Gujarati
Hindi
10-2.Transmission of Heat
normal

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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