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MHT CET 2025 (20 Apr, Shift 1) · Newton's Law of Cooling

Newton's Law of Cooling question from MHT CET 2025 (20 Apr, Shift 1)

Q9MHT CET 2025 · 20 APR · SHIFT 1Medium

Newton's Law of Cooling

A body cools from 60∘C60^\circ\text{C} to 40∘C40^\circ\text{C} in 6 minutes. After next 6 minutes its temperature will be (Temperature of the surroundings is 10∘C10^\circ\text{C})

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Answer: (b)

  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. First 6 min: 206=k(50−10)⇒k=112\frac{20}{6} = k(50 - 10) \Rightarrow k = \frac{1}{12}
  3. Next 6 min: 40−T6=112(40+T2−10)⇒40−T=T+204\frac{40 - T}{6} = \frac{1}{12}\left(\frac{40 + T}{2} - 10\right) \Rightarrow 40 - T = \frac{T + 20}{4}
  4. 160−4T=T+20⇒T=28∘160 - 4T = T + 20 \Rightarrow T = 28^\circC
  5. So the answer is (b) 28∘28^\circC.
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