A hot liquid kept in a breaker cools from `80^(@)C` to `70^(@)C` in two minutes. If the surrounding temperature is `30^(@)C`, then the time of cooling of the same liquid from `60^(@)C` to `50^(@)C` is :

Options

A. 240 s
B. 480 s
C. 360 s
D. 216 s.

Answer

D

Explanation

(d) : According to Newton.s law of cooling
`(theta_(1)-theta_(2))/(t)=K[(theta_(1)+theta_(2))/(2)-theta_(0)]`
where `theta_(0)` is temperature of currounding.
`:.(80-70)/(2)=K[(80+70)/(2)-30]` . . . (i)
Also `(60-50)/(t)=K[(60+50)/(2)-30]` . . . (ii)
Dividing (i) by (ii) we have :
`t/2=(75-30)/(55-30)rArr(t)/(2)=(45)/(25)`
`rArrt=(45)/(25)xx2=(90)/(25)=(18)/(5)=3*6` minutes.
`=216` seconds
correct choice is (d).