When 100 g of a liuqid A at `100^(@)C` is added to 50 g of a liquid B at temperature `75^(@)C`, the temperature of the mixture becomes 90 C, if 100g of liquid A at `100^(@)C` is added to 50 g of liquid B at `50^(@)C` will be.

Options

A. `60^(@)C`
B. `70^(@)C`
C. `85^(@)C`
D. `80^(@)C`

Answer

D

Explanation

By the priciple of calorimetry
`100S _(A) (90-100) + 50S _(B) (90-75) =0" "(i)`
and ` 100S_(A) (T-100) + 50 S_(B) (T-50) =0" "(ii)`
where `S _(A) and S_(B)` are specific heats of A and B respectively and T is final temperature in second case. From (i)
`4S_(A)=3S_(B)`
From (ii)
`(S_(A))/(S_(B)) =- (50(t-50))/(100 (T-100))or T = 80^(@)C.`