A liquidtakes 5 minutes to cool from80^(@)Cto50^(@)C . Howmuchtimw will ittake to cool from 60^(@)Cto 30^(@)C? Thetemperatuer of thesurroundingis 20^(@)C .

A liquidtakes 5 minutes to cool from80^(@)Cto50^(@)C . Howmuchtimw wil

1.	A liquidtakes 5 minutes to cool from80^(@)Cto50^(@)C . Howmuchtimw will ittake to cool from 60^(@)Cto 30^(@)C? Thetemperatuer of thesurroundingis 20^(@)C .
Answer» SOLUTION :According to Newton.s law of cooling `(dT)/(dt) = k (T - T_(0))` If the BODYCOOLS by radiationfrom `T_(1)`to `T_(2)""^(@)C`in time T , then `(dT)/(dt) = (T_(1) - T_(2))/(t) and T = T_(ave) = (T_(1) + T_(2))/(2)` `therefore((T_(1) - T_(2))/(t)) = k[((T_(1) + T_(2))/(2))-T_(0)]` Substitutingthe values , we get . `((80 - 50)/(5)) = k[((80 + 50)/(2))-20] to (1) and ((60 -30)/(t)) = k [((60+30)/(2))-20 ] to (2) ` Solvingequations (1) and (2) , we gett = 9 MINUTES .