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. 5 min
B. 9 min
C. 4 min
D. 12 min

Answer

B

Explanation

According to Newton's law of cooling
`((theta_1-theta_2)/(t))=K[((theta_1+theta_2)/(2)-theta_0)]`
So that `((80-50)/(5))=K[((80+50)/(2)-20)]`
And `((60-30)/(t))=K[((60+30)/(2))-20]`
Solving these for t, we get `t=9min`.