Question

A beaker contains $200$ gm of water. The heat capacity of the beaker is equal to that of $20$ gm of water. The initial temperature of water in the beaker is ${20}^{\circ }C$. If 440 gm of hot water at ${92}^{\circ }C$ is poured in it, the final temperature, neglecting radiation loss, will be nearest to

• A. ${58}^{\circ }C$
• B. ${68}^{\circ }C$
• C. ${73}^{\circ }C$
• D. ${78}^{\circ }C$

Answer 1

Answer: (B)

${68}^{\circ }C$

Let the final temperature of the system be $T$.

Thus the heat absorbed by the cold water and beaker system=$200\times 1\times (T-20)+20\times 1\times (T-20)$

Heat lost be the hot water=$440\times 1\times (92-T)$

From conservation of heat energy,

$200\times 1\times (T-20)+20\times 1\times (T-20)$ $=440\times 1\times (92-T)$

$⟹T={68}^{\circ }C$