A liquid takes 5 minutes to cool from `80^@C` to` 50^@C` . How much time will it take to cool from `60^@C` to `30^@C`? The temperature of surroundings is `20^@C`.

Explanation

According to Newton.s law of cooling `(dT)/(dt) = k (T - T_(0))`
If the body cools by radiation from `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)]`
Substituting the values , we get .
`((80 - 50)/(5)) = k[((80 + 50)/(2))-20] to (1) and ((60 -30)/(t)) = k [((60+30)/(2))-20 ] to (2) `
Solving equations (1) and (2) , we get t = 9 minutes .