Equal masses of three liquids A, B and C have temperature `10^(@)C, 25^(@)C` and `40^(@)c` respectively. If A and B are mixed, the mixture has a temperature of `15^(@)C`. If B and C are mixed, the mixture has a temperature of `30^(@)C`, if A and C are mixed will have a temperature of

Options

A. `16^(@)C`
B. `20^(@)C`
C. `25^(@)C`
D. `29^(@)C`

Answer

A

Explanation

When A and B are mixed.
`mc_(A)(15-10) = mc_(B)(25-15)` or `c_(A)/c_(B) = 2`...............(i)
When B and C are mixed
`mc_(B)(30-25) = mc_(cC(40-30)` or `c_(B)/c_(C) = 2`...........(ii)
From Eqs. (i) and (ii), we have
`c_(A)/c_(C) = 4`
Now let `theta` be the equilibrium temperature when A and C are mixed. Then
`mc_(A)(theta-10) = mc_(C)(40-theta)`
or `4(theta-10)=40-theta` (Using `c_(A)/c_(C) = 4`)
or `theta=16^(@)`C