Q.
A calorimeter contains 0.2 kg of water at 30°C. 0.1 kg of water at 60°C is added to it, the mixture is well stirred and the resulting temperature is found to be 35°C. The thermal capacity of the calorimeter is
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a
6300 J/K
b
1260 J/K
c
4200 J/K
d
None of these
answer is B.
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Detailed Solution
Let X be the thermal capacity of calorimeter and specific heat of water = 4200 J/kg-K.Heat lost by 0.1 kg of water = Heat gained by water in calorimeter + Heat gained by calorimeter0.1 × 4200 × (60 - 35) = 0.2 × 4200 × (35 - 30) + X(35 - 30) 10500 = 4200 + 5X ⇒ X = 1260 J/K
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