A calorimeter of heat capacity `100 J//K` is at room temperature of `30^(@)C`. `100g` of water at `40^(@)C` of specific heat `4200 J//kg-K` is poured into the calorimeter. What is the temperature of water is calorimeter?

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Let the temperature of water in calorimeter is t.
Then heat lost by water = heat gained by calorimeter
`(0.1) xx4200xx(40-t) = 100(t-30) implies 420xx40-420t-300 implies t=32.88^(@)C`

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