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.

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., 2 ln 2 , 8 ln 2, ln 2 , 4 ln 2, ln 2 , ln 2, ln 2 , 2 ln 2

MARKS App · Accepted Answer

Change in entropy, d S = d Q T Δ Q = heat supplied = C Δ T { C = J ° C - 1 } d Q = C d t d S = C d T T Intergrating both sides ∫ S i S f d S = C ∫ d T T ⇒ S f − S i = Δ S = C . ln T | 100 200 ⇒ Δ S = C [ ln 200 − ln 100 ] = C ln 2 and C = J ° C - 1 ⇒ Δ S = ln 2 Entrpoy change in same for both cases as C is constant, and temperature change (i .e . from 100 to 200) is same .

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