Entropy
By: S K Mondal
Chapter 7
T1 to T2. The body is then brought back to its initial state by contact with a single reservoir at temperature T1. Calculate the changes of entropy of the body and of the reservoirs. What is the total change in entropy of the whole system? If the initial heating were accomplished merely by bringing the body into contact with a single reservoir at T2, what would the various entropy changes be? Solution:
TRY PLEASE
Q7.14
A body of finite mass is originally at temperature T1, which is higher than that of a reservoir at temperature T2. Suppose an engine operates in a cycle between the body and the reservoir until it lowers the temperature of the body from T1 to T2, thus extracting heat Q from the body. If the engine does work W, then it will reject heat Q–W to the reservoir at T2. Applying the entropy principle, prove that the maximum work obtainable from the engine is W (max) = Q – T2 (S1 – S2) Where S1 – S2 is the entropy decrease of the body.
Solution:
If the body is maintained at constant volume having constant volume heat capacity Cv = 8.4 kJ/K which is independent of temperature, and if T1 = 373 K and T2 = 303 K, determine the maximum work obtainable. (Ans. 58.96 kJ) Final temperature of the body will be T2 ∴
S2 – S1 =
T2
∫ mc
v
T1
dT ⎛T ⎞ = m cv ln ⎜ 2 ⎟ T ⎝ T1 ⎠ [ cv = heat energy CV]
(ΔS) reservoir =
Q−W T2
∴ (ΔS) H.E. = 0
or
Q−W ≥0 T2 T2 (S2 – S1) + Q – W ≥ 0
or
W ≤ Q + T2 (S2 – S1)
or
W ≤ [Q – T2 (S1 – S2)]
∴
Wmax = [Q – T2 (S1 – S2)]
∴
(ΔS) univ. = (S2 − S1 ) +
Wmax = Q – T2 (S1 – S2) ⎛T ⎞ = Q + T2Cv ln ⎜ 2 ⎟ ⎝ T1 ⎠ ⎛T ⎞ = Cv (T1 – T2) + T2 CV ln ⎜ 2 ⎟ ⎝ T1 ⎠ ⎡ ⎛ 303 ⎞ ⎤ = 8.4 ⎢373 − 303 + 303 ln ⎜ ⎟⎥ ⎝ 373 ⎠ ⎦ ⎣ Page 79 of 265