#### Question

A body is heated at 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?

#### Solution

Let θ be the temperature of the body at any time t.

Temperature of air is given to be 10^{o} C.

According to Newton's law of cooling, we have

`(d theta)/dt prop theta-10^@`

`(d theta)/dt =-k( theta-10^@),k>0`

`(d theta)/(theta-10^@)=-kdt`

Integrating both sides, we get

`int(d theta)/(theta-10^@)=-kintdt`

`ln(theta-10^@)=-kt+C`

`theta=10^@+e^(-k(t)+c).........(1)`

When t = 0, θ = 110^{o}

Substituting in the equation, we get

`110^@=10^@+e^(-k(0)+c)`

`e^c=100^@`

Substituting the above in (1), we get

`theta=10^@+100^@e^(-kt).........(2)`

As per the data in the question,

`60^@=10^@+100^@e^(-k(1))`

`50^@=100^@e^(-k(1)).........(3)`

`e^(-k/2)=1/2`

k=ln2

`35^@=10^@+100^@e^(-kt)`

`25^@=100^@e^(-kt)...........(4)`

Dividing (4) by (3), we get

`2=e^(-k(1-t))`

ln2=-k(1-t)

`t-1=ln2/k=ln2/ln2=1`

Hence, additional time required for cooling from 60^{o} to 35^{o} is 1 hour.