A body cools from 80 to 50 in 5 min. In the next time of ' t ' in, the body continues to cool from 50 to 30. The total time taken by the body to cool from 80 to 30 is [The temperature of the surroundings is 20.] (a) 10 min (b) 7.5 min (c) 15.0 min (d) 12.5 min

Answer: (d). Newton's law: (T₁ - T₂)/(t) = k((T₁ + T₂)/(2) - T₀). 80 → 50: 30/5 = k(65 - 20) ⇒ k = 2/15 50 → 30: 20/t = 2/15(40 - 20) ⇒ t = 7.5 min Total time = 5 + 7.5 = 12.5 min So the answer is (d) 12.5 min.

A body cools from 80 to 50 in 5 min. In the next time of ' t ' in… | MHT CET 2025 (25 Apr, Shift 1)

Q6MHT CET 2025 · 25 APR · SHIFT 1Medium

Newton's Law of Cooling

A body cools from $80^\circ\text{C}$ to $50^\circ\text{C}$ in 5 min. In the next time of ' t ' in, the body continues to cool from $50^\circ\text{C}$ to $30^\circ\text{C}$ . The total time taken by the body to cool from $80^\circ\text{C}$ to $30^\circ\text{C}$ is [The temperature of the surroundings is $20^\circ\text{C}$ .]

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Answer: (d)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
$80 \to 50$ : $\frac{30}{5} = k(65 - 20) \Rightarrow k = \frac{2}{15}$
$50 \to 30$ : $\frac{20}{t} = \frac{2}{15}(40 - 20) \Rightarrow t = 7.5$ min
Total time $= 5 + 7.5 = 12.5$ min
So the answer is (d) 12.5 min.

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Newton's Law of Cooling question from MHT CET 2025 (25 Apr, Shift 1)

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