A body cools from `70^(@)C "to"40^(@)C` in 5 minutes . Calculate the time it takes to cool from `50^(@)C " to " 20^(@)C` . The temperature of the surroundings is `10^(@)C`.

Explanation

We are given :
Both cools from `70^(@)C"to"40C` in 5 minutes .
`T_(1)=70^(@)C,T_(2)=40^(@)C,t=5` minutes =300s .
Also temperature of surrounding , `T_(o)=10^(@)C`
According to Newton .s law of cooling.
`(ms(T_(1)-T_(2)))/(t)=k((T_(1)+T_(2))/(2)-T_(0))thereforems((70-40))/(300)=k((70+40)/(2)-10))`
`(msxx30)/(300)=45k`
Also in the second case body coolsfrom `50^(@)C" to " 20^(@)C`
`thereforems((50-20))/(t)=k((50+20)/(2)-10)`
`(30ms)/(t)=25k`
Dividing equation (i) and (ii) we get , `((msxx30)/(300))/((msxx30)/(t))=(45k)/(25k)rArrt=300xx(45)/(25)=540s=9`min.