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50 g of an alloy containing 80% copper a...

50 g of an alloy containing `80%` copper and `20%` silver is heated up to `t_(1)=80^(@)C` and then dropped in a calorimeter of water equivalent W = 10 g containing m = 90 g of water at `t_(2)=20^(@)C`. What will be the final temerature of the mixture ? Specific heat capacity of copper and silver are `s_(c)=0.09" cal.g"^(-1).^(@)C^(-1) and s_(s)="0.15 cal.g"^(-1).^(@)C^(-1)` respectively.

Text Solution

Verified by Experts

Mass of copper in the alloy, `m_(c)=(50xx80)/(100)=40g`
`therefore" Mass of silver, "m_(s)=50-40=10g`
Let the final temperature of the mixture be t.
Applying the formula, H = mst
Heat lost by copper `=40xx0.09xx(80-t)=(288-3.6t)cal`
Heat lost by silver `=10xx0.05xx(80-t)=(40-0.5t)cal`
Total heat lost by copper and silver `=(328-4.1t)cal`
Heat gained by water and the calorimeter
`=(10+90)xx1xx(t-20)=(100t-2000)cal`
`because" Heat lost = heat gained"`
`therefore" "328-4.1t=100t-2000`
`"or "104.1t=2328`
`"or, "t=(2328)/(104.1)=22.36^(@)C`
`therefore" Final temperature of the mixture is "22.36^(@)C.`
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