Question

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

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

Answer 1

The correct option is C

${68}^{\circ }C$

Let the final temperature of the system be $T$

$\text{Heat lost by hot water}=\text{Heat gained by cold water in beaker}+\text{Heat absorbed by beaker}$
or, $(mc\Delta T{)}_{\text{hot water}}=(mc\Delta T{)}_{\text{cold water}}+(mc\Delta T{)}_{\text{beaker}}$
where, $m$ is the mass of body, $c$ is the specific heat of body and $\Delta T$ is the change in temperature of the body.
or, $440(92-T)=200\times (T-20)+20\times (T-20)$
or, $T={68}^{\circ }C$