`0.1 m^(3)` of water at `80^(@)C` is mixed with `0.3m^(3)` of water at `60^(@)C`. The final temperature of the mixture is

Options

A. `65^@ C`
B. `70^@` C
C. `60^@ C`
D. `75^@ C`

Answer

A

Explanation

Heat lost = Heat gain
`rArr m_1 S Delta T_1 = m_2S Delta T_2`
`rArr V_1 rho S Delta T_1 = V_2 rho S Delta T_2`
`rArr 0.1 (80-T) = 0.3(T-60)`
`rArr 8-0.1 T = 0.3 T - 18`
`rArr 26 - 0.4 T rArr T = (26)/(0.4) = 65^(@)C`.