A body cools from $80^{\circ}\,C$ to $60^{\circ}\,C$ in $5$ minutes. The temperature of the surrounding is $20^{\circ} C$. The time it takes to cool from $60^{\circ}\,C$ to $40^{\circ}\,C$ is........... $s$
JEE MAIN 2023, Medium
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Rate of cooling $\alpha$ Temperature difference
$\frac{80-60}{5}=k\{70-20\}$
$\frac{60-40}{t}=k[50-20]$
$\frac{4 t}{20}=\frac{50}{30}$
$t=\frac{25}{3} min =500\,sec$
$\Rightarrow t=500 \text { seconds }$
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