KCET · Physics · Thermal Properties of Matter
A hot body is allowed to cool. The surrounding temperature is constant at \(30^{\circ} \mathrm{C}\). The takes time \(t_{1}\) to cool from \(70^{\circ} \mathrm{C}\) to \(68^{\circ} \mathrm{C}\) and time \(t_{2}\) to cool from \(60^{\circ} \mathrm{C}\) to \(59.5^{\circ} \mathrm{C}\). Then
- A \(\mathrm{t}_{2}=\mathrm{t}_{1}\)
- B \(\mathrm{t}_{2}=2 \mathrm{t}_{1}\)
- C \(\mathrm{t}_{2}=\frac{1}{2} \mathrm{t}_{1}\)
- D \(t_{2}=4 t_{1}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{t}_{2}=2 \mathrm{t}_{1}\)
Step-by-step Solution
Detailed explanation
By Newton's law of cooling
\(-\frac{\mathrm{dT}}{\mathrm{dt}}=\alpha\left(\mathrm{T}-\mathrm{T}_{0}\right)\)
\(\therefore \frac{2}{t_{1}}=\alpha(90-30)=60 \alpha\)
and \(\frac{0.5}{t_{2}}=\alpha(60-30)=30 \alpha\)
From Eqs. (i) and (ii),
\(\mathrm{t}_{2}=2 \mathrm{t}_{1}\)
\(-\frac{\mathrm{dT}}{\mathrm{dt}}=\alpha\left(\mathrm{T}-\mathrm{T}_{0}\right)\)
\(\therefore \frac{2}{t_{1}}=\alpha(90-30)=60 \alpha\)
and \(\frac{0.5}{t_{2}}=\alpha(60-30)=30 \alpha\)
From Eqs. (i) and (ii),
\(\mathrm{t}_{2}=2 \mathrm{t}_{1}\)
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