MHT CET · Physics · Thermal Properties of Matter
A metal sphere cools at the rate of \(1.5^{\circ} \mathrm{C} / \mathrm{min}\) when its temperature is \(80^{\circ} \mathrm{C}\). At what rate will it cool when its temperature falls to \(50^{\circ} \mathrm{C}\). [Temperature of surrounding is \(30^{\circ} \mathrm{C}\) ]
- A \(0.9^{\circ} \mathrm{C} / \mathrm{min}\)
- B \(0.6^{\circ} \mathrm{C} / \mathrm{min}\)
- C \(1.5^{\circ} \mathrm{C} / \mathrm{min}\)
- D \(1.2^{\circ} \mathrm{C} / \mathrm{min}\)
Answer & Solution
Correct Answer
(B) \(0.6^{\circ} \mathrm{C} / \mathrm{min}\)
Step-by-step Solution
Detailed explanation
When temperature is \(80^{\circ} \mathrm{C}\), by Newton's law of cooling, we have
\(1.5=\mathrm{k}\left(\frac{80+30}{2}-30\right)=\mathrm{k}(55-30)=25 \mathrm{~K}\)
When temperature is \(50^{\circ} \mathrm{C}\), let \(\mathrm{r}\) be the rate of cooling.
Then
\(\begin{aligned}
& \mathrm{r}=\mathrm{k}\left(\frac{50+30}{2}-30\right)=\mathrm{k}(40-30)=10 \mathrm{~K} \\
& \therefore \frac{\mathrm{r}}{1.5}=\frac{10}{25} \\
& \mathrm{r}=0.6^{\circ} \mathrm{C} / \mathrm{min}
\end{aligned}\)
\(1.5=\mathrm{k}\left(\frac{80+30}{2}-30\right)=\mathrm{k}(55-30)=25 \mathrm{~K}\)
When temperature is \(50^{\circ} \mathrm{C}\), let \(\mathrm{r}\) be the rate of cooling.
Then
\(\begin{aligned}
& \mathrm{r}=\mathrm{k}\left(\frac{50+30}{2}-30\right)=\mathrm{k}(40-30)=10 \mathrm{~K} \\
& \therefore \frac{\mathrm{r}}{1.5}=\frac{10}{25} \\
& \mathrm{r}=0.6^{\circ} \mathrm{C} / \mathrm{min}
\end{aligned}\)
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