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

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