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When 210 g of water at 80^(@)C is kept i...

When 210 g of water at `80^(@)C` is kept in a calorimeter of water equivalent 90 g, its temperatue decreases to `60^(@)C` in 10 min. If the water is replaced by a liquid of mass 100 g, the same fall in temperature takes place in 5 min. If the rate of cooling is the same in both cases, what is the specific heat capacity of the liquid ?

Text Solution

Verified by Experts

Heat lost by the calorimeter and water in 10 min
`=90xx(80-60)+210xx1xx(80-60)="6000 cal"`
`therefore" Rate of cooling "=("6000 cal")/("10 min")="600 cal.min"^(-1)`
Let the specific heat of the liquid be s.
Heat lost by the calorimeter and liquid in 5 min
`=90xx(80-60)+100xxsxx(80-60)`
`=(1800+2000s)"cal"`
Hence, rate of cooling `=((1800+2000s))/(5)" cal.min"^(-1)`
As per given condition,
`((1800+2000s))/(5)=600`
`"or, "1800+2000s=3000`
`"or, "s="0.6 cal.g"^(-1).^(@)C^(-1)`
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