A liquid takes 5 minutes to cool from `80^@C` to` 50^@C` . How much time will it take to cool from `60^@C` to `30^@C`? The temperature of surroundings is `20^@C`.

Options

A. 40 min
B. 9 min
C. 30 min
D. 20 min

Answer

B

Explanation

According to Newton's law of cooling
Case I
`(80^(@)-50^(@))/(5)`
`=K[(80^(@)+50^(@))/(2)-20^(@)]" "...(i)`
Case II
`((60^(@)-30^(@))/(t))`
`=K[(60^(@)+30^(@))/(2)-20^(@)]" "...(ii)`
Dividing Eq. (i) by Eq. (ii), we get
`(t)/(5)=(45)/(25)`
`implies t=9` min