Properties Of Solids And Liquids Question 536
Question: 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
Options:
A) $ 58{}^\circ C $
B) $ 68{}^\circ C $
C) $ 73{}^\circ C $
D) $ 78{}^\circ C $
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Answer:
Correct Answer: B
Solution:
Heat lost by hot water = Heat gained by cold water in + Heat absorbed by beaker
$ \Rightarrow 440( 92-\theta )=200\times ( \theta -20 )+20\times ( \theta -20 ) $
$ \Rightarrow \theta\text{ = 68}{}^\circ \text{C} $
BETA
