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

Options

A. `(C_(1)T_(1)+C_(2)T_(2)+C_(3)T_(3))/(m_(1)C_(1)+m_(2)C_(2)+m_(3)C_(3))`
B. `(m_(1)C_(1)T_(1)+m_(2)C_(2)T_(2)+m_(3)C_(3)T_(3))/(m_(1)C_(1)+m_(2)C_(2)+m_(3)C_(3))`
C. `(m_(1)C_(1)T_(1)+m_(2)C_(2)T_(2)+m_(3)C_(3)T_(3))/(m_(1)T_(1)+m_(2)T_(2)+m_(3)T_(3))`
D. `(m_(1)T_(1)+m_(2)T_(2)+m_(3)T_(3))/(C_(1)T_(1)+C_(2)T_(2)+C_(3)T_(3))`

Answer

B

Explanation

Net heat gain would be zero,
So , `m_(1)c_(1)(T-T_(1))+m_(2)c_(2)(T-T_(2))+m_(3)c_(3)(T-T_(3))=0or, T=(m_(1)c_(1)T_(1)+m_(2)c_(2)T_(2)+m_(3)c_(3)T_(3))/(m_(1)c_(1)+m_(2)c_(2)+m_(3)c_(3))`