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JEE Mains · Physics · STD 11 - Transmission of heat
  • JEE Mains 2023
  • Medium
  • Numerical

A body cools from \(60^{\circ} C\) to \(40^{\circ} C\) in \(6\) minutes. If, temperature of surroundings is \(10^{\circ} C\). Then, after the next 6 minutes, its temperature will be \(.........{ }^{\circ} C\).

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Answer & Solution

Correct Answer

(A) \(28\)

Step-by-step Solution

Detailed explanation

By average form of Newton's law of cooling

\(\frac{20}{6}= k (50-10)\)

\(\frac{40- T }{6}= K \left(\frac{40+ T }{2}-10\right)\)

From equation \((i)\) and \((ii)\)

\(\frac{20}{40- T }=\frac{40}{10+ T / 2}\)

\(10+\frac{ T }{2}=80-2\,T\)

\(\frac{5 T }{2}=70 \Rightarrow T =28^{\circ}\,C\)

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