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

The correct option is B ${68}^{\circ }C$

Heat lost by hot water = Heat gained by cold water in beaker + Heat absorbed by beaker
$\Rightarrow 440(92-T)=200\times (T-20)+20\times (T-20)\Rightarrow T={68}^{\circ }C$