Stefan Hendricks

2021-12-29

A metal is heated up to a temperature of ${500}^{\circ }C$. It is then exposed to a temperature of ${38}^{\circ }C$. After 2 minutes, the temperature of the metal becomes ${190}^{\circ }C$. When will the temperature be ${100}^{\circ }C$?

puhnut1m

Expert

Initial temp $={500}^{\circ }C$
${T}_{a}=\text{Ambient temp}={38}^{\circ }C$
After 2 min, temp $={190}^{\circ }C$
Calculate: Time when temp reaches to ${100}^{\circ }C$
$\frac{dT}{dt}=-K\left(T-{T}_{a}\right)$

${\int }_{500}^{190}\frac{dT}{T-{T}_{a}}=-K{\int }_{0}^{2}dt$
$t=0,T={500}^{\circ }C$
$t=2,T={190}^{\circ }C$
${\left[\mathrm{ln}\left(T-{T}_{a}\right)\right]}_{500}^{190}=-K\left(2-0\right)$
$-\mathrm{ln}\frac{231}{76}=-2K⇒k=\frac{1}{2}\mathrm{ln}\frac{231}{76}$
$\int \frac{dT}{T-{T}_{a}}=-K\int dt$
$t=0,T={500}^{\circ }C$
$T=t,T={100}^{\circ }C$
${\int }_{500}^{100}\frac{dT}{T-{T}_{a}}=-k{\int }_{0}^{t}dt$
${\left[\mathrm{ln}\left(T-{T}_{a}\right)\right]}_{500}^{100}=-k\left(t-0\right)$
$\mathrm{ln}\left(100-38\right)-\mathrm{ln}\left(500-38\right)=-kt$
$\mathrm{ln}\left(\frac{100-38}{500-38}\right)=-\frac{1}{2}\mathrm{ln}\left(\frac{231}{76}\right)t$

Neunassauk8

Expert

The mean temperature of the metal as it cools from is,
$\theta =\frac{500+190}{2}$
$\theta =\frac{690}{2}={345}^{\circ }C$
The rate of decrease of temperature is,

According to the Newtons

karton

Expert