JEE Main 2015PhysicsThermodynamicsHardMCQ

JEE Main 2015Thermodynamics Question with Solution

JEE Main 2015 (04 Apr)

Question

A solid body of constant heat capacity 1 J -1 is being heated by keeping it in contact with reservoirs in two ways: (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies the same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies the same amount of heat. In both, cases the body is brought from initial temperature 100 K to final temperature 200 K . Entropy change of the body in the two cases respectively is: Note: This question was awarded as a bonus since temperatures were given in centigrade instead of in Kelvin. Proper corrections are made in the question to avoid it.

Choose an option

Show full solutionCorrect option: C
Correct answer
Cln2, ln2

Step-by-step explanation

Change in entropy, 

dS=dQTΔQ=heatsupplied=CΔT{C=J °C-1}dQ=CdtdS=CdTTIntergratingbothsidesSiSfdS=CdTTSfSi=ΔS=C.lnT|100200ΔS=C[ln200ln100]=Cln2andC=J °C-1ΔS=ln2Entrpoy change in same for both cases as C is constant, andtemperature change (i.e. from 100 to 200) is same.

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About this question

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