InterviewSolution
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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 isA. `(c_1T_1+c_2T_2+c_3T_3)/(m_1c_1+m_2c_2+m_3c_3)`B. `(m_1c_1 T_1+m_2c_2t_2+m_3c_3T_3)/(m_1c_1+m_2c_2+m_3c_3)`C. `(m_1c_1T_1+m_2c_2T_2+m_3c_3T_3)/(m_1T_1+m_2T_2+m_3T_3)`D. `(m_1T_1+m_2T_2+m_3T_3)/(c_1T_1+c_2T_2+c_3T_3)` |
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Answer» Correct Answer - B Let the final temperature be `T^@C` Total heat supplied by the three liquids in coming down to `0^@C` `=m_1c_1T_1+m_2c_2T_2+m_3c_3T_3` ..(i) Total heat used by three liquids in raising temperature from `0^@C` to `T^@C` `=m_1c_1T+m_2c_2T+m_3c_3T` ..(ii) By equating Eqs. (i) and (ii) we get `(m_1c_1+m_2c_2+m_3c_3)T=m_1c_1T_1+m_2c_2T_2+m_3c_3T_3` `impliesT=(m_1c_1T_1+m_2c_2T_2+m_3c_3T_3)/(m_1c_1+m_2c_2+m_3c_3)` |
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