A liquid at 30^{circ} C is poured very slowly | vedclass

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(C) Let the mass of the calorimeter be $m$,the specific heat of the calorimeter be $x$,the specific heat of the liquid be $s$,and the latent heat of t

Vedclass · Accepted Answer

$270$

Vedclass · Answer

(C) Let the mass of the calorimeter be $m$,the specific heat of the calorimeter be $x$,the specific heat of the liquid be $s$,and the latent heat of the liquid be $L$.When $5 \ gm$ of liquid at $30^{\circ} C$ is added to the calorimeter at $110^{\circ} C$,the calorimeter cools down to $80^{\circ} C$ and the liquid evaporates:$m \cdot x \cdot (110 - 80) = 5 \cdot s \cdot (80 - 30) + 5 \cdot L$$30 \cdot mx = 250s + 5L$ ... $(i)$Now,$80 \ gm$ of liquid at $30^{\circ} C$ is added to the calorimeter at $80^{\circ} C$ and the final equilibrium temperature becomes $50^{\circ} C$:$m \cdot x \cdot (80 - 50) = 80 \cdot s \cdot (50 - 30)$$30 \cdot mx = 1600s$ ... $(ii)$Comparing equations $(i)$ and $(ii)$:$250s + 5L = 1600s$$5L = 1350s$$\frac{L}{s} = \frac{1350}{5} = 270$

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