Hot water cools from ${60^o}C$ to ${50^o}C$ in the first $10$ minutes and to ${42^o}C$ in the next $10$ minutes. The temperature of the surrounding is ......... $^oC$
Diffcult
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(b) According to Newton's law of cooling
$\frac{{{\theta _1} - {\theta _2}}}{t} = K\left[ {\frac{{{\theta _1} + {\theta _2}}}{2} - {\theta _0}} \right]$
In the first case, $\frac{{(60 - 50)}}{{10}} = K\,\left[ {\frac{{60 + 50}}{2} - {\theta _0}} \right]$
$1 = K\,(55 - \theta )$….$(i)$
In the second case, $\frac{{(50 - 42)}}{{10}} = K\,\left[ {\frac{{50 + 42}}{2} - {\theta _0}} \right]$
$0.8 = k\,(46 - {\theta _0})$….$(ii)$
Dividing $(i)$ by $ (ii)$ , we get $\frac{1}{{0.8}} = \frac{{55 - \theta }}{{46 - \theta }}$
or $46 - {\theta _0} = 44 - 0.8\theta $ ==> ${\theta _0} = {10^o}C$
$\frac{{{\theta _1} - {\theta _2}}}{t} = K\left[ {\frac{{{\theta _1} + {\theta _2}}}{2} - {\theta _0}} \right]$
In the first case, $\frac{{(60 - 50)}}{{10}} = K\,\left[ {\frac{{60 + 50}}{2} - {\theta _0}} \right]$
$1 = K\,(55 - \theta )$….$(i)$
In the second case, $\frac{{(50 - 42)}}{{10}} = K\,\left[ {\frac{{50 + 42}}{2} - {\theta _0}} \right]$
$0.8 = k\,(46 - {\theta _0})$….$(ii)$
Dividing $(i)$ by $ (ii)$ , we get $\frac{1}{{0.8}} = \frac{{55 - \theta }}{{46 - \theta }}$
or $46 - {\theta _0} = 44 - 0.8\theta $ ==> ${\theta _0} = {10^o}C$
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