Question 4.4: Given that the latent heat of vaporization of water at 100°C...

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}