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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 \(.........{ }^{\circ} C\).
Pick an option to check your answer.
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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