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A cup of coffee at temperature $100^{\circ}$$C$ is placed in a room whose temperature is $15^ {\circ}$$C$ and it cools to $60^{\circ}$$C$ in $5$ minutes . Find its temperature after a further interval of $5$ minutes.

$\begin{array}{1 1} (A) \; 18.82^{\circ}C \\ (B) \; 23.82^{\circ}C\\ (C) \; 45.82^{\circ}C\\ (D) \; 38.82^{\circ}C \end{array} $

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Toolbox:
  • First order,first degree DE variable separatable : Variable of a DE are rearranged to separate there (i.e.) $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
  • It can be written as $\large\frac{g_1(y)}{g_2(y)}$$dy=\large\frac{f_1(x)}{f_2(x)}$$dx$
  • The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
Step 1:
The initial temperature of the coffee $T_0=100^{\circ}c$
The temperature of the room =$15^{\circ}c$
Let $T$ be the temperature at time $t$ in minutes.
Step 2:
By Newtons law of cooling,
$\large\frac{dT}{dt}$$\propto(T-15)$
$\therefore\large\frac{dT}{dt}$$=k(T-15)$
Step 3:
$\large\frac{dT}{t-15}$$=kdt$
$\int \large\frac{dT}{T-15}=$$\int kdt+\log c$
$\log T-15=kdt +\log c$
$\log \large\frac{T-15}{c}=$$kt$
$T-15=ce^{kt}$
Step 4:
At $t=0,T_0=100$
$\therefore 100-15=c\Rightarrow c=85$
We have $T-15=85e^{kt}$
Step 5:
Where $t=5minutes,T=60^{\circ}$
$\therefore 60-15=85e^{5k}$
$e^{5k}=\large\frac{45}{85}=\frac{9}{17}$
Step 6:
Temperature after 5min is the temperature at $t=10min$
$\therefore T-15=85e^{10k}$
$\qquad\quad\;\;\;=85(e^{5k})^2$
$e^{5k}=\frac{9}{17}$
$\qquad\quad\;\;\;=85\times \large\frac{81}{289}$
$\qquad\quad\;\;\;=23.82$
$T=38.82^{\circ}c$
The temperature after 10 minutes =$38.82^{\circ}c$
answered Sep 8, 2013 by sreemathi.v
 

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