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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........... \(s\)

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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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