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A body initially at $80^o C$ cools to $64^o C$ in $5$ minutes and to $52^o C$ in $10 $ minutes. The temperature of the body after $15$ minutes will be ...... $^oC$
$42.7$
$35$
$47$
$40$
Solution
(a) According to Newton law of cooling
$\frac{{{\theta _1} – {\theta _2}}}{t} = K\,\left[ {\frac{{{\theta _1} + {\theta _2}}}{2} – {\theta _0}} \right]$
For first process : $\frac{{(80 – 64)}}{5} = K\left[ {\frac{{80 + 64}}{2} – {\theta _0}} \right]$…$(i)$
For second process : $\frac{{(80 – 52)}}{{10}} = K\left[ {\frac{{80 + 52}}{2} – {\theta _0}} \right]$…$(ii)$
For third process : $\frac{{(80 – \theta )}}{{15}} = K\left[ {\frac{{80 + \theta }}{2} – {\theta _0}} \right]$…$(iii)$
On solving equation $(i)$ and $(ii)$ we get $K = \frac{1}{{15}}$ and ${\theta _0} = 24^\circ C$.
Putting these values in equation $(iii)$ we get $\theta = 42.7^\circ C$
$80\,^oC$ $\xrightarrow{5\,\,\min }64{{\,}^{o}}C$ …… $(1)$
$80\,^oC$ $\xrightarrow{10\,\,\min }52{{\,}^{o}}C$…… $(2)$
$80\,^oC$ $\xrightarrow{15\,\,\min }\,\theta \,=$ ?…… $(3)$
