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Three liquids with masses m1,m2,m3 are t...

Three liquids with masses `m_1`,`m_2`,`m_3` are throughly mixed. If their specific heats are `c_1`,`c_2`,`c_3` and their temperature `T_1`,`T_2`,`T_3`, respectively, then the temperature of the mixture is

A

`(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

`(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

`(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

`(m_(1)theta_(1)+m_(2)theta_(2)+m_(3)theta_(3))/(S_(1)theta_(1)+S_(2)theta_(2)+S_(3)theta_(3))`

Text Solution

Verified by Experts

The correct Answer is:
B
Let `theta_(1) gt theta_(2) gt theta_(3)` and `theta` = resultant temperature. From principle of colorimetry
`m_(3)S_(3)(theta_(3)-theta) = m_(1)S_(1) (theta -theta_(1))+m_(2)S_(2)(theta-theta_(2))`.
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