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A calorimeter of negligible heat capacity contains 100 gm water at 40^(@)C. The water cools to 35^(@)C. in 5 minutes. If the water isnow replaced bya liquid ofsame volume as that of water at same initial temperature, it cools to 35^(@)C in 2 mintues. Given specific heats of water and that liquid are 4200J//Kg""^(@)C. and 2100J//Kg""^(@)C respectively. Findthe density of the liquid. |
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Answer» Solution :Using average form of Newton's law of cooling, we USE For water `(40-35)/(5) =(K)/(0.1 XX 4200)((40+35)/(2)-T_(s))`…(1) For liquid `(40-35)/(2) =(k)/(m xx 2100)((40+35)/(2)-T_(s))` ..(2) `((1))/((2))` gives `(2)/(5) = (mxx 2100)/(0.1 xx 4200)` `rArr m =(2 xx 420)/(5 xx 2100) = 0.08 kg = 80 gm` As the volume of liquid is same that of water `100 cm^(3)`, then density of liquid is `P =(m)/(V) =(80 xx 10^(-3))/(100 xx 10^(-6)) = 800 kg//m^(3)` |
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