A body is heated to `110^(@)C` and placed in air at ` 10^(@)C`. After 1 hour its temperature is `60^(@)C` . How much additional time is required for it to cool to `35^(@)C ` ?

Explanation

Let `theta` be the temperature of body at time t. Temperature of air is given to be `10^(@)C.` Then, according to Newton's law of cooling, we have
`(d theta)/(dt)prop(theta-10^(@))`
`(d theta)/(dt)=-k(theta-10^(@)),kgt0`
` implies(d theta)/(theta-10^(@))=-k dt`
On integrating both sides, we get
`int (d theta)/(theta-10^(@))=-kintdt`
In `(theta-10^(@))=-kt +c `
`theta=10^(@)+e^(kt+c)" "...(1)`
went `t=0, theta=110^(@)`
Substituting in the equation in we get
`110^(@)=10^(@)+e^(-k(0)+c)`
`e^(c)=100^(@)`
Substituting the above equation, we get (1), we get
`theta=10^(@)+100^(@)e^(-kt)" "...(2)`
Given that `theta=60^(@)C,` when `t=1,` we get
`60^(@)=10^(@)+100^(@)e^(-k)`
`50^(@)=100^(@)e^(-k)`
`e^(-k)=1/2`
`impliese^(k)=2`
`k=log2`
Now, we have to fing t, when `theta=35^(@)`
From equation (2). we have
`theta=10^(@)+100^(@)e^(-kt)`
`35^(@)=10^(@)+100^(@)e^(-kt)`
` 25^(@)=100^(@)e^(-kt)`
`e^(-kt)=1/4impliese^(kt)=4`
`kt=log4`
`t=(log4)/(log2)`
`therefore` Additional time `=(log34)/(log2)-1=1` hour.