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गर्म पानी $60^{o} C$ से $50^{\circ} C$ पहले $10$ मिनट में ठंडा होता है और $42^{o} C$ तक दूसरे $10$ मिनट में ठंडा होता है। वातावरण का तापमान $\dots^oC$ है
$25$
$10$
$15$
$20$
Solution
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
$\frac{1}{{0.8}} = \frac{{55 – {\theta _0}}}{{46 – {\theta _0}}}$
$\frac{{10}}{8} = \frac{{55 – {\theta _0}}}{{46 – {\theta _0}}}$
$460 – 10{\theta _0} = 440 – 8{\theta _0}$
$2{\theta _0} = 20$
${\theta _0} = {10^ \circ }c$
