Properties Of Solids And Liquids Question 452
Question: Three liquids of equal volumes are thoroughly mixed. If their specific heats are $ {{s} _{1}},s _{2},s _{3} $ and their temperatures $ {\theta _{1}},{\theta _{2}},{\theta _{3}} $ and their masses $ {{m} _{1}},m _{2},m _{3} $ respectively, then the final temperature of the mixture is
Options:
A) $ \frac{s _{1}{\theta _{1}}+s _{2}{\theta _{2}}+s _{3}{\theta _{3}}}{m _{1}s _{1},m _{2}s _{2},m _{3}s _{3}} $
B) $ \frac{m _{1}s _{1}{\theta _{1}}+m _{2}s _{2}{\theta _{2}}+m _{3}s _{3}{\theta _{3}}}{m _{1}s _{1},m _{2}s _{2},m _{3}s _{3}} $
C) $ \frac{m _{1}s _{1}{\theta _{1}}+m _{2}s _{2}{\theta _{2}}+m _{3}s _{3}{\theta _{3}}}{m _{1}{\theta _{1}},m _{2}{\theta _{2}},m _{3}{\theta _{3}}} $
D) $ \frac{m _{1}{\theta _{1}}+m _{2}{\theta _{2}}+m _{3}{\theta _{3}}}{{{s} _{1}}{\theta _{1}},{{s} _{2}}{\theta _{2}},{{s} _{3}}{\theta _{3}}} $
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Answer:
Correct Answer: B
Solution:
[b] Let lost temperature is $ T{}^\circ C $
then $ m _{1}s _{1}T _{1}+m _{2}s _{2}T _{2}+m _{3}s _{3}T _{3} $ = released energy =$ m _{1}s _{1}T _{1}+m _{2}s _{2}T _{2}+m _{3}s _{3}T _{3} $ = energy taken $ \Rightarrow T=\frac{m _{1}s _{1}T _{1}+m _{2}s _{2}T _{2}+m _{3}s _{3}T _{3}}{m _{1}s _{1}+m _{2}s _{2}+m _{3}s _{3}} $
