What mass of steam at 100^@ C must be mixed with 150 g of ice at its melting point, in a thermally insulated container to produce water of 50^@ C?

1 Answer
Apr 25, 2018

33.45 grams of steam.

Explanation:

Let it be x grams of steam. As latent heat of steam to water is 533 calories per gram, when it converts to water at 50^@C, it looses heat given by

533x+50x=583x

and as latent heat of ice to water is 80 calories per gram, when it converts to water at 50^@C, it gains heat given by

150xx80+150xx50=12000+7500=19500

Therefore, we must have 583x=19500 - assuming that there is no loss of heat to environment and

hence x=19500/583=33.45g