Question

A calorimeter of negligible heat capacity contains $100 \,g$ 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 $4200 \,J / kg ^{\circ} C$ and $2100 \,J / kg ^{\circ} C$ respectively. The density of the liquid in $kg / m ^{3}$ is ________.

Answer 1

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 $\frac{2}{5}=\frac{m \times 2100}{0.1 \times 4200}$
$\Rightarrow m=\frac{2 \times 420}{5 \times 2100}=0.08\, kg =80\, g$
As volume of liquid is same that of water $100\, cm ^{3}$, then density of liquid is
$P=\frac{m}{V}=\frac{80 \times 10^{-3}}{100 \times 10^{-6}}=800 \,kg / m ^{3}$