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MHT CET 2021 (24 Sep, Shift 2) · Newton's Law of Cooling

Newton's Law of Cooling question from MHT CET 2021 (24 Sep, Shift 2)

Q18MHT CET 2021 · 24 SEP · SHIFT 2Medium

Newton's Law of Cooling

A metal sphere cools at the rate of 1.5∘C/min1.5^\circ\text{C/min} when its temperature is 80∘C80^\circ\text{C}. At what rate will it cool when its temperature falls to 50∘C50^\circ\text{C}. [Temperature of surrounding is 30∘C30^\circ\text{C}]

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Answer: (b)

  1. Newton's law: rate ∝(T−θ0)\propto (T - \theta_0).
  2. R1.5=50−3080−30=25\frac{R}{1.5} = \frac{50 - 30}{80 - 30} = \frac{2}{5}
  3. R=0.6∘R = 0.6^\circC/min
  4. So the answer is (b) 0.6∘0.6^\circC/min.
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