Gujarati
Hindi
10-1.Thermometry, Thermal Expansion and Calorimetry
normal

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

A

$58$

B

$68$

C

$73$

D

$78$

Solution

Heat lost by hot water $=$ (Heat gained by cold

water in beaker $+$ Heat absorbed by beaker)

$\Rightarrow 440(92-\theta)=200 \times(\theta-20)+20 \times(\theta-20)$

$\Rightarrow \theta=68^{\circ} \mathrm{C}$

Standard 11
Physics

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