what will be the time when temperature reach $30◦ C$?

Question

Assume that the rate at which a body cools is proportional to the difference in temperature between the body and its surroundings. A body is heated to $110◦C$ and is placed in air at $10◦C$. After one hour, its temperature is $60◦C.$ At what time will its temperature reach $30◦C$?

My attempt :By the formula $$T(t)-T(s)=(T(0)-T(s))e^{-kt}$$

$\color{orange}{\text{Given condition}}$: It takes $t=60\ minutes $ for temperature fall $\color{red}{110^\circ \ C \to 60^\circ \ C}$

Setting the corresponding values, final temperature $T(t)=60^\circ\ C$ , initial temperature $T(0)=110^\circ\ C$ & surrounding body temperature $T(s)=10^\circ\ C$ we get $$60-10=(110-10)e^{-k(60)}$$

$$e^{-k}=\frac{1}{2}$$ $$-60k=\log(\frac{1}{2})$$ $$k=-\log(\frac{1}{2})=\log 2\tag 1$$

Now, for temperature fall $\color{red}{110^\circ \ C \to 30^\circ \ C}$ , setting $T(t)=30^\circ \ C$, $T(0)=110^\circ\ C$ & $T(s)=10^\circ\ C$ we get $$30-10=(110-10)e^{-kt}$$ $$e^{-kt}=\frac{20}{100}=\frac{1}{5}$$ $$\implies -kt=\ln\left(\frac{1}{5}\right)$$ Setting the value of $k$ from (1), we get time $t$ as follows $$t \log 2=-\frac{1}{5}$$

that is $$t= \frac{\log 5}{\log 2} Hrs $$

Is my answer is correct or not ?

Answer 1

Your result is correct. But I thought it could be helpful to show a short cut.

• The starting model is $T(t) = 10 + 100e^{-kt}$.
• After $1$ hour $T(1)= 60 \Leftrightarrow 50 = 100\cdot \color{blue}{\frac 12}$
• $\Rightarrow t_o = 1$ is the $\color{blue}{halftime}$ of the cooling process $$\Rightarrow T(t) = 10+100\cdot 2^{-t}\Rightarrow 30 = 10+100\cdot 2^{-t}$$ $$\Leftrightarrow \frac 15 = 2^{-t} \Rightarrow t=\log_2 5 $$ ($t$ measured in hours)