Question

A ball of thermal capacity $10\text{cal}{/}^{\circ }C$ 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 $200\text{gm}$. The temperature of the vessel and its contents rises from ${10}^{\circ }C$ to ${40}^{\circ }C$. What is the temperature of furnace?

• A. ${640}^{\circ }C$
• B. ${64}^{\circ }C$
• C. ${600}^{\circ }C$
• D. ${100}^{\circ }C$

Answer 1

The correct option is A ${640}^{\circ }C$

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