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Standard XII

Physics

Calorimetry

A calorimeter...

Question

A calorimeter of water equivalent $5$ g has water of mass $55$ g upto a certain level. Another identical calorimeter has a liquid of mass $38$ g up to same level. As both of them cool in the same surroundings from $50$ $^{o}$ C to $46$ $^{o}$ C, water takes $80$ s whereas the liquid takes $32$ s to cool. If the specific heat of water is $1$ cal/g- $^{o}$ C , the specific heat of the liquid in cal/g- $^{o}$ C is :

0.8

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0.4

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0.5

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0.2

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Solution

The correct option is C 0.5
Let heat loss rate of calorimeter be $a c a l / s$
Heat lost in 1st case $= 80 a = 60 \times 1 \times (50 - 46) = 240$
=> $a = 3 c a l / s$

Let specific heat of liquid be $C c a l / g - C$

Heat lost in second case $= 32 a = 5 \times 1 \times 4 + 38 \times 4 C$
=> $C = \frac{96 - 20}{38 \times 4} = 0.5$

Specific heat of the liquid $= 0.5 c a l / g -^{\circ} C$

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The correct option is C 0.5Let heat loss rate of calorimeter be a cal/sHeat lost in 1st case =80a=60×1×(50−46)=240=> a=3 cal/sLet specific heat of liquid be Ccal/g−CHeat lost in second case =32a=5×1×4+38×4C=>C=96−2038×4=0.5Specific heat of the liquid =0.5cal/g−∘C