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?

1 Answer
Jul 14, 2018

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

If water reaches at 0^@C final temperature then heat lost by water would be

=78.7xx1xx62.1=4887.27cal

The heat required to melt ice at0^@C is 22.4.xx80=1792cal which is much less than the heat lost.

So final temperature will be higher than 0^@C. Let it be t^@C. So heat gained by ice cold water to reach at this temperature will be
=22.4xx1xxt=22.4t Cal.

Again the heat lost by water at 62.1^@C to reach the final temperature will be

78.7xx1xx(62.1-t) Cal.

By calorimetric principle

22.4t+1792=78.7xx1xx(62.1-t)

=>22.4t+78.7t=4887.27-1792

=>101.1t=4887.27-1792

=>t~~30.6^@C