A liquid contained in a beaker cools from `80^(@)C` to `60^(@)C` in 10 minutes. If the temperature of the surrounding is `30^(@)C`, find the time it will take to cool further to `48^(@)C` . Here cooling take place ac cording to Newton's law of cooling.

Explanation

The average temperature of `80^(@)C and 60^(@)C` is , `T = (80+60)/(2) = 70^(@)C` ltbr. The average temperature of `60^(@)C` and `48^(@)C` is ltbr. `T = (60+48)/(2) = 54^(@)C`
For first condition, `dT = 80^(@) - 60^(@) = 20^(@)C, dt = 10 mi n, T_(0)=30^(@)C`
`(dT)/(dt) = -K[T-T_(0)]`
`:. 20/(10 mi n) = -K[70-30] = -40 K`...(i)
For second condition, `dt = 60^(@) -48^(@) = 12^(@)C, dt = ? T_(0)=30^(@)C`
`:. (12)/(dt) = -K [ 54-30 ] = -K xx 24` ...(ii)
Dividing (i) by (ii) , we get
`(20//10)/(12//dt) = 40/24 or dt = 40/24 xx (10 xx 12)/(20) = 10 mi n`.