A body cools from `80^(@)C` to `50^(@)C` in 5 min-utes Calculate the time it takes to cool from `60^(@)C` to `30^(@)C` The temperature of the surroundings is `20^(@)C` .

Explanation

In first case : Change in temperature, `dT = 80^@ - 50^@ = 30^@C`
Time interval, dt = 5 min. Average temperature, `T = (80^@ + 50^@)/(2) = 65^@C`
Temperature difference ` =T - T_0 = 65^@ - 20^@ = 45^@C`
As ` (dT)/(dt) = K[T - T_0] :. (30)/(5) = - K xx 45 ....(i)`
In second case : Change in temperature, `dT = 60^@ - 30^@ = 30^@C`
Time interval, dt = ?, Average temperature, `T = (60^@ + 30^@)/(2) = 45^@C`
Temperature difference ` = T - T_0 = 45^@ - 20^@ = 25^@C`
As ` (dT)/(dt) = - K (T - T_0) :. (30)/(dt) = - K xx 25` .....(ii)
Dividing (i) by (ii), we get
`(dt)/(5) = (45)/(25) = (9)/(5) or dt = 9 min`