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`.

Explanation

According to Newtons law of cooling
`K=C/(ms)`
`((theta_(1)-theta_(2))/t)=K[((theta_(1)+theta_(2))/2)-theta_(0)]`
Substituting the values, er get
`((80-50)/5)=K[((80+50)/2)-20]to(1)`
and `((60-30)/t)=K[((60+30)/2)-20]to(2)`
Solving equations (1) and (2),
we get t = 9 minutes