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A calorimeter of water equivalent ‘5g’ has water of mass 55 g upto a certain level. Another identical calorimeter has a liquid of mass ‘38g’ upto same level. As both of them cool in the same surroundings from 50°C to 46°C, water takes 80 s where as the liquid takes 32 s to cool. If the specific heat of water is 1 cal/g-⁰C, the specific heat of the liquid in cal/g-⁰C is

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0.8

0.2

0.5

0.4

answer is C.

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Detailed Solution

From Newton’s law of cooling
$\frac{dθ}{dt} \propto \frac{1}{ms} \Rightarrow t \propto ms$
$∴ \frac{W + m_{1} s_{1}}{W + m_{2} s_{2}} = \frac{t_{1}}{t_{2}}; W =$ thermal capacity of calorimeter

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