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.

Explanation

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)`