Question

In a heating experiment, it was noted that the temperature of liquid in a beaker rose at $4.0\text{K}$ per minute just before it began to boil, and $40$ minutes later, all the liquid had boiled away.
For this liquid, what is the value of ratio
$\frac{\text{specific heat capacity}}{\text{specific latent heat of vapourisation}}$ (in SI units)?

• A. $\frac{1}{10}$
• B. $\frac{1}{40}$
• C. $\frac{1}{160}$
• D. $\frac{1}{640}$

Answer 1

The correct option is C $\frac{1}{160}$

The rate of heat capacity $P$ remains constant through out the experiment.
If $m$ is the mass of the liquid,
$c$ is its specific heat capacity,
$L$ is its specific latent heat of vapourisation,
then, just before the liquid starts to boil,
$P=\frac{Q}{t}=mc\frac{\Delta \theta }{\Delta t}=mc\times 4$
Where $Q$ is the heat absorbed, $\frac{\Delta \theta }{\Delta t}=4{\text{K min}}^{-1}$
During next $40$ minute,
$P=\frac{Q}{t}=\frac{mL}{40}$
$\therefore 4mc=\frac{mL}{40}$ or $\frac{c}{L}=\frac{1}{4\left(40\right)}=\frac{1}{160}$