In an experiment to determine the specific heat capacity of a solid, following observations were made:
Mass of calorimeter + stirrer= $x k g$
Mass of water= $y k g$
Initial temperature of water= $t_{1}^{o} C$
Mass of solid= $z k g$
Temperature of solid= $t_{2}^{o} C$
Temperature of mixture= $t^{o} C$
Specific heat capacity of calorimeter and water are $c_{1}$ and $c_{2}$ respectively. Express the specific heat capacity c of the solid in terms of the above data

\frac{(x c_{1} + y c_{2}) (t - t_{1})}{z (t_{2} - t)}

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\frac{(x c_{1} + y c_{2}) (t - t_{2})}{z (t_{1} - t)}

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\frac{(x c_{1} + y c_{2}) (t + t_{2})}{z (t_{1} + t)}

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\frac{(x c_{1} + y c_{2}) (t + t_{1})}{z (t_{2} + t)}

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Solution

The correct option is A $\frac{(x c_{1} + y c_{2}) (t - t_{1})}{z (t_{2} - t)}$
Let final temperature of mixture is $t$ .
Let the specific heat of solid is $c_{3}$
Heat lost by (calorimeter + water) $=$ heat gained by solid
$z c_{1} (t_{1} - t) + y c_{2} (t_{1} - t) = z c_{3} (t - t_{2})$
Specific heat of solid, $c_{3} = \frac{(x c_{1} + y c_{2}) (t - t_{1})}{z (t_{2} - t)}$

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Q. A solid of mass 250 g at

90^{0} C

is dropped in a calorimeter containing 100 g of water at

15^{0} C

. The resulting temperature of mixture becomes

40^{0} C

. The mass of calorimeter is 100 g and its specific heat capacity is 0.1 cal/

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. Find the specific heat capacity of the solid.

A mass $m_{1}$ of a substance of specific heat capacity $c_{1}$ at temperature $t_{1}$ is mixed with a mass $m_{2}$ of other substance of specific heat capacity $c_{2}$ at a lower temperature $t_{2}$ .Deduce the expression for the temperature t of the mixture.State the assumption made,if any.