1.

A body at temperature `40^(@)C` is kept in a surrounding of constant temperature `20^(@)C`. It is observed that its temperature falls to `35^(@)C` in `10` minutes. Find how much more time will it taken for the body to attain a temperature of `30^(@)C`.

Answer» from equation `(14.3)`
`Delta theta_(1) = Delta theta_(i)e^(-kt)`
for the interval in which temperature falls from `40 to 35^(@)C`
`(30 -20) = (40 -20)e^(-k.10)`
`rArr e^(-10k) = (3)/(4)`
`rArr k =("ln"(4)/(3))/(10)`
for the next interval
`(30 - 20) =(35 - 20)e^(-kt)`
`rArr e^(-kt) = (2)/(3) rArr kt = "ln"(3)/(2)`
`rArr (("ln"(4)/(3))t)/(10) ="ln"(3)/(2)`
`rArr t = 10 (("ln"(4)/(3)))/(("ln"(4)/(3)))` minute `- 14.096 min`
Aliter: (by approximate method)
for the interval in whcih temperature falls form `40 to 35^(@)C`
`lt theta gt = (40 +35)/(2) = 37.5^(@)C`
from equation `(14.4) ltlt(d theta)/(dt)gtgt =-k(lt thetagt -theta_(0))`
`rArr ((35^(@)C-40^(@)C))/(10(min)) =- K(37.5^(@)C-20^(@)C)`
`rArr K = (1)/(35) (min^(-1))`
for the interval in which temperature falls from `35^(@) to 30^(@)C`
`lt theta gt = (35 +30)/(2) = 32.5^(@)C`
from equation `(14.4)`
`((30^(@)C-35^(@)C))/(t) =- (32.5^(@)C-20^(@)C)K`
`rArr` required time, `t = (5)/(12.5) xx 35 min = 14 min`


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