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  • A body cools from   to  <img alt="60^{\circ} \mathrm{C}" src

A body cools from 80^{\circ} \mathrm{C}  to  60^{\circ} \mathrm{C} in 5 minutes. The temperature of the surrounding is 20^{\circ} \mathrm{C}. The time it takes to cool from 60^{\circ} \mathrm{C}  to  40^{\circ} \mathrm{C} is:

Option: 1

450 \mathrm{~s}


Option: 2

500 \mathrm{~s}


Option: 3

420 \mathrm{~s}


Option: 4

\frac{25}{3} \mathrm{~s}


Answers (1)

best_answer

 

When body Cools by Radiation from   \ \theta_1^0 C to theta \ \theta_2^0 C in time t                                                                                                                                     

 Then   \left[\frac{\theta_{1}-\theta_{2}}{t} \right ]=k\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0} \right ]                                                                                                                                                                             

 Where \theta_{av}=\frac{\theta_{1}+\theta_{2}}{2}
\begin{aligned} & \frac{20}{5}=\mathrm{k}[70-20] \, \, \, \, \, \, \, \, ....(1)\\ & \frac{20}{\mathrm{t}}=\mathrm{k}[50-20]\, \, \, \, \, \, \, .....(2) \end{aligned}\\

After \, \, solving \, \, \mathrm{t}=\frac{25}{3} \mathrm{~min}\\

In \, \, second \, \, \mathrm{t}=\frac{25}{3} \times 60$ \\$=500 \, \, \mathrm{sec}.

Posted by

Divya Prakash Singh

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