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`

Explanation

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