JEE Mains · Physics · STD 11 - Transmission of heat
- JEE Mains 2025
- Medium
- MCQ
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 :
Pick an option to check your answer.
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
\(\begin{aligned}
& \frac{90-80}{\mathrm{t}}=\mathrm{k}\left(\frac{90+80}{2}-20\right) \\
& \frac{80-60}{\mathrm{t}^{\prime}}=\mathrm{k}\left(\frac{80+60}{2}-20\right)
\end{aligned}\)
\(\begin{aligned}
& \text { (i)/(ii) } \\
& \frac{10 \times \mathrm{t}^{\prime}}{\mathrm{t} \times 20}=\frac{65}{50} \\
& \mathrm{t}^{\prime}=\frac{65}{50} \times 2 \mathrm{t}=\frac{65}{25} \mathrm{t}=\frac{13}{5} \mathrm{t}
\end{aligned}\)
\(\begin{aligned}
& \frac{90-80}{\mathrm{t}}=\mathrm{k}\left(\frac{90+80}{2}-20\right) \\
& \frac{80-60}{\mathrm{t}^{\prime}}=\mathrm{k}\left(\frac{80+60}{2}-20\right)
\end{aligned}\)
\(\begin{aligned}
& \text { (i)/(ii) } \\
& \frac{10 \times \mathrm{t}^{\prime}}{\mathrm{t} \times 20}=\frac{65}{50} \\
& \mathrm{t}^{\prime}=\frac{65}{50} \times 2 \mathrm{t}=\frac{65}{25} \mathrm{t}=\frac{13}{5} \mathrm{t}
\end{aligned}\)
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