Gujarati
10-1.Thermometry, Thermal Expansion and Calorimetry
medium

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

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