1. 13/10t 2. 13/5t 3. 10/13t 4. 5/13t

Answer: 2. 13/5t Key idea: Newton's law of cooling in the average-temperature form: Δ T/t=k(T_avg-T₀) . First case: 10/t=k(85-20)=65k⇒ k=2/13t . Second case: 20/t'=k(70-20)=50k⇒ t'=20/50k=2/5·13t/2=13/5t . Option 2.

Topic: Heat transfer: conduction, convection, radiation· Properties of Solids and Liquids

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 a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is :

1. \(\frac{13}{10}t\)

2. \(\frac{13}{5}t\)

3. \(\frac{10}{13}t\)

4. \(\frac{5}{13}t\)

Doubtly Asked 30 Oct 2026 NEET UG 2021 neet Save WhatsApp Report

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Answer: 2. \(\frac{13}{5}t\)

Key idea: Newton's law of cooling in the average-temperature form: \(\dfrac{\Delta T}{t}=k(T_{avg}-T_0)\).

First case: \(\dfrac{10}{t}=k(85-20)=65k\Rightarrow k=\dfrac{2}{13t}\).
Second case: \(\dfrac{20}{t'}=k(70-20)=50k\Rightarrow t'=\dfrac{20}{50k}=\dfrac{2}{5}\cdot\dfrac{13t}{2}=\dfrac{13}{5}t\).

Option 2.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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 a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is :

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