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Calorimeter

A ball of hea...

Question

A ball of heat capacity $10$ cal $^{o} C^{- 1}$ is heated to a certain temperature and then transferred into a vessel containing water and having a total water equivalent $200$ g. The temperature of water and vessel rises from $10^{o}$ C to $40^{o}$ C. Then find the initial temperature of the ball.

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Solution

Given: A ball of heat capacity $10 c a l^{o} C^{- 1}$ is heated to a certain temperature and then transferred into a vessel containing water and having a total water equivalent 200g. The temperature of water and vessel rises from $10^{o} C$ to $40^{o} C$
To find the initial temperature of the ball.
Thermal capacity of ball = $10 c a l /^{o} C$
Let $T$ be the furnace temperature.
Water equivalent of vessel and contents = $200 g m$
Resultant temperature = $40^{o} C$
According to principle of calorimetry,
Heat lost by hot body = heat gained by cold body
$10 (T - 40^{\circ} C) = 200 \times (40^{\circ} C - 10^{\circ} C) ⟹ T - 40^{\circ} C = \frac{200}{10} \times 30^{\circ} C$
$⟹ T = (600 + 40)^{\circ} C = 640^{o} C$
is the initial temperature of the ball.

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