Question

$5$g of steam at ${100}^o$C is mixed with $5$g of the ice at $0^o$C. The final temperature of the mixture is?

• A. ${100}^o$C
• B. ${95}^o$C
• C. ${90}^o$C
• D. ${80}^o$C

Answer 1

The correct option is A ${100}^o$C

Heat energy required for$5gm$ 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 ${100}^{\circ }=(80\cdot 5)+(5\cdot 1\cdot 100)=900$calories.

Now,heat liberated by $5gm$ 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 $5gm$ of ice to get converted to $5gm$ 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 $900537=1.66g$

So,the final temperature of the mixture will be ${100}^{\circ }C$