A body cools from 80^{circ},C to 60^{circ},C in 5 minutes. temperature of surrounding is 20^{circ} C

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10-2.Transmission of Heat

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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........... $s$

A

$500$

B

$\frac{25}{3}$

C

$450$

D

$420$

(JEE MAIN-2023)

Solution

Rate of cooling $\alpha$ Temperature difference

$\frac{80-60}{5}=k\{70-20\}$

$\frac{60-40}{t}=k[50-20]$

$\frac{4 t}{20}=\frac{50}{30}$

$t=\frac{25}{3} min =500\,sec$

$\Rightarrow t=500 \text { seconds }$

Standard 11

Physics

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