# 5g of ice at 0^@C is mixed with 5g of steam at 100^@C. what would be the final temp.?

Mar 16, 2018

Heat energy required for $5 g$ of water at ${0}^{\circ} C$ to get converted to water at ${100}^{\circ} C$ is latent heat required + heat required to change its temperature by ${100}^{\circ} C$=$\left(80 \cdot 5\right) + \left(5 \cdot 1 \cdot 100\right) = 900$ calories.

Now,heat liberated by $5 g$ of steam at ${100}^{\circ} C$ to get converted to water at ${100}^{\circ} C$ is $5 \cdot 537 = 2685$ calories

So,heat energy is enough for $5 g$ of ice to get converted to $5 g$ of water at ${100}^{\circ} C$

So,only $900$ calories of heat energy will be liberated by steam,so amount of steam that will be converted to water at the same temperature is $\frac{900}{537} = 1.66 g$

So,the final temperature of the mixture will be ${100}^{\circ} C$ in which $5 - 1.66 = 3.34 g$ of steam and $5 + 1.66 = 6.66 g$ of water will coexist.

values used during solving this problem are,
Latent heat for melting of ice =$80$calorie ${g}^{-} 1$,latent heat for vaporisation of water=$537$ calorie ${g}^{-} 1$ and specific heat of water =$1$ calorie ${g}^{-} 1$ $\circ {C}^{-} 1$