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
P dilip_k
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 at#0^@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#

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