JEE Main 2022PhysicsThermal Properties of MatterHardNumerical

JEE Main 2022 — Thermal Properties of Matter Question with Solution

JEE Main 2022 (24 Jun Shift 2)

Question

In an experiment to verify Newton's law of cooling. a graph is plotted between. the temperature difference ΔT of the water and surroundings and time as shown in figure. The initial temperature of water is taken as 80°C. The value of t2 as mentioned in the graph will be

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Show full solutionCorrect answer: 16
Correct answer
16

Step-by-step explanation

According to the Newton's law of cooling, -dTdt=KT-T0.

Applying approximation, -∆T∆t=KT1+T22-T0 

For first 6 min, T1=80°C, T2=40+20=60°C, T0=20°C

⇒-60-806=K70-20⇒K=-60-806×70-20

⇒K=206×50=115

For 6 to t2, T1=60°C, T2=20+20=40°C, T0=20°C

-60-80t2-6=K50-20⇒t2-6=15×2030⇒t2=16 min

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About this question

This is a previous-year question from JEE Main 2022, covering the Thermal Properties of Matter chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.