0.05 kg of steam at 373 K and 0.45 kg of ice at 253K are mixed in an insulated vessel. Find the equilibrium temperature of the mixture in Kelvin scale.

Given, $L_{\text{fusion}}=80\text{\hspace{0.17em}cal}/g=336J/g,$$L_{\text{vaporization}}=540\text{\hspace{0.17em}cal}/g=2268J/g,$$S_{ice}=2100J/kgK=0.5\text{\hspace{0.17em}cal}/gK$ and  $S_{water}=4200J/kgK=1\text{\hspace{0.17em}cal}/gK$
 

Heat lost by steam at  $100°C$ to change to $100°C$  water  $Q_1=m{L}_{vap}=0.05\times 2268\times 1000=113400J$
Heat lost by water at $100°C$  to change to  $0°C$ water  $Q_2=m_C\Delta T=0.05\times 4200\times 100=21000J$
Heat required by 0.45 kg ice from  $253kto273k$
 $Q_3=m\times C_{ice}\times \Delta T=0.45\times 2100\times 20=18900J$
Similarly,   $Q_4=m{L}_{fusi}=0.45\times 336\times 1000=151200J$
 $Q_1+Q_2>Q_{3}but{Q}_1+Q_2<Q_3+Q_4$
So whole ice will not melt.
 Final temp will be  $273kor0°C$