Question

derive an expression for specific heat of a substance using method of mixture .

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Answer 1

Explanation:

Calculations : Let’s’ J/ gm/0C be the specific heat of the solid.

(i) Fall in temperature of hot solid = (t2 – t)0C

(ii) Rise in the temperature of calorimeter as well as water =

(t – t2)0C

(iii) Heat taken up by calorimeter = m1 x 0.4 x (t – t1)0C

(iv) Heat taken up by water = (m2 – m1) x 4.2 x (t – t1)0C

(v) Heat given out by hot solid = M x s x (t2 – t)0C

Now, heat given out by hot solid = Heat taken up by calorimeter + Heat taken up by water.

Or Ms (t2 - t) = m1 x 0.4 (t – t1) + (m2 – m1) x (4.2 x (t – t1)

From this heat equation, ‘s’ can be calculated where all other values are known.