A Carnot refrigerator extracts heat from water at 0⁰ C and rejects it to room at 24.4⁰ C. The work required by the refrigerator for every 1 kg of water converted into ice (latent heat of ice = 336 kJ kg^-1)  

Given, m = 1 kg,

$T_1=24.4+273=297.4\mathrm{K},T_2=0+273=273\mathrm{K}$

$L=336\mathrm{kJ}{\mathrm{kg}}^{-1}$

Total heat produced per kg of mass,

$\text{ i.e. }{Q}_2=mL=1\times 336\mathrm{kJ}$

We know that,  $\frac{Q_1}{Q_2}=\frac{T_1}{T_2}$

$\Rightarrow Q_1=\left(\frac{T_1}{T_2}\right)Q_2=\frac{297.4}{273}\times 336$

$=\frac{99926.4}{273}\mathrm{kJ}=366\mathrm{kJ}$

Work required by refrigerator,

$W=Q_1-Q_2=366-336=30\mathrm{kJ}$