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JEE Mains · Physics · STD 11 - Transmission of heat
- JEE Mains 2023
- Medium
- Numerical
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........... \(s\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(500\)
Step-by-step Solution
Detailed explanation
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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