A body cools from 80^(@)C to 60^(@)C in 2 minutes. In how much time it cools from 60^(@) to 40^(@)C ? The temperature of the surroundings is 10^(@)C

A body cools from 80^(@)C to 60^(@)C in 2 minutes. In how much time it

1.	A body cools from 80^(@)C to 60^(@)C in 2 minutes. In how much time it cools from 60^(@) to 40^(@)C ? The temperature of the surroundings is 10^(@)C
Answer» SOLUTION :I case : Mean temperature of the body `(80+60)/2=70^(@)C` Mean excess temperature = 70 - 10 = `60^(@)C` `(d THETA)/dt=K(theta-theta_(0))rArr20/2=K(60)to(1)` II case: Mean temperature of the body `(60+40)/2=50^(@)C` Mean excess temperature = (50 - 10) = `40^(@)C` Let .t. minutes be the time to cool down from `60^(@)C` to `40^(@)C` Then `20/t=K(40)to(2)` Dividing equation (1) by (2) `t/2=60/40` i.e., t = 3 minutes