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

Options

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

Answer

D

Explanation

By the principle of calorimetry `100S_(A)(90-100)+50S_(B)theta0-75)=0` …(i)
and `100S_(A)(T-100)+50S_(B)(T-50)=0` ...(ii)
where `S_(A) and S_(B)` are specific heats of A and Brespectively 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))orT=80^(@)C`