JEE Mains · Physics · STD 11 - 10.2 transmission of heat
A cup of coffee cools from \(90^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) in t minutes when the room temperature is \(20^{\circ} \mathrm{C}\). The time taken by the similar cup of coffee to cool from \(80^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) at the same room temperature is :
- A \(\frac{13}{10} \mathrm{t}\)
- B \(\frac{10}{13} \mathrm{t}\)
- C \(\frac{5}{13} \mathrm{t}\)
- D \(\frac{13}{5} \mathrm{t}\)
Answer & Solution
Correct Answer
(D) \(\frac{13}{5} \mathrm{t}\)
Step-by-step Solution
Detailed explanation
By using average form of Newton's law of cooling \(\begin{aligned} & \frac{90-80}{\mathrm{t}}=\mathrm{k}\left(\frac{90+80}{2}-20\right) \\ & \frac{80-60}{\mathrm{t}^{\prime}}=\mathrm{k}\left(\frac{80+60}{2}-20\right) \end{aligned}\)…
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