JEE Main 2020PhysicsThermal Properties of MatterMediumMCQ

JEE Main 2020 — Thermal Properties of Matter Question with Solution

JEE Main 2020 (04 Sep Shift 1)

Question

The specific heat of water =4200 J kg–1 K–1 and the latent heat of ice =3.4×105 J kg–1. 100 grams of ice at 0 oC is placed in 200 g of water at 25 oC. The amount of ice that will melt as the temperature of water reaches 0 oC is close to (in grams)

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Show full solutionCorrect option: A
Correct answer
A61.7

Step-by-step explanation

MSΔT=Ml

2001000×4200×25=m×340×103

m=61.7

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About this question

This is a previous-year question from JEE Main 2020, 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.