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10-2.Transmission of Heat
easy
Liquid cools from $50^oC$ to $45^oC$ in $5\ minutes$ and from $45^oC$ to $41.5^oC$ in the next $5\ minutes$.The temperature of the surrounding is ......... $^oC$
A
$27$
B
$40.3$
C
$23.3$
D
$33.3$
Solution
From Newton's law of cooling
$\operatorname{ms}\left(\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}\right) \propto\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]$
in the first case $\mathrm{ms}\left(\frac{50-40}{5}\right) \propto\left[\frac{50+40}{2}-\theta_{0}\right]$ $\ldots .(\mathrm{i})$
in second case $\left(\frac{45-41.5}{5}\right) \propto\left[\frac{45+41.5}{2}-\theta_{0}\right]$ $. . .(\text { ii })$
From $(i)$ and $(ii)$
$\theta_{0}=33.3^{\circ} \mathrm{C}$
Standard 11
Physics
