The mass, specific heat capacity and the temperature of a solid are $1000 g$ , $\frac{1}{2}$ $cal {(g . {}^{\circ}C)}^{- 1}$ and $80 ° C$ respectively. The mass of the liquid and the calorimeter are $900 g$ and $200 g$ . Initially, both are at room temperature $20 ° C$ . Both calorimeter and the solid are made of same material. In the steady state, temperature of mixture is $40 ° C$ , then find the specific heat capacity of the unknown liquid.

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$100 cal / g \cdot {}^{\circ}C$

$1 cal / g \cdot {}^{\circ}C$

$0.1 cal / g \cdot {}^{\circ}C$

$10 cal / g \cdot {}^{\circ}C$

answer is C.

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Detailed Solution

$m_{1} = m a s s o f s o l i d = 1000 g$

$S_{1} = s p e c i f i c h e a t o f s o l i d = \frac{1}{2} c a l / g - {}^{\circ}C$

= $S_{2}$ or specific heat of calorimeter

$m_{2}$ = mass of calorimeter = $200 g$

$m_{3}$ = mass of unknown liquid = $900 g$

$S_{3}$ = specific heat of unknown liquid

From law of heat exchange,

Heat given by solid = Heat taken by calorimeter + Heat taken by unknown liquid

$m_{1} S_{1} |Δ θ_{1}| = m_{2} S_{2} |Δ θ_{2}| + m_{3} S_{3} |Δ θ_{3}|$

$1000 \times \frac{1}{2} \times (80 - 40) = 200 \times \frac{1}{2} (40 - 20) + 900 \times S_{3} (40 - 20)$

Solving this equation we get, $S_{3} = 1 cal / g \cdot {}^{\circ}C$

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