A ball of thermal capacity $10 cal /^{\circ} C$ is heated to the temperature of furnace. It is then transferred into a vessel containing water. The water equivalent of vessel and the contents is $200 gm$ . The temperature of the vessel and its contents rises from $10^{\circ} C$ to $40^{\circ} C$ . What is the temperature of furnace?

640^{\circ} C

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64^{\circ} C

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600^{\circ} C

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100^{\circ} C

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Solution

The correct option is A $640^{\circ} C$
Thermal capacity of ball $= m c = 10 cal /^{\circ} C$
Let T be the furnace temperature
Water Eq. of vessel and contents $= m c = 200 gm$ .
Resultant temperature $= 40^{\circ} C$
According to principle of calorimetry,
Heat lost by hot body = heat gained by cold body
$m c (T - 40) = m c (40 - 10)$
$10 (T - 40) = 200 \times 30$
$∴ T = 640^{\circ} C$
Hence, the correct answer is option (a).

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