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Specific Heat Capacity

Two liquids ...

Question

Two liquids $A$ and $B$ are at $32^{\circ} C$ and $24^{\circ} C$ . When mixed in equal masses the temperature of the mixture is found to be $28^{\circ} C$ . Their specific heats are in the ratio of

3 : 2

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2 : 3

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1 : 1

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4 : 3

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Solution

The correct option is D $1 : 1$
Let specific heats of A and B be $C_{a}$ and $C_{b}$ respectively.
Also there masses are m and final temperature $t = 28^{o} C$
Heat released by A=heat absorbed by B
$\Rightarrow m C_{a} (32 - t) = m C_{b} (t - 24)$
$\Rightarrow \frac{C_{a}}{C_{b}} = \frac{4}{4} = 1$
Thus required ratio is $1 : 1$

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