A calorimeter of negligible heat capacity contains 100 g water 40oC. The water cools to 35oC in 5 minutes. If the water is now replaced by a liquid of same volume as that of water at same initial temperature, it cools to 35oC in 2 minutes. Given specific heats of water and that liquid are 4200 J/kgoC and 2100 J/kgoC respectively. The density of the liquid in kg/m3 is _________
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answer is 800.
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Detailed Solution
Using average form of Newton's law of cooling, we use For water 40−355=k0.1×420040+352−Ts ……….(i) For liquid 40−355=km×210040+352−Ts ………(ii) (ii) gives 25=m×21000.1×4200⇒m=2×4205×2100=0.08kg=80gAs volume of liquid is same that of water 100 cm3, then density of liquid is P=mV=80×10−3100×10−6=800kg/m3