Properties Of Solids And Liquids Question 408
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 440 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 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let the final temperature be T.
Then $ 200\times 1\times ( T-20 )+20\times ( T-20 ) $ =440(92 - T) Solving it, we get $ T=68{}^\circ C. $ .
