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A calorimeter holds 200 g of water at 10...

A calorimeter holds 200 g of water at `10^(@)C`. When 50 g of water at `100^(@)C` is added to it, the temperature of the mixture becomes `27^(@)C`. A metal ball of mass 100 g at `10^(@)C` is now added to water, and the final temperature reaches `26^(@)C`. Find the specific heat of the metal.

Text Solution

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Let water equivalent of the calorimeter be W and speficic heat of the metal be s.
In the first case,
heat hained by the calorimeter and 200 g of water = heat lost by 50 g of water
`"i.e., "(W+200)xx1xx(27-10)=50xx1xx(100-27)`
`"or, "W+200=(50xx73)/(17)" ...(1)"`
Now the amount of water in the calorimeter becomes `(200+50)=250g.`
Hence, in the second case,
heat lost by the calorimeter and 250 g of water = heat gained by the metal ball
`"i.e., "(W+250)xx1xx(27-26)=100xxsxx(26-10)`
`"or, "W+250=1600s" ...(2)"`
(2) - (1) gives,
`50=1600s-(50xx73)/(17)`
`"or, 1600s"=50+(50xx73)/(17)=50(1+(73)/(17))=(50xx90)/(17)`
`therefore" "s=(50xx90)/(17xx1600)="0.165 cal.g"^(-1).^(@)C^(-1)`
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