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JEE Mains · Physics · STD 11 - 10.2 transmission of heat
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 \(.........{ }^{\circ} C\).
- A \(28\)
- B \(22\)
- C \(20\)
- D \(21\)
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 \((ii)\) \(\frac{20}{40- T }=\frac{40}{10+ T / 2}\) \(10+\frac{ T }{2}=80-2\,T\)…
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