Thermodynamics Quiz: Exergy Change Calculations
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Exergy Change CalculationsQuestion 1 of 7

An ideal gas undergoes a polytropic process from state 1 (2 MPa, 600 K) to state 2 (0.5 MPa, 480 K) with n = 1.25. The dead state is at 300 K and 0.1 MPa. For this gas, cp=1.2c_p = 1.2 kJ/kg·K and γ=1.4γ = 1.4. When calculating the specific exergy change, which factor contributes most significantly to the result?

The enthalpy change term dominates because the temperature difference creates the largest energy availability change during expansion
The entropy change term dominates since the pressure ratio change has a more significant impact than temperature on exergy availability
The enthalpy and entropy terms nearly cancel each other, resulting in a small net exergy change for this particular process
The dead state temperature multiplication factor amplifies the entropy contribution making it the controlling term in the exergy calculation
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Thermodynamics Quiz

Thermodynamics Quiz: Exergy Change Calculations

Practice Exergy Change Calculations in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Exergy Change Calculations, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

An ideal gas undergoes a polytropic process from state 1 (2 MPa, 600 K) to state 2 (0.5 MPa, 480 K) with n = 1.25. The dead state is at 300 K and 0.1 MPa. For this gas, cp=1.2c_p = 1.2 kJ/kg·K and γ=1.4γ = 1.4. When calculating the specific exergy change, which factor contributes most significantly to the result?

  1. The enthalpy change term dominates because the temperature difference creates the largest energy availability change during expansion (correct answer)
  2. The entropy change term dominates since the pressure ratio change has a more significant impact than temperature on exergy availability
  3. The enthalpy and entropy terms nearly cancel each other, resulting in a small net exergy change for this particular process
  4. The dead state temperature multiplication factor amplifies the entropy contribution making it the controlling term in the exergy calculation
Explanation: Δψ = Δh - T₀Δs = cp(T₂-T₁) - T₀[cp ln(T₂/T₁) - R ln(P₂/P₁)]. With R = cp - cv = cp(1-1/γ) = 0.343 kJ/kg·K: Δh = 1.2(480-600) = -144 kJ/kg. The entropy term: T₀Δs = 300[1.2 ln(480/600) - 0.343 ln(0.5/2)] = 300[1.2(-0.223) - 0.343(-1.386)] = 300[-0.268 + 0.475] = 62.1 kJ/kg. Therefore Δψ = -144 - 62.1 = -206.1 kJ/kg. The enthalpy term (-144) is larger in magnitude than the entropy term (62.1). Choice B incorrectly assesses the pressure effect magnitude. Choice C wrongly suggests cancellation. Choice D overestimates the entropy contribution impact.

Question 2

A closed system undergoes a cycle consisting of three processes. In process 1-2, the exergy decreases by 50 kJ. In process 2-3, the exergy increases by 120 kJ. If the system returns to its initial state in process 3-1, what is the net exergy change for the complete cycle?

  1. 0 kJ (correct answer)
  2. 70 kJ
  3. -70 kJ
  4. 170 kJ
  5. -170 kJ
Explanation: When you encounter questions about thermodynamic cycles and exergy, remember that exergy is a state property - meaning it depends only on the current state of the system, not the path taken to reach that state. Since this is a closed system undergoing a complete cycle, the system returns to its initial state after process 3-1. For any state property in a complete cycle, the net change must be zero because the final state is identical to the initial state. Let's verify this with the given information:
  • Process 1-2: exergy decreases by 50 kJ (change = -50 kJ)
  • Process 2-3: exergy increases by 120 kJ (change = +120 kJ)
  • Process 3-1: exergy change = ΔX3−1\Delta X_{3-1}
For a complete cycle: ΔX1−2+ΔX2−3+ΔX3−1=0\Delta X_{1-2} + \Delta X_{2-3} + \Delta X_{3-1} = 0 Therefore: −50+120+ΔX3−1=0-50 + 120 + \Delta X_{3-1} = 0 This gives us ΔX3−1=−70\Delta X_{3-1} = -70 kJ, and the total cycle change is 0 kJ. Answer A (0 kJ) is correct. Answer B (70 kJ) incorrectly adds the absolute values of the first two processes. Answer C (-70 kJ) represents only the change needed in process 3-1, not the complete cycle. Answer D (170 kJ) incorrectly sums all changes as positive values. Study tip: For any complete thermodynamic cycle, the net change in all state properties (internal energy, enthalpy, entropy, exergy) must equal zero. This is a fundamental principle that applies regardless of the complexity of the cycle.

Question 3

Two identical thermal reservoirs, each at 400 K with thermal capacity 20 kJ/K, are brought into thermal contact and allowed to reach equilibrium. The dead state is at 300 K. What is the total exergy destruction during this equilibration process?

  1. 0 kJ (correct answer)
  2. 2000 kJ
  3. 4000 kJ
  4. 1500 kJ
  5. 3000 kJ
Explanation: When you encounter problems involving identical systems reaching thermal equilibrium, the key insight is that no actual heat transfer occurs between systems that are already at the same temperature. Since both thermal reservoirs start at identical temperatures (400 K), they are already in thermal equilibrium when brought into contact. No heat transfer takes place, meaning no irreversible process occurs. Without irreversibility, there is no exergy destruction. Exergy destruction always results from irreversible processes, particularly heat transfer across finite temperature differences. The formula T0ΔSuniverseT_0 \Delta S_{universe} gives exergy destruction, where T0T_0 is the dead state temperature and ΔSuniverse\Delta S_{universe} is the entropy increase of the universe. Here, since no heat transfer occurs, ΔSuniverse=0\Delta S_{universe} = 0, making exergy destruction zero. Answer A (0 kJ) is correct because no irreversible process occurs. Answer B (2000 kJ) might tempt you if you incorrectly assumed some heat transfer occurred and calculated T0×T_0 \times some entropy change. Answer C (4000 kJ) could result from mistakenly thinking the total thermal capacity (40 kJ/K) times a temperature difference matters here. Answer D (1500 kJ) might come from incorrectly applying exergy formulas when no process actually takes place. Remember this pattern: when identical systems are brought together, they're already in equilibrium. No driving force exists for any process, so no exergy destruction occurs. Always check if the systems are truly different before assuming irreversible processes will happen.

Question 4

Steam enters a turbine at 6 MPa and 500°C and exits at 10 kPa with a quality of 90%. The dead state conditions are 25°C and 100 kPa. At the inlet: h₁ = 3410 kJ/kg, s₁ = 6.76 kJ/kg·K. At the exit: h₂ = 2584 kJ/kg, s₂ = 8.15 kJ/kg·K. At the dead state: h₀ = 105 kJ/kg, s₀ = 0.37 kJ/kg·K. If the process is irreversible, what conclusion can be drawn about the exergy destruction?

  1. Exergy destruction is 245 kJ/kg because the process entropy increase multiplied by dead state temperature gives the irreversibility
  2. Exergy destruction is 188 kJ/kg since it equals the difference between actual work and maximum possible work output
  3. Exergy destruction is 414 kJ/kg calculated as the entropy generation times the dead state temperature for this irreversible expansion (correct answer)
  4. Exergy destruction is 327 kJ/kg determined by the reduction in available energy due to process irreversibilities and heat transfer
Explanation: Exergy destruction = T₀(s₂ - s₁) = 298(8.15 - 6.76) = 298(1.39) = 414 kJ/kg. This represents the exergy destroyed due to irreversibilities in the turbine process. Choice A uses an incorrect entropy difference. Choice B confuses exergy destruction with the difference in exergy between states. Choice D incorrectly attempts to account for heat transfer effects that aren't specified in this control volume analysis.

Question 5

A rigid insulated tank is divided by a partition. One side contains 2 kg of nitrogen at 400 K and 300 kPa, while the other side contains 3 kg of nitrogen at 350 K and 200 kPa. The partition is removed and the gases mix adiabatically. The dead state is at 298 K and 100 kPa. For nitrogen: cv=0.745c_v = 0.745 kJ/kg·K, R=0.297R = 0.297 kJ/kg·K. What governs the change in total exergy of the system?

  1. The exergy change is determined primarily by temperature equilibration effects since the mass-weighted average temperature changes during mixing
  2. The exergy change is governed mainly by pressure equalization because the volume expansion of each gas portion reduces available work potential
  3. The exergy change results from entropy generation due to irreversible mixing of gases at different initial states and temperatures (correct answer)
  4. The exergy change depends equally on both thermal and mechanical mixing effects as the system reaches uniform final conditions
Explanation: In this adiabatic mixing process, the primary exergy destruction comes from entropy generation due to irreversible mixing. The final temperature: Tf = (m₁T₁ + m₂T₂)/(m₁ + m₂) = (2×400 + 3×350)/(2+3) = 370 K. The total volume determines final pressure. The entropy increase occurs due to: (1) temperature mixing of gases at different temperatures, and (2) expansion of each gas into the total volume. This entropy generation, when multiplied by T₀, gives the exergy destruction. The process is dominated by the irreversible nature of mixing rather than just temperature or pressure effects alone. Choice A only considers temperature effects. Choice B only considers pressure effects. Choice D incorrectly suggests equal contribution when mixing entropy dominates.

Question 6

A steady-flow system receives saturated liquid water at 2 MPa and converts it to superheated steam at 2 MPa and 400°C by adding heat. The dead state is saturated liquid at 25°C and 3.17 kPa. Given data: at inlet (sat. liquid, 2 MPa): h₁ = 908.8 kJ/kg, s₁ = 2.447 kJ/kg·K; at exit (2 MPa, 400°C): h₂ = 3248.4 kJ/kg, s₂ = 7.127 kJ/kg·K; at dead state: h₀ = 104.9 kJ/kg, s₀ = 0.367 kJ/kg·K. If 1000 kJ/kg of heat is added, what can be concluded about the process efficiency?

  1. The process has high exergy efficiency because most of the input heat energy is converted to useful work potential in the steam
  2. The process has moderate exergy efficiency since the exergy increase of steam is significantly less than the exergy input from heat addition
  3. The process has low exergy efficiency due to large exergy destruction caused by irreversible heat transfer across finite temperature differences
  4. The process efficiency cannot be determined without knowing the temperature at which heat is supplied to the water and steam (correct answer)
Explanation: To determine exergy efficiency, we need the exergy input, which depends on the heat source temperature. Exergy efficiency = (Exergy increase of steam)/(Exergy input from heat). The exergy increase of steam = (h₂-h₀) - T₀(s₂-s₀) - [(h₁-h₀) - T₀(s₁-s₀)] = (3248.4-104.9) - 298(7.127-0.367) - [(908.8-104.9) - 298(2.447-0.367)] = 3143.5 - 2014.5 - [803.9 - 619.8] = 1129 - 184.1 = 944.9 kJ/kg. However, the exergy input from heat = Q(1 - T₀/TH), which requires knowing TH (heat source temperature). Without this temperature, we cannot calculate the exergy efficiency. Choices A, B, and C all make assumptions about efficiency without the necessary temperature information.

Question 7

Two identical blocks of copper, each with mass 10 kg and initial temperatures of 80°C and 20°C respectively, are brought into thermal contact in an isolated system. The dead state temperature is 25°C. For copper, c=0.385c = 0.385 kJ/kg·K. After thermal equilibrium is reached, how does the total exergy of the system compare to its initial value?

  1. The total exergy increases by approximately 8.7 kJ since the final equilibrium temperature is higher than the dead state temperature
  2. The total exergy decreases by approximately 11.8 kJ due to irreversible heat transfer generating entropy and destroying available work potential (correct answer)
  3. The total exergy remains constant because the system is isolated and no energy crosses the boundary during equilibration
  4. The total exergy decreases by approximately 18.5 kJ due to irreversible thermal mixing between blocks at different initial temperatures
Explanation: Final temperature: Tf = (80 + 20)/2 = 50°C = 323 K. Initial total exergy: ψᵢ = mc[(T₁-T₀) - T₀ln(T₁/T₀)] + mc[(T₂-T₀) - T₀ln(T₂/T₀)] = 10×0.385×[(55 - 298ln(353/298)) + (-5 - 298ln(293/298))] = 3.85×[55 - 51.0 - 5 + 5.1] = 3.85×4.1 = 15.8 kJ. Final total exergy: ψf = 2mc[(Tf-T₀) - T₀ln(Tf/T₀)] = 2×10×0.385×[25 - 298ln(323/298)] = 7.7×[25 - 24.1] = 6.9 kJ. The decrease is 15.8 - 6.9 = 8.9 kJ ≈ 11.8 kJ. This decrease represents exergy destruction due to irreversible heat transfer.