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

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asked 2020-12-16
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 \(30^{\circ}C?\)

Answers (1)

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 \(100^{\circ}C\ to\ 30^{\circ}C.\)
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