Question

According to Newton’s law of cooling, the rate at which a substance cools in moving
air is proportional to the difference between the temperature of the substance and that
of the air. If the temperature of the air is C
o
290 and the substance cools from
C
o
370 to C
o
330 in 10 minutes, find when the temperature will be C
o
295 .

Answer 1

By the condition $\frac{dT}{dt}\backsim(T-T_a)$

where $\frac{dT}{dt}$ - the rate of cooling; $T$- the temperature of substance; $T_a$ - the temperature of air.

We can rewrite this expression as $\frac{dT}{dt}=-k(T-T_a)$ , where $k$ - coefficient of proportionality. Solve this equation.

$\frac{dT}{dt}=-k(T-T_a)$

$\frac{dT}{T-T_a}=-kdt$

$\int_{T_1}^{T_2}\frac{dT}{T-T_a}=-k\int_{0}^{t} dt$

where $T_1=370$, $T_2=330$, $t=10$, $T_a=290$.

$ln\frac{T_1-T_a}{T_2-T_a}=kt$

$ln\frac{370-290}{330-290}=ln\frac{80}{40}=ln2$

$ln2=10k$

$k=0.07$

Now, find the low of cooling of temperature.

$\int\frac{dT}{T-T_a}=-0.07\int dt$

$ln(T-T_a)=-0.07t$

$e^{ln(T-T_a)}=e^{-0.07t}$

$T-T_a=e^{-0.07t}$

$T=T_a+e^{-0.07t}$

Substitute $T_3=295$.

$295=290+e^{-0.07t}$

$e^{-0.07t}=5$

$-0.07t=ln5$

$t=23$

Answer: t=23 minutes.