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5 g of ice at 0 c is mixed with 5g of steam at 100 c what is final temp |
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Answer» ergy 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 caloriesSo,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 900537=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.I hope this will HELP youIf not then comment me |
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