$Δ H_{2}$ = latent heat at 300° C

$T _{r 1}$ = 373.15 / 647.1 = 0.577

$T_{r 2}$ =573.15 / 647.1 = 0.886

Then by Eq. (4.14),

$\frac{\Delta H_{2} }{\Delta H_{1}} =\left(\frac{1- T_{r2} }{1- T_{r1}} \right) ^{0.38}$    (4.14)

$\Delta H_{2} = (2257)\left(\frac{1-0.886}{1 – 0.577} \right) ^{0.38}= (2257) ( 0.270 )^{0.38} = 1371 J⋅ g^{-1}$

The value given in the steam tables is $1406 J⋅ g^{-1}$