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Three bodies $A, B$ and $C$ of masses $m, m$ and $\sqrt 3\,m$ respectively are supplied heat at a constant rate. The change in temperature $\theta $ versus time $t$ graph for $A, B$ and $C$ are shown by $I, II$ and $III$ respectively. If their specific heat capacities are $S_A, S_B$ and $S_C$ respectively then which of the following relation is correct ? (Initial temperature of body is $0\,^oC$ )
$S_A > S_B > S_C$
$S_B = S_C < S_A$
$S_A = S_B = S_C$
$S_B = S_C > S_A$
Solution
$\frac{\mathrm{dQ}}{\mathrm{dt}}=$ constant say $\mathrm{K}$
${\therefore \frac{\mathrm{dQ}}{\mathrm{dt}} \times \mathrm{t}=\mathrm{ms}\left(\theta-\theta_{0}\right) \quad \text { as } \theta_{0}=0}$
${\therefore \mathrm{Kt}=\mathrm{ms} \theta \quad \text { or } \quad \theta=\frac{\mathrm{K}}{\mathrm{ms}} \mathrm{t}}$
$[y=m x+C]$
$\therefore \frac{\mathrm{K}}{\mathrm{ms}}=$ slope $=\tan \alpha$
$\Rightarrow s=\frac{\mathrm{K}}{\mathrm{m} \tan \alpha}$
For $A: S_{A}=\frac{K}{m \tan 60^{\circ}}=\frac{K}{m \sqrt{3}}$
For $\mathrm{B}: \mathrm{S}_{3}=\frac{\mathrm{K}}{\mathrm{m} \tan 45^{\circ}}=\frac{\mathrm{K}}{\mathrm{m}}$
For $\mathrm{C}: \mathrm{S}_{\mathrm{C}}=\frac{\mathrm{K}}{\mathrm{m} \sqrt{3} \tan 30^{\circ}}=\frac{\mathrm{K}}{\mathrm{m} \sqrt{3} \times \frac{1}{\sqrt{3}}}=\frac{\mathrm{K}}{\mathrm{m}}$
$\therefore S_{B}=S_{c}>S_{A}$
