A hot liquid kept in a beaker cools from 80,^oC to 70,^oC in 2, min . If the surrounding

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10-2.Transmission of Heat

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A hot liquid kept in a beaker cools from $80\,^oC$ to $70\,^oC$ in $2\, min$. If the surrounding temperature is $30\,^oC$, then the time of cooling of the same liquid from $60\,^oC$ to $50\,^oC$ ........ $\sec$

A

$240$

B

$360$

C

$480$

D

$216$

Solution

$\frac{\theta_{1}-\theta_{2}}{t}=-K\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right)$

$t=2 \min =120 \mathrm{s}$

$\frac{80-70}{120}=-\mathrm{K}\left(\frac{80+70}{2}-30\right)$ $…(i)$

$\frac{60-50}{t}=-K\left(\frac{60+50}{2}-30\right) \ldots(\text { ii })$

$(i)$ $/(\text { ii }) \Rightarrow(t=216 \mathrm{s})$

Standard 11

Physics

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