Calculate the entropy change for the melting of $1 \, kg$ of ice at $0 \, ^\circ C$ in $SI$ units. (Latent heat of fusion of ice = $80 \, cal \, g^{-1}$) (in $.67$)

$A$ reversible adiabatic process is one in which:

One mole of an ideal gas is expanded isothermally and reversibly to half of its initial pressure. $\Delta S$ for the process in $J \ K^{-1} \ mol^{-1}$ is $[\ln \ 2 = 0.693$ and $R = 8.314 \ J \ mol^{-1} \ K^{-1}]$

Mixing of non-reacting gases is generally accompanied by

Two blocks of the same metal having same mass and at temperatures $T_1$ and $T_2$ respectively,are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy,$\Delta S,$ for this process is

One mole of ice is converted into water at $273 \, K$. The entropy of $H_2O_{(s)}$ and $H_2O_{(l)}$ are $38.20 \, J \, mol^{-1} \, K^{-1}$ and $60.01 \, J \, mol^{-1} \, K^{-1}$ respectively. The enthalpy change for the conversion is $....... \, J \, mol^{-1}$.