JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Topic Test - Thermometry, Calorimetry & Thermal Expansion

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 40 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) \[78{}^\circ C\]

Correct Answer: B

Solution :

Heat lost by hot water = Heat gained by cold water in + Heat absorbed by beaker

\[\Rightarrow 440\left( 92-\theta \right)=200\times \left( \theta -20 \right)+20\times \left( \theta -20 \right)\]

\[\Rightarrow \text{ }\!\!\theta\!\!\text{ = 68}{}^\circ \text{C}\]