A calorimeter of negligible heat capacity contains 100 g water at `40^(@)C`. The capacity contains 100 g water at `40^(@)C`. The water cools to `35^(@)C` is 5 min. If water is now replaced by a liquid of the same volume as that of water at the same initial temperature it cols to `35^(@)C` in 2 min. Given specific heats of water and liquid are
`4200Jk^(-1)""^(@)C^(-1)` and `2100Jkg^(-1)""^(@)C^(-1)`
respectively. Find the density of liquid give the answer in `kgm^(-3)`. [Assume Newton law of cooling is applicable]

Answer

`800`

Explanation

`-(dT)/(dt)=K/(100xxS_(w))(T-T_(0))`
`int_(40)^(35)(-dT)/(T-T_(0))=int_(0)^(5)K/(100xxS_(w))dt`
`int_(40)^(35)(-dT)/((T-T_0))=int_(0)^(2)(dt)/(100xxrho_(l)s_(l))`
`(5K)/(100S_(w))=(2K)/(100xxrho_(l)s_(l))`
`rho_(l)=4/5g//cm^(3)=(4xx10^(-3)kg)/(5xx10^(-6)m^(3))=4/5xx10^(3)`
`=40/5xx10^(2)=800kg//m^(3)`