$\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} $

- 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 =$38.82^{\circ}c$

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