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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

`(c_1T_1 + c_2 T_2 + c_3T_3)/(m_1c_1 + m_2c_2 + m_3c_3)`

B

`(m_1c_1T_1 + m_2c_2T_2 + m_3c_3T_3)/(m_1c_1 + m_2c_2 + m_3c_3)`

C

`(m_1c_2T_1 + m_2c_2T_2 +m_3c_3T_3)/(m_1T_2 + 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)`

Text Solution

Verified by Experts

The correct Answer is:
B
Heat loss = heat gain
`m_1c_1T_1 + m_2c_2T_2 +m_3c_3T_3 = [m_1c_1 + m_2c_2 + m_3c_3]T`
`T = (m_1c_1T_1 + m_2c_2 + m_3c_3T_3)/(m_1c_1 + m_2c_2 + m_3 c_3)`
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