JEE Main 2020PhysicsThermal Properties of MatterEasyMCQ

JEE Main 2020 — Thermal Properties of Matter Question with Solution

JEE Main 2020 (03 Sep Shift 2)

Question

A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. m''' grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of m'' is close to (Latent heat of water =540 cal g-1, specific heat of water=1 cal g-1°C-1)

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Show full solutionCorrect option: A
Correct answer
A2

Step-by-step explanation

20031-25=m×540+m169

1200=m609

m≈2

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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.