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 the surroundings is 20^(@)C.

A liquid takes 5 minutes to cool from 80^(@)C" to "50^(@)C. How much t

1.	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 the surroundings is 20^(@)C.
Answer» Solution :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