# 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}$$C in 5 minutes . Find its temperature after a further interval of 5 minutes. \begin{array}{1 1} (A) \; 18.82^{\circ}C \\ (B) \; 23.82^{\circ}C\\ (C) \; 45.82^{\circ}C\\ (D) \; 38.82^{\circ}C \end{array} ## 1 Answer Toolbox: • First order,first degree DE variable separatable : Variable of a DE are rearranged to separate there (i.e.) f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0 • It can be written as \large\frac{g_1(y)}{g_2(y)}$$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 =$38.82^{\circ}c$