ગરમ પાણીનું તાપમાન \(60\,^oC\) થી \(50\,^oC\) થતાં \(10\,minutes\)  અને પછી \(42\,^oC\) થતાં બીજી \(10\,minutes\) લાગે તો વાતાવરણનું તાપમાન......... \(^oC\) હશે?

   By \(Newton's\,law\) of cooling

\(\frac{{{\theta _1} - {\theta _2}}}{t} =  - K\left[ {\frac{{{\theta _1} + {\theta _2}}}{2} - {\theta _0}} \right]\)

where \({\theta _0}\) is the temperature of surrounding. Now, hot water cools from \({60^ \circ }C\,to\,{50^ \circ }C\) in \(10\,minutes,\)

\(\frac{{60 - 50}}{{10}} =  - K\left[ {\frac{{60 + 50}}{2}{\theta _0}} \right]\,\,\,\,\,\,\,\,\,\,...\left( i \right)\)

Agian, it cools from \({50^ \circ }C\,to\,{42^ \circ }C\) in next \(10\,minutes.\)

\(\frac{{50 - 42}}{{10}} =  - K\left[ {\frac{{50 + 42}}{2} - {\theta _0}} \right]\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\)

Dividing equations \((i)\) by \((ii)\) we get

\(\frac{1}{{0.8}} = \frac{{55 - {\theta _0}}}{{46 - {\theta _0}}}\)

\(\frac{{10}}{8} = \frac{{55 - {\theta _0}}}{{46 - {\theta _0}}}\)

\(460 - 10{\theta _0} = 440 - 8{\theta _0}\)

\(2{\theta _0} = 20\)

\({\theta _0} = {10^ \circ }c\)