A liquid cools from `70^(@)C` to `60^(@)C` in `5` minutes. Calculate the time taken by the liquid to cool from `60^(@)C` to `50^(@)C` , If the temperature of the surrounding is constant at `30^(@)C` .

Explanation

The average temperature of the liquid in the first case is
`theta_(1)=(70^(@)C+60^(@)C)/2=65^(@)C` . The average tempareture difference from the surrounding is
`theta_(1)-theta_(0)=65^(@)C-30^(@)C=35^(@)C` . The rate of fall of temperature is
`-(d theta_(1))/(dt)=(70^(@)C-60^(@)C)/(5min)=2^(@)Cmin^(-1)` . From newton's law of cooling,
`2^(@)Cmin^(-1)=bA(35^(@)C)` or, `bA=2/35min` . In the second case, the average temperature of the liquid is
`theta_(2)=(60^(@)C+50^(@)C)/2=55^(@)C` . so that, `theta_(2)-theta_(0)=55^(@)C-30^(@)C=25^(@)C` . If it taken a time t to cool down from `60^(@)C to 50^(@)C, the rate of fall of temperature is
`-(d theta_(2))/(dt)=(60^(@)C-50^(@)C)/(t)=(10^(@)C)/(t)` . From newton's law of cooling and (i),
`(10^(@)C)/(t)=(2)/(35min)xx25^(@)C` . or, `t=7min.