Gibbs will get steamed

To solve these kind of questions , it is better to define a energy reference level , from where all energies are to be measured.

For this particular one it will be water at $373K$

Let $C_p$ for water be $C_{p2}$ and steam be $C_{p1}$ The reaction is:

$\ce{H_2O}_{(l)} \to \ce{H_2O}_{(g)}$

We know that:

$\Delta G = \Delta H - T \Delta S$

$\Delta H_{\text{reaction} } = \Delta H_{\text{products}} - \Delta H_{\text{reactants}}$

$\Delta H_{\text{reaction}} = ( H_{v} + \Delta T C_{p1} ) - ( \Delta T C_{p2} )$

$\Delta H_{\text{reaction}} = 39899.86 \text{J/mol}$

Now similarly:

$\Delta S_{\text{reaction} } = \Delta S_{\text{products}} - \Delta S_{\text{reactants}}$

$\Delta S_{\text{reaction} } = ( \dfrac{ H_v }{T} + C_{p1} \ln{\dfrac{T_2}{T_1} } ) - ( C_{p2} \ln{ \dfrac{T_2}{T_1} } )$

$\Delta S_{\text{reaction} } = ( \dfrac{40657}{373} + C_{p1} \ln{ \dfrac{393}{373} } ) - ( C_{p2} \ln{\dfrac{393}{373} } )$

$\Delta S_{\text{reaction} } = 107.023$

Putting values

$\Delta G = -2106.17$