Will ice at $273 \ K$ placed in a surrounding at $298 \ K$ melt to form water at $273 \ K$? Prove this statement. The molar enthalpy of fusion of ice is $6.025 \ kJ \ mol^{-1}$.

(A) The process of melting ice at $273 \ K$ to water at $273 \ K$ is spontaneous because the surrounding temperature $(298 \ K)$ is higher than the melting point of ice $(273 \ K)$.
For the process $H_2O(s) \rightarrow H_2O(l)$ at $273 \ K$:
$\Delta H = 6.025 \ kJ \ mol^{-1} = 6025 \ J \ mol^{-1}$.
$\Delta S_{sys} = \frac{\Delta H}{T} = \frac{6025 \ J \ mol^{-1}}{273 \ K} = 22.07 \ J \ K^{-1} \ mol^{-1}$.
For the surrounding,$\Delta S_{surr} = \frac{-\Delta H}{T_{surr}} = \frac{-6025 \ J \ mol^{-1}}{298 \ K} = -20.22 \ J \ K^{-1} \ mol^{-1}$.
Total entropy change $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} = 22.07 - 20.22 = 1.85 \ J \ K^{-1} \ mol^{-1}$.
Since $\Delta S_{total} > 0$,the process is spontaneous.