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

1 Answer
Mar 16, 2018

Heat energy required for 5g of water at 0^@C to get converted to water at 100^@C is latent heat required + heat required to change its temperature by 100^@C=(80*5)+(5*1*100)=900 calories.

Now,heat liberated by 5g of steam at 100^@C to get converted to water at 100^@C is 5*537=2685 calories

So,heat energy is enough for 5g of ice to get converted to 5g of water at 100^@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 900/537=1.66g

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

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