$C p . m = 30.5$ $J K^{- 1} m o l^{- 1} + T (0.0102 J K^{- 2} m^{- 1})$ .
Find heat required to raise temperature of 1 mol water vaporate from 100 $^{\circ} C$ to 50 $^{\circ} C$ .
$T_{1}$ $= 100 + 273 = 373 K$
$T_{2}$ $= 500 + 273 = 773 K$

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Solution

Given: $T_{1} = 373 K, T_{2} = 773 K, C_{p} = 30.5 J K^{- 1} + 0.0102 T J K^{- 2}$
Using the relation, $d q = n C_{p} d T$ [ $n = 1$ ]
On integrating both sides,
$Q = \int d q = \int_{T_{1}}^{T_{2}} C_{p} d T = \int_{373 K}^{773 K} (30.5 J K^{- 1} + 0.0102 T J K^{- 2}) d T$
$Q = 30.5 J K^{- 1} [T]_{373 K}^{773 K} + (0.0102 J K^{- 2} {[\frac{T^{2}}{2}]}_{373 K}^{773 K}$
$Q = [(30.5 J K^{- 1} (773 - 373) K] + (0.0102 J K^{- 2}) \frac{1}{2} [(773 K)^{2} - (373 K)^{2}]$
$Q = [(30.5 J K^{- 1} (400 K)] + (\frac{0.0102}{2} J K^{- 2}) (458400 K^{2})$
$Q = 12200 J + 2337.84 J$
$Q = 14.54 K J$

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