# A mixture containing 22.4 g of ice (at exactly 0.00 ∘C) and 78.7 g of water (at 62.1 ∘C) is placed in an insulated container. Assuming no loss of heat to the surroundings, what is the final temperature of the mixture?

Jul 14, 2018

Taking sp.heat of water to be 1calg^-1""^@C^-1

If water reaches at ${0}^{\circ} C$ final temperature then heat lost by water would be

$= 78.7 \times 1 \times 62.1 = 4887.27 c a l$

The heat required to melt ice at${0}^{\circ} C$ is $22.4 . \times 80 = 1792 c a l$ which is much less than the heat lost.

So final temperature will be higher than ${0}^{\circ} C$. Let it be ${t}^{\circ} C$. So heat gained by ice cold water to reach at this temperature will be
$= 22.4 \times 1 \times t = 22.4 t$ Cal.

Again the heat lost by water at ${62.1}^{\circ} C$ to reach the final temperature will be

$78.7 \times 1 \times \left(62.1 - t\right)$ Cal.

By calorimetric principle

$22.4 t + 1792 = 78.7 \times 1 \times \left(62.1 - t\right)$

$\implies 22.4 t + 78.7 t = 4887.27 - 1792$

$\implies 101.1 t = 4887.27 - 1792$

$\implies t \approx {30.6}^{\circ} C$