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If accidently the calorimeter remained open to atmosphere for some time during the experiment, due to which the steady state temperature comes out to be 30^(@)C, then total heat loss to surrounding during the experiment, is (Use the specific heat capacity of the liquid from previous questions). |
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Answer» `20 kcal` `=(1000)(1//2)(80-30) = 25000 cal` Heat absorbed by the water calomiter SYSTEM `=(900)(1)(30-20)+(200)(1//2)(30-20)` `=10,000 cal` So heat loss to surrounding =`15000 cal`. |
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