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_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)`

Answer

B

Explanation

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)`