JEE Main 2015PhysicsThermal Properties of MatterMediumMCQ

JEE Main 2015 — Thermal Properties of Matter Question with Solution

JEE Main 2015 (11 Apr Online)

Question

An experiment takes 10 min to raise the temperature of water in a container from 0oC to 100oC and another 55 min to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of the water to be 1 cal g∘C-1 , the heat of vaporization according to this experiment will come out to be:

Choose an option

Show full solutionCorrect option: B
Correct answer
B550 cal g-1

Step-by-step explanation

Let the constant state of supply of heat per unit time =λ. So according to the first condition. λt1=mc×100 ⇒λm=100Ct1   ...(i), and according to the second condition, λt2=mLf⇒λm=Lft2 ...(ii). By (i) and (ii) 100Ct1=Lft2⇒1cal×10010=Lf55⇒Lf=550 cal g-1

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About this question

This is a previous-year question from JEE Main 2015, 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.