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10-1.Thermometry, Thermal Expansion and Calorimetry
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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 $20°C.$ If $440\, gm$ of hot water at $92°C$ is poured in it, the final temperature (neglecting radiation loss) will be nearest to........ $^oC$
A
$58$
B
$68$
C
$73$
D
$78$
Solution
(b) Heat lost by hot water $=$ Heat gained by cold water in beaker $+$ Heat absorbed by beaker
$\Rightarrow$ $440(92 – \theta ) = 200 \times (\theta – 20) + 20 \times (\theta – 20)$
$\Rightarrow$ $\theta$ $= 68^\circ C$
Standard 11
Physics