A metal sphere cools at a rate of 1.5/min when its temperature is 80. When the temperature of the sphere is 40, its rate of cooling is 0.3/min. The temperature of the surrounding (θ₀) is (a) 30 (b) 35 (c) 25 (d) 27

Answer: (a). Newton's law: rate ∝ (T - θ₀). (1.5)/(0.3) = (80 - θ₀)/(40 - θ₀) = 5 80 - θ₀ = 200 - 5θ₀ ⇒ θ₀ = 30° C So the answer is (a) 30° C.

A metal sphere cools at a rate of 1.5/min when its temperature is 80… | MHT CET 2025 (20 Apr, Shift 2)

Q8MHT CET 2025 · 20 APR · SHIFT 2Medium

Newton's Law of Cooling

A metal sphere cools at a rate of $1.5^\circ\text{C/min}$ when its temperature is $80^\circ\text{C}$ . When the temperature of the sphere is $40^\circ\text{C}$ , its rate of cooling is $0.3^\circ\text{C/min}$ . The temperature of the surrounding $(\theta_0)$ is

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Answer: (a)

Newton's law: rate $\propto (T - \theta_0)$ .
$\frac{1.5}{0.3} = \frac{80 - \theta_0}{40 - \theta_0} = 5$
$80 - \theta_0 = 200 - 5\theta_0 \Rightarrow \theta_0 = 30^\circ$ C
So the answer is (a) $30^\circ$ C.

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