Prove that when 'x' grams of steam at 100°C is mixed with 'y' grams of ice at 0°C and allowed to attain thermal equilibrium, the final temperature of mixture is 

Explanation:

Heat lost by the steam equals heat gained by the ice.

The steam first condenses, then cools.  The amount of heat lost is:

q = x Lvapor + x C (100 − T)

q = 2254x + 4.18x (100 − T)

The ice first melts, then warms.  The amount of heat gained is:

q = y Lfreeze + y C (T − 0)

q = 334y + 4.18y T

Setting the expressions equal:

2254x + 4.18x (100 − T) = 334y + 4.18y T

2254x + 418x − 4.184x T = 334y + 4.18y T

2672x − 334y = 4.18x T + 4.18y T

334 (8x − y) = 4.18 T (x + y)

T = 80 (8x − y) / (x + y)

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