Q81 / 90 · JEE Main 2015 (04 Apr) · Physics

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.

  • A
    2ln2, 8ln2
  • B
    ln2, 4ln2
  • Correct answer
    C
    ln2, ln2
  • D
    ln2, 2ln2

Solution

Change in entropy, 

dS=dQTΔQ= heat supplied=CΔT      {C=J °C-1}dQ=CdtdS=CdTTIntergrating both sides∫SiSfdS=C∫dTT  ⇒Sf−Si=ΔS=C.lnT|100200⇒ΔS=C[ln200−ln100]=Cln2and  C=J °C-1⇒  ΔS=ln2Entrpoy change in same for both cases as C is constant, andtemperature change (i.e. from 100 to 200) is same.

Hence, the correct answer is option: 'C'.

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