Calorimetry Mixture Type question

Question

100 gm of ice at 100°C is mixed with 100gm of water at 100°C then final temperature of the mixture will be

Answer 1

Let the final temperature of the mixture be T.

Then 100 x 80 + 100(T - 0) x \(\frac {1}{2100}\) x 80 + 100(T - 0) x 1/2

(as specific heat of ice is 0.5 cal/g/℃ and specific heat of water is 1 cal/g°C)

= 100 x 1 x (100−T)

Solving, we get T = 13.33℃