Two liquids A and B are at `30^@C` and `20^@C`, respectively When they are mixied in equal masses, the temperature of the mixture is found to be `26^@C`. The ratio of their specific heat is

Explanation

Let the specific heat of A and B be `c_(A)` and `c_(B)` respectively. Now heat lost by A in cooling from `30^(@)`C to `26^(@)`C will be same as liquid B warming up from `20^(@)`C to `26^(@)`C. Since, the masses are same then from the principle of calorimeter,
Heat lost by A = Heat gained by B
`therefore` `m.c_(A)(30^(@)C - 26^(@)C) = m.c_(B)(26^(@)C-20^(@)C)`
`4_(c_(A)) = 6_(C_(B)` or `(c_(A))/(c_(B))` = `6/4` = `3/2`