A body cools from `50^(@)C` to `40^(@)C` in 5 min. The surroundings temperature is `20^(@)C`. In what further times (in minutes) will it cool to `30^(@)C` ?

Options

A. 5
B. 15/2
C. 25/3
D. 10

Answer

C

Explanation

Let instantaneous temperature of the cooling body be `theta` and the surrounding temperature be `theta_(0)` Let the body take time `t_(12)` to cool from temperature `theta_(1)totheta_(2)`
By Newton's Law of cooling
`(d theta)/(dt)=-k(theta-theta_(0))`
Where k is a positive constant `rArrint_(theta_(1))^(theta_(2))(d theta)/(theta-theta_(0))=-kint_(0)^(t)dtrArrt_(12)=(1)/(k)log_(e)((theta_(1)-theta_(0))/(theta_(2)-theta_(0)))`
Let time to cool from 50°C to 40°C be `t_(a)` and time to cool from 40°C to 30°C be `t_(b)`
`rArr(t_(b))/(t_(a))=(log_(e)((40-20)/(30-20)))/(log_(e)((50-20)/(40-20)))=(log_(e)2)/(log_(e)3-log_(e)2)`
Given `t_(a)=5m " inutes" rArrt_(b)~~8.5m " inutes"`