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. 30 s
B. 60 s
C. 90 s
D. 48 s

Answer

D

Explanation

According to Newton's law of cooling ltBrgt `(theta_1-theta_2)/(t)prop[(theta_1+theta_2)/(2)-theta]`
for the first condition
`(80-60)/(60)prop[(80-60)/(2)-30]` ..(i)
and for the second condition
`(60-50)/(t)prop[(60+50)/(2)-30]` ..(ii)
By solving Eqs. (i) and (ii), we get `t=48s`.