Entropy

Problem:

An ice tray contains 500 g of water.  Calculate the change in entropy of the water as it freezes completely and slowly at 0 oC.
latent heat of fusion for water/ice = 333000 J/kg

Solution:

• Concepts:
Change in entropy: dS = dQ/T
ΔS = ∫if dS = ∫if dQr/T
The subscript r denotes a reversible path.
• Reasoning:
We are asked calculate the change in entropy ΔS = ΔQ/T.   While the water changes phase, the temperature stays constant.
• Details of the calculation:
ΔS = ΔQ/T.  ΔQ = -mL, m = mass of water, L= latent heat of fusion = 333000 J/kg.
ΔS = -(0.5 kg)(333000J /kg)/273 K = -610 J/K.

Problem:

A cup contains 0.2 kg of tea and is heated to 90 oC.  The specific heat capacity of tea is ~ 1100 cal/(kg K).  If the tea is allowed to cool to 20 oC calculate the change in entropy (J/K).  Also calculate the entropy change in the room (Troom = 20 oC) and show that the total change in entropy is positive.

Solution:

• Concepts:
Change in entropy: ΔS = ∫if dS = ∫if dQr/T
• Reasoning:
We are asked calculate the change in entropy along a reversible path.
• Details of the calculation:
For the tea
ΔStea = ∫363K293 mcdT/T = mc ln(293/363) = -0.2 kg * 1100 cal/(kg K) * 0.214 = -47 cal/K.
ΔStea = -47.1*4.186 J/K = -197 J/K.
For the room
ΔSroom = (0.2 kg *1100*4.186 J/K * 70 K)/(293 K) = 220 J/K.
ΔStotal = ΔStea + ΔSroom = 23 J/K (positive)

Problem:

100 g of water at 20 oC is mixed with 100 g of water at 80 oC.  What is the net change in entropy?

Solution:

• Concepts:
Change in entropy: ΔS = ∫if dS = ∫if dQr/T
The subscript r denotes a reversible path.
• Reasoning:
Although this is an irreversible process, we could get the same final result (200 g of water at 50 oC) by slowly warming the cool water from 20 oC to 50 oC and slowly cooling the warm water from 80 oC to 50 oC.
• Details of the calculation:
ΔS = ΔShot + ΔScold = ∫293323 dQ/T + ∫353323 dQ/T =  ∫293323 mcdT/T + ∫353323 mcdT/T
= (100 g)(1 cal/( g oC))[ln(323/293) + ln(323/353)] = 0.87 cal/oC.

Problem:

The surface of the Sun is approximately at 5700 K, and the temperature of the Earth's surface is approximately 290 K.  What entropy changes occur when 1000 J of thermal energy is transferred from the Sun to the Earth?

Solution:

• Concepts:
Change in entropy: ΔS = ∫if dS = ∫if dQr/T
The subscript r denotes a reversible path.
• Reasoning:
We are asked calculate the change in entropy ΔS = ΔQ/T.
• Details of the calculation:
During the process the temperatures of the sun and the earth do not change appreciably.  The change in the entropy of the sun is therefore ΔS = -1000 J/5700 K = -0.175 J/K.  The change in the entropy of the earth is ΔS = 1000J/290 K = 3.448 J/K.  The entropy of the sun-earth system increases by 3.27 J/K.

Problem:

A certain amount of water of heat capacity C is at a temperature of 0oC.  It is placed in contact with a heat reservoir at 100oC and the two come into thermal equilibrium.
(a)  What is the entropy change of the universe?
(b)  The process is now divided into two stages: first the water is brought into contact with a heat reservoir at 50oC and comes into thermal equilibrium; then it is placed in contact with the heat reservoir at 100oC.  What is the entropy change of the universe?
(c)  What is the entropy change of the universe in the limit of infinitely many stages with infinitely small temperature differences?

Solution:

• Concepts:
Entropy
Reasoning:
To calculate the change in entropy of a system for a finite process, when T changes appreciably, we use ΔS = ∫ifdS = ∫ifdQr/T, where the subscript r denotes a reversible path.  To calculate the change in entropy, we find some reversible path that can take the system from its initial to its final state and evaluate the integral along that path.  The actual path of the system from the initial to the final state may be very different and not reversible.  But the change in entropy depends only on the initial and final state, not on the path.
• Details of the calculation:
(a)  Change in entropy of the water:
ΔSW = ∫ifdS = ∫TiTfCdT/T = Cln(Tf/Ti) = C*ln(373/273)
Change in entropy of the reservoir: ΔSR = -ΔQ/(373 K) = -C*100/373
Total entropy change: C[ln(373/273) - 100/373) = C*0.044
(b)  ΔSW = C*ln(373/273)
ΔSR = -C*50/323 - C*50/373
Total entropy change: C[ln(373/273) - C*50/323 - C*50/373) = C*0.023
(c)  In this limit we have a reversible process and ΔS = 0.

Problem:

Consider two identical blocks of material with heat capacity C, which is temperature independent. The "hot" one is initially at temperature TH and the "cold" one at temperature TC < TH.   The two blocks act as reservoirs for a ideal heat engine.  As the engine works, the reservoirs gradually equilibrate and reach a common temperature Tf
(a)  What is the temperature Tf?
(b)  How much work does the engine do?
(c)  Compute the change in entropy ΔS for each of the two blocks.

Solution:

• Concepts:
Efficiency of a heat engine, entropy
• Reasoning:
For a  Carnot engine operating between Ta and Tb  we have dQa/Ta = dQb/Tb, dS = 0.
• Details of the calculation:
(a)  ΔSH = ∫THTfCdT/T = Cln(Tf/TH), ΔSC = ∫TCTfCdT/T = Cln(Tf/TC).
ΔSH + ΔSC = 0 --> ln(Tf/TH) + ln(Tf/TC) = 0, Tf = (THTC)½.
(b)  W =  (QH - QC) = C(TH - Tf) - C(Tf - TC) = C((TH - TC - 2(THTC)½).
ΔSH = ∫THTfCdT/T = Cln(Tf/TH) = ½Cln(TC/TH).
ΔSC = ∫TCTfCdT/T = Cln(Tf/TC) = ½Cln(TH/TC) = -ΔSH.

Problem:

An ideal heat engine is powered by two reservoirs of equal heat capacity C, which is temperature independent.  As the engine works, the reservoirs gradually equilibrate.
(a)  Find the overall efficiency of the engine from the starting point where the reservoirs are at temperatures T1 = 90 oC and T2 = 30 oC to the moment of complete equilibration.
(b)  Now assume that a real heat engine is powered by the same reservoir with initial temperatures T1 = 90 oC and T2 = 30 oC.  Let C = 105 J/oC.  The real engine's overall efficiency is 20% of the overall efficiency of the ideal engine.  Find the change in entropy of the system once the reservoirs have equilibrated.

Solution:

• Concepts:
Efficiency of a heat engine, entropy
• Reasoning:
The efficiency of a heat engine is e = W/Qhigh = (Qhigh - Qlow)/Qhigh.  For a  Carnot engine operating between Ta and Tb  we have dQa/Ta = dQb/Tb, dS = 0.
• Details of the calculation:
e = W/Qhigh = (Qhigh - Qlow)/Qhigh = (C(T1 - Tf) - C(Tf - T2))/(C(T1 - Tf)).
ΔS1 = ∫T1TfCdT/T = Cln(Tf/T1), ΔS2 = ∫T2TfCdT/T = Cln(Tf/T2).
ΔS1 + ΔS2 = 0 --> ln(Tf/T1) + ln(Tf/T2) = 0, Tf = (T1T2)½ = 331.6459 K.
e = 1 - ((T1T2)½- T2))/((T1 - (T1T2)½)) = 1 - (T2/T1)½.
e = 1 - (303/363)½ = 0.0864.
(b)  ΔQ = Q1 - Q2 = 0.2*Q1*(1 - (T2/T1)½),  Q2 = 0.8*Q1 + 0.2*Q1(T2/T1)½.
Q2/Q1 = C(Tf - T2)/(C(T1 - Tf)) = 0.8 + 0.2*(T2/T1)½ = 0.982725 = α.
Tf = (αT1 + T2)/(1 + α) = 332.7386 K.
ΔS1 + ΔS2 = C(ln(Tf/T1) + ln(Tf/T2)) = 658 J/K is the change in the entropy of the system.

Problem:

A Carnot heat engine has the following entropy-temperature diagram.

(a)  Describe the cycle.  For each segment identify the process, say whether work is done by the working system or on it and whether heat is added to the system or extracted from it.
(b)  How much work is done by the system?