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MHT CET 2024 (10 May, Shift 2) · Newton's Law of Cooling

Newton's Law of Cooling question from MHT CET 2024 (10 May, Shift 2)

Q14MHT CET 2024 · 10 MAY · SHIFT 2Medium

Newton's Law of Cooling

An ordinary body cools from ' 4θ′4\theta' ' to ' 3θ′3\theta' ' in ' t ' minutes. The temperature of that body after next 't' minutes is (Assume Newton's law of cooling and room temperature is θ\theta)

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Answer: (d)

  1. Newton's law: T1−T2t=k(T1+T22−θ)\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - \theta\right).
  2. First tt: θt=k(3.5θ−θ)=2.5kθ\frac{\theta}{t} = k(3.5\theta - \theta) = 2.5k\theta
  3. Next tt: 3θ−Tt=kθ+T2\frac{3\theta - T}{t} = k\frac{\theta + T}{2}
  4. Dividing: 15θ−5T=θ+T⇒T=7θ315\theta - 5T = \theta + T \Rightarrow T = \frac{7\theta}{3}
  5. So the answer is (d) 7θ3\frac{7\theta}{3}.
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