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In similar calorimeters, equal volumes of water and alcohol, when poured, take $100 s$ and $74 s,$ respectively, to cool from $50^{°} C$ to $40^{°} C.$ If the thermal capacity of each calorimeter is numerically equal to volume of either liquid, then calculate the specific heat capacity of alcohol (given: the relative density of alcohol as 0.8 and specific heat capacity of water as $1 cal / g /^{°} C$ ).

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$0.8 cal / g^{°} C$

$0.6 cal / g^{°} C$

$0.9 cal / g^{°} C$

$1 cal / g^{°} C$

answer is B.

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Detailed Solution

Let V be volume of either liquid

Mass of water $= V \times 1 g$

Mass of alcohol $= V \times 0.8 = 0.8 V g$

Rate of cooling of the water calorimeter $= \frac{1}{100} [V \times (50^{°} - 40^{°}) + V \times 1 \times (50^{°} - 40^{°})] = (1 / 5) Vcal / s$

Rate of cooling of alcohol calorimeter $= \frac{1}{74} [V \times (50^{°} - 40^{°}) + 0.5 V \times s (50^{°} - 40^{°})] = (1 / 74) (10 V + 8 V s) cal / s$

As, rate of cooling of both is same $5 V = (1 / 74) (10 V + 8 V s)$

$s = 0.6 cal / g^{°} C$

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