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10-2.Transmission of Heat
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Liquid cools from $50^oC$ to $45^oC$ in $5$ minutes and from $5^oC$ 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
$\operatorname{ms}\left(\frac{\theta_{1}-\theta_{2}}{t}\right) \propto\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]$
in the first case $\operatorname{ms}\left(\frac{50-40}{5}\right) \propto\left[\frac{50+40}{2}-\theta_{0}\right]$ $\ldots .(i)$
in second case $\left(\frac{45-41.5}{5}\right) \propto\left[\frac{45+41.5}{2}-\theta_{0}\right]$ $…(ii)$
From $(i)$ and $(ii)$
$\theta_{0}=33.3^{\circ} \mathrm{C}$
Standard 11
Physics
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