Question

A calorimeter of negligible heat capacity contains $100g$ water $40^{\circ }C$. The water cools to $35^{\circ }C$ 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 $35^{\circ }C$ in $2$ minutes. Given specific heats of water and that liquid are $4200J/k{g}^{\circ }C$ and $2100J/k{g}^{\circ }C$ respectively. The density of the liquid in $kg/m^3$ is ________.

Answer 1

Answer: 800

Using average form of Newton's law of cooling, we use
For water $\frac{40-35}{5}=\frac{k}{0.1\times 4200}\left(\frac{40+35}{2}-T_s\right)$.....(i)
For liquid $\frac{40-35}{5}=\frac{k}{m\times 2100}\left(\frac{40+35}{2}-T_s\right)$....(ii)
$\frac{\text{ (ii) }}{\text{ (i) }}$ gives $\frac{2}{5}=\frac{m\times 2100}{0.1\times 4200}$
$\Rightarrow m=\frac{2\times 420}{5\times 2100}=0.08kg=80g$
As volume of liquid is same that of water $100c{m}^3$, then density of liquid is
$P=\frac{m}{V}=\frac{80\times 10^{-3}}{100\times 10^{-6}}=800kg/m^3$