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JEE Main 2025 (29 Jan, Shift 2) · Newton's Law of Cooling

Newton's Law of Cooling question from JEE Main 2025 (29 Jan, Shift 2)

Q10JEE Main 2025 · 29 JAN · SHIFT 2Medium

Newton's Law of Cooling

A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:

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Answer: (a)

  1. Newton's law: T1−T2t=k(T1+T22−T0)\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right).
  2. 90→8090 \to 80: 10t=k(85−20)=65k\frac{10}{t} = k(85 - 20) = 65k
  3. 80→6080 \to 60: 20t′=k(70−20)=50k\frac{20}{t'} = k(70 - 20) = 50k
  4. Dividing: t′t=2010×6550=135\frac{t'}{t} = \frac{20}{10}\times\frac{65}{50} = \frac{13}{5}
  5. So the answer is (a) 135t\frac{13}{5}t.
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