By using average form of Newton's law of cooling $\begin{aligned}
& \frac{90-80}{\mathrm{t}}=\mathrm{k}\left(\frac{90+80}{2}-20\right) \
& \frac{80-60}{\mathrm{t}^{\prime}}=\mathrm{k}\left(\frac{80+60}{2}-20\right)
\end{aligned}\begin{aligned}
& \text { (i)/(ii) } \
& \frac{10 \times \mathrm{t}^{\prime}}{\mathrm{t} \times 20}=\frac{65}{50} \
& \mathrm{t}^{\prime}=\frac{65}{50} \times 2 \mathrm{t}=\frac{65}{25} \mathrm{t}=\frac{13}{5} \mathrm{t}
\end{aligned}$
A cup of coffee cools from 90° C to 80° C in t minutes when the room temperature is 20° C. The time taken by the similar cup of coffee to cool from…
JEE Main 2025 — Physics Thermodynamics
2025mcqmedium
A cup of coffee cools from 90∘C to 80∘C in t minutes when the room temperature is 20∘C. The time taken by the similar cup of coffee to cool from 80∘C to 60∘C at the same room temperature is :
Official previous-year question
Held on 29 Jan 2025 · Verified 6 Jul 2026.
Options
- A
1013t
- B
1310t
- C
135t
- D
513t
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