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Question
Explain briefly, how you would determine the specific heat capacity of a liquid?
Solution
In this case, take any solid of known specific heat and a liquid, whose specific heat capacity is to be determined. The solid chosen should be insoluble and non reactive with the liquid. The observations are to be noted as below:
(i) Mass of empty calorimeter with strirrer = m1 gm.
(ii) Mass of calorimeter + Liquid (or unknown sp. heat) = m2 gm.
(iii) Mass of solid = M gm.
(iv) Specific heat of solid (J/ gm/ °C) = 'S1'
(v) Initial temp. of liquid (as well as calorimeter) = t1 °C
(vi) Temperature of hot solid = t2 °C
(vii) Final temperature of the mixture = t°C.
Calculation: Let 's' be the specific heat capacity of the liquid.
(i) Fall in temperature of the hot solid = (t2 - t1)°C
(ii) Rise in temperature of the liquid = (t - t1)°C
(iii) Rise in temperature of calorimeter = (t - t1)°C
(iv) Heat given out by hot solid = M × s1 × (t2 - t)
(v) Heat taken up by liquid = (m2 - m1) × s × (t - t1)
(vi) Heat taken up by calorimeter = m1 × 0.4 × (t - t1)
Hence, heat taken up by liquid + Heat taken up by calorimeter = Heat given out by solid
or (m2 - m1) × s × (t - t1) + m1 × 0.4 × (t - t1) = M × s1 × (t2 - t)
From this heat equation, we get
s = `("M" xx "s"_1 xx ("t"_2 - "t") - "m"_1 xx 0.4 ("t" - "t"_1))/(("m"_2 - "m"_1)("t" - "t"_1))` J/gm/°C
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