\(t = \frac{1}{k}{\log _e}\left( {\frac{{{\theta _2} - {\theta _0}}}{{{\theta _1} - {\theta _0}}}} \right)\)

From question and above equation,

\(5 = \frac{1}{k}{\log _e}\frac{{\left( {40 - 30} \right)}}{{\left( {80 - 130} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\)

\(And,t = \frac{1}{k}{\log _e}\frac{{\left( {32 - 30} \right)}}{{\left( {62 - 30} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)\)

Dividing equation \((2)\) by \((1)\),

\(\frac{t}{5} = \frac{{\frac{1}{k}{{\log }_e}\frac{{\left( {32 - 30} \right)}}{{\left( {62 - 30} \right)}}}}{{\frac{1}{k}{{\log }_e}\frac{{\left( {40 - 30} \right)}}{{\left( {80 - 30} \right)}}}}\)

On solving we get, time taken to cool down from \({62^ \circ }C\,to\,{32^ \circ }C,\)

\(t=8.6\,minutes.\)