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Standard X

Physics

Calorimeter

Two identical...

Question

Two identical calorimeters P and Q contain equal quantity of the water at $12^{o}$ C. A $5$ g piece of metal x of specific heat $0.2$ cal $g^{- 1}^{o} C^{- 1}$ is dropped into P and a $5$ g piece of metal y into Q. The equilibrium temperature in P is $22^{o}$ C and in Q $23^{o}$ C. If the initial temperature of both the metals is $40^{o}$ C. Find the specific heat of metal y.

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Solution

Let the mass of water be $m$ and specific heat of metal $y$ be $S$ :
$m \times 1 \times (22 - 12) = 5 \times 0.2 \times (40 - 22) m = \frac{5 \times 0.2 \times 18}{10} m = 1.8 g$
Now, for metal $y$
$1.8 \times 1 \times (23 - 12) = 5 \times S \times (40 - 23) \Rightarrow S = \frac{1.8 \times 11}{5 \times 17} S = 0.233 c a l g^{- 1} C^{- 1}$

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Let the mass of water be m and specific heat of metal y be S:m×1×(22−12)=5×0.2×(40−22)m=5×0.2×1810m=1.8gNow, for metal y1.8×1×(23−12)=5×S×(40−23)⇒S=1.8×115×17S=0.233calg−1C−1