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

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 ${62^ \circ }C\,to\,{32^ \circ }C,$

$t=8.6\,minutes.$