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0.1 m^(3) of water at 80^(@)C is mixed ...

`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

A

`65^@ C`

B

`70^@` C

C

`60^@ C`

D

`75^@ C`

Text Solution

Verified by Experts

The correct Answer is:
A
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`.
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