A solid body of constant heat capacity 1 J//^(@)C 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 same amount of heat . (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat . In both the cases body is brought from initial temperature 100^(@)C to final temperature 200^(@)C. Entropy change of the body in the two cases respectively is

A solid body of constant heat capacity 1 J//^(@)C is being heated by k

1.	A solid body of constant heat capacity 1 J//^(@)C 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 same amount of heat . (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat . In both the cases body is brought from initial temperature 100^(@)C to final temperature 200^(@)C. Entropy change of the body in the two cases respectively is
Answer» 1n 2, 4 1n 2 1n 2, 1n 2 1n 2, 21n 2 21n 2, 81n 2 Solution :`100^(@)C = 373 K, 200^(@)C = 473 K` Increase in temperature = `100^(@)C = 100 K` Heat CAPACITY, `C = 1 J//K` (i) For each heat reservoir, increase in temperature = `100/2 = 50 K` So, entropy change, `Delta S_(1) = C[int_(373)^(423) (DT)/(T) + int_(423)^(473) (dT)/(T)]` `= 1n (423)/(373) + 1n (473)/(423)` `= 1n(473)/(373)` (ii) For each heat reservoir, increase in temperature = `100/8 = 12.5 K` So, entropy change, `Delta S_(2) = C[int_(385.5)^(373) (dT)/(T) + .....+ int_(473)^(460.5) (dT)/(T)]` ` = 1n(473)/(373)` [Note: The VALUES of `Delta S_(1) and Delta S_(2)` does not match any of the given answers. if we take the temperatures to be 100 K and 200 K instead of `100^(@)C and 200^(@)C`, then `Delta S_(1) = Delta S_(2) = 1n2 J`.