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.

A body at temperature 40^(@)C is kept in a surrounding of constant tem

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» Solution :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`