Hot water cools from 80 to 60 in 1 minutes. In cooling from 60 to 50 it will take (room temperature = 30) (a) 48 s (b) 42 s (c) 50 s (d) 45 s

Answer: (a). Newton's law: (T₁ - T₂)/(t) = k((T₁ + T₂)/(2) - T₀). Case 1: 20/1 = k(70 - 30) ⇒ k = 1/2 per min Case 2: 10/t = 1/2(55 - 30) = 12.5 ⇒ t = 0.8 min 0.8 min = 48 s So the answer is (a) 48 s.

Hot water cools from 80 to 60 in 1 minutes. In cooling from 60 to 50… | MHT CET 2025 (21 Apr, Shift 2)

Q7MHT CET 2025 · 21 APR · SHIFT 2Medium

Newton's Law of Cooling

Hot water cools from $80^\circ\text{C}$ to $60^\circ\text{C}$ in 1 minutes. In cooling from $60^\circ\text{C}$ to $50^\circ\text{C}$ it will take (room temperature $= 30^\circ\text{C}$ )

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Answer: (a)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
Case 1: $\frac{20}{1} = k(70 - 30) \Rightarrow k = \frac{1}{2}$ per min
Case 2: $\frac{10}{t} = \frac{1}{2}(55 - 30) = 12.5 \Rightarrow t = 0.8$ min
$0.8$ min $= 48$ s
So the answer is (a) 48 s.

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