Two identical calorimeters P and Q contain an equal quantity of the water at $12^{o} C$ . A 5 g piece of metal $x$ of specific heat $0.2 c a l g^{- 1}^{o} C^{- 1}$ is dropped into P and a 5g 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

P
Q
$x = P$
$y = Q$
m
1
12 $^{o} C$
m
1
12 $^{o} C$ 5g
0.2
$22^{o} C$

5g
Q
$23^{o} C$
$\frac{5 \times 0.2 (40 - 22)}{5 \times s (40 - 23)} = \frac{m \times (22 - 12)}{m (23 - 12)}$

$S = \frac{0.2 \times (18) \times (11)}{17 \times (10)} = 0.233 g$

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