A liquid of specific heat 0.5 cal/`g^(0)`c at `60^(0)`C, is mixed with another liquid of specific heat 0.3 cal/`gm^(0)`c at `20^(0)`C. After mixing, the temperature of the mixture becomes `30^(0)`C. In what proportion by weight are the liquids mixed ?

Explanation

Heat lost by the liquid A = `m_(1) S_(1) (t_(1) -t)`
= `m_(1) xx 0.5 xx 30 ` cal
Heat gained by the liquid B = `m_(2)S_(2) (t - t_(2))`
= `m_(2) xx 0.3 xx 10` cal
Heat lost by the liquid A = heat gained by the liquid B
`m_(1) xx 0.5 xx 30 = m_(2) xx 0.3 xx 10` (or)
`(m_(1))/(m_(2)) = (0.3 xx 10)/(0.5 xx 30) = (1)/(5) = 1 :5 `