A thermos flask contains 250 g coffee at `90^(@)C`. To this 20 g of milk at `5^(@)C` is added. After equilibrium is established, the temperature of the liquid is
(Assume no heat loss to the thermos bottle . Take specific heat of coffee and milk as `1.00 cal//g^@C`)

Options

A. `3.23^@C`
B. `3.17^@C`
C. `83.7^@C`
D. `37.8^@C`

Answer

C

Explanation

`T_(m) = 5^(@)C,T_(c)=90^(@)C,m_(m)= 20g, m_(c) = 250g` (given)
At equilibrium, heat lost by Hot body (coffee) = Heat gained by cold body (milk)
`Q_(coffee) = Q_(milk)`
`m_(c)s_(c)(T_(c)-T) = m_(m)s_(m)(T-T_(m))`
`implies` `(0.25 kg) (90-T) = (0.02) (T - 5)` `[because S_(c) = S_(m)]`
`therefore` `T = 83.704^(@)C`