Question # Solve the differential equation for Newton’s Law of Cooling to find the temperature function in the following case. A pot of boiling soup (100^{circ}C

Decimals
ANSWERED Solve the differential equation for Newton’s Law of Cooling to find the temperature function in the following case.
A pot of boiling soup $$(100^{\circ}C)$$ is put in a cellar with a temperature of
$$10^{\circ}C.$$
After 30 minutes, the soup has cooled to $$80^{\circ}C$$.
When willthe temperature of the soup reach $$30^{\circ}C?$$ 2020-12-17
Data analysis
Given,
Initial temperature of soup, $$T_{0} = 100^{\circ}C$$
Surrounding temperature, $$T_{s} = 10^{\circ}C$$
Final temperature, $$T_{f} = 80^{\circ}C$$
Time taken, $$t = 30\ min$$
To find the time taken to reach to $$30^{\circ}C$$ with the same conditions.
Solution
The Newton's law of cooling formula is,
$$T(t) = T_{s}\ +\ (T_{0}\ -\ T_{s}) e^{-kt}$$
Substituting the values to find ' k ',
$$80 = 10\ +\ (100\ -\ 10)e^{-k(30)}$$
$$\Rightarrow\ 70 = 90e^{-30k}$$
$$\Rightarrow\ e^{30k} = 9/7$$
$$\Rightarrow\ 30k = \ln(9/7)$$
$$\Rightarrow\ k = 0.008377$$
(Rounded to 6 decimals)
Now for final temperature of $$30^{\circ}C,$$ the time is,
$$\Rightarrow\ 30 = 10\ +\ (100\ -\ 10)e^{-(0.008377(t))}$$
$$\Rightarrow\ e^{0.008377(t)} = 9/2$$
$$\Rightarrow\ (0.008377) t = \ln (9/2)$$
$$\Rightarrow t = 179.55$$
(Rounded to two decimals)
Hence it takes 179.55 minutes to cool from $$100^{\circ}C\ to\ 30^{\circ}C.$$