Thermodynamics Quiz: Internal Energy Enthalpy And Total Energy
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Internal Energy Enthalpy And Total EnergyQuestion 1 of 17

For a closed system at constant pressure, which statement correctly describes the relationship between enthalpy and internal energy when the system expands and does work against the external pressure?

Enthalpy change equals internal energy change because pressure is constant throughout the process
Enthalpy change is greater than internal energy change due to the additional PV work term
Enthalpy change is less than internal energy change because work is done by the system
Enthalpy change is independent of internal energy change at constant pressure conditions
Enthalpy change equals internal energy change plus the work done against external pressure
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Thermodynamics Quiz

Thermodynamics Quiz: Internal Energy Enthalpy And Total Energy

Practice Internal Energy Enthalpy And Total Energy 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 Internal Energy Enthalpy And Total Energy, 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.

All questions

Question 1

For a closed system at constant pressure, which statement correctly describes the relationship between enthalpy and internal energy when the system expands and does work against the external pressure?

  1. Enthalpy change equals internal energy change because pressure is constant throughout the process
  2. Enthalpy change is greater than internal energy change due to the additional PV work term (correct answer)
  3. Enthalpy change is less than internal energy change because work is done by the system
  4. Enthalpy change is independent of internal energy change at constant pressure conditions
  5. Enthalpy change equals internal energy change plus the work done against external pressure
Explanation: When you encounter questions about enthalpy and internal energy in expanding systems, focus on the fundamental relationship between these state functions and the work involved in the process. For any process, enthalpy (H) and internal energy (U) are related by the equation H=U+PVH = U + PV. When we consider changes in these quantities, we get ΔH=ΔU+Δ(PV)\Delta H = \Delta U + \Delta(PV). At constant pressure, this simplifies to ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta V. Since the system expands (ΔV>0\Delta V > 0), the term PΔVP\Delta V is positive, making the enthalpy change greater than the internal energy change. Option B correctly captures this relationship. The enthalpy change includes not only the internal energy change but also the additional PΔVP\Delta V work term associated with expansion against constant external pressure. Option A incorrectly assumes that constant pressure means the changes are equal, missing the crucial PΔVP\Delta V contribution. Option C gets the relationship backwards—it suggests enthalpy change is smaller, which would require the system to contract rather than expand. Option D wrongly claims independence between these inherently related thermodynamic properties. Remember this key pattern: at constant pressure, ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta V. For expansion processes (ΔV>0\Delta V > 0), enthalpy change will always exceed internal energy change by exactly the PΔVP\Delta V work term. This relationship is fundamental to understanding why enthalpy is the preferred function for constant-pressure processes.

Question 2

Which combination of energy transfers would result in an increase in internal energy but a decrease in total energy for a closed system?

  1. Heat addition with the system performing work while losing kinetic energy
  2. Heat removal with work done on the system while gaining potential energy
  3. Heat addition with work done on the system while losing potential energy (correct answer)
  4. Heat removal with the system performing work while gaining kinetic energy
  5. Heat addition with the system performing work while gaining both kinetic and potential energy
Explanation: When analyzing energy changes in closed systems, you need to distinguish between internal energy (the system's microscopic energy) and total energy (which includes macroscopic kinetic and potential energy). The first law of thermodynamics states that ΔU=Q−W\Delta U = Q - W, where internal energy change equals heat added minus work done by the system. For internal energy to increase while total energy decreases, you need heat addition (positive Q) combined with work done on the system (negative W, making -W positive), while simultaneously losing macroscopic energy. Option C achieves this: heat addition increases internal energy, work done on the system further increases internal energy, but the loss of potential energy reduces total energy more than internal energy gains. Option A is incorrect because performing work (positive W) decreases internal energy change, and losing kinetic energy alone might not overcome the internal energy gain from heat addition. Option B fails because heat removal (negative Q) decreases internal energy, contradicting the requirement for internal energy increase. Option D is wrong because heat removal decreases internal energy, and performing work further reduces internal energy change. The key insight is that macroscopic energy changes (kinetic and potential) affect total energy directly, while heat and work primarily influence internal energy. Only when substantial macroscopic energy is lost can total energy decrease despite internal energy gains. Study tip: Always separate internal energy (governed by the first law) from total energy (including macroscopic motion). Look for scenarios where microscopic energy increases while macroscopic energy decreases significantly.

Question 3

A rigid container holds a gas at pressure P₁ and temperature T₁. Heat Q is added at constant volume until the pressure doubles. Which expression correctly represents the change in enthalpy of the gas?

  1. ΔH = Q because enthalpy change equals heat added at constant volume
  2. ΔH = Q + P₁V because enthalpy includes the pressure-volume work term
  3. ΔH = Q + P₁V + P₂V because enthalpy accounts for both initial and final PV terms
  4. ΔH = Q - P₁V because the pressure-volume work opposes the heat addition
  5. ΔH = Q + (P₂ - P₁)V because enthalpy change includes the change in pressure-volume product (correct answer)
Explanation: When analyzing constant volume processes involving heat addition, you need to distinguish between internal energy changes (governed by the first law) and enthalpy changes (which include pressure-volume effects). For this rigid container at constant volume, the first law gives us ΔU=Q\Delta U = Q since no work is done (W=0W = 0 because ΔV=0\Delta V = 0). However, enthalpy is defined as H=U+PVH = U + PV, so the change in enthalpy becomes: ΔH=ΔU+Δ(PV)=Q+Δ(PV)\Delta H = \Delta U + \Delta(PV) = Q + \Delta(PV) Since volume remains constant, Δ(PV)=V⋅ΔP=V(P2−P1)=V(2P1−P1)=P1V\Delta(PV) = V \cdot \Delta P = V(P_2 - P_1) = V(2P_1 - P_1) = P_1V Therefore, ΔH=Q+P1V\Delta H = Q + P_1V, making B correct. A incorrectly applies the constant pressure relationship (ΔH=Q\Delta H = Q) to a constant volume process. At constant volume, enthalpy change always exceeds the heat added because of the Δ(PV)\Delta(PV) term. C adds both P1VP_1V and P2VP_2V terms, which would be ΔH=Q+P1V+2P1V=Q+3P1V\Delta H = Q + P_1V + 2P_1V = Q + 3P_1V. This incorrectly treats Δ(PV)\Delta(PV) as P1V+P2VP_1V + P_2V instead of (P2−P1)V(P_2 - P_1)V. D subtracts P1VP_1V, giving ΔH=Q−P1V\Delta H = Q - P_1V. This reverses the sign of the pressure-volume contribution, which would only make sense if pressure decreased rather than doubled. Key strategy: Remember that ΔH=ΔU+Δ(PV)\Delta H = \Delta U + \Delta(PV). At constant volume, ΔU=Q\Delta U = Q and Δ(PV)=V⋅ΔP\Delta(PV) = V \cdot \Delta P. The enthalpy change in constant volume heating always exceeds the heat added due to the positive pressure increase term.

Question 4

For a flowing stream of gas entering a control volume with velocity 100 m/s at elevation 50 m and exiting with velocity 150 m/s at elevation 30 m, which term represents the change in total energy per unit mass if the internal energy change is 200 J/kg?

  1. ΔE = 200 + ½(150² - 100²) + g(30 - 50) = 7,050 J/kg
  2. ΔE = 200 + ½(150² - 100²) + g(50 - 30) = 7,446 J/kg
  3. ΔE = 200 + ½(150² - 100²) + g(30 - 50) = 6,054 J/kg (correct answer)
  4. ΔE = 200 + ½(100² - 150²) + g(30 - 50) = -5,446 J/kg
  5. ΔE = 200 + ½(150² + 100²) + g(30 + 50) = 29,584 J/kg
Explanation: When analyzing energy changes in flowing streams, you need to account for three forms of energy per unit mass: internal energy, kinetic energy, and potential energy. The total energy change is ΔE=Δu+Δke+Δpe\Delta E = \Delta u + \Delta ke + \Delta pe, where each term represents the change from inlet to outlet conditions. The correct approach requires careful attention to the direction of each energy change. For kinetic energy: Δke=12(Vout2−Vin2)=12(1502−1002)=12(22,500−10,000)=6,250 J/kg\Delta ke = \frac{1}{2}(V_{out}^2 - V_{in}^2) = \frac{1}{2}(150^2 - 100^2) = \frac{1}{2}(22,500 - 10,000) = 6,250 \text{ J/kg}. For potential energy: Δpe=g(zout−zin)=9.81(30−50)=−196.2 J/kg\Delta pe = g(z_{out} - z_{in}) = 9.81(30 - 50) = -196.2 \text{ J/kg}. Therefore: ΔE=200+6,250−196.2=6,053.8 J/kg\Delta E = 200 + 6,250 - 196.2 = 6,053.8 \text{ J/kg}, which rounds to answer C. Answer A incorrectly calculates the kinetic energy term, getting 6,250 J/kg but then mysteriously arriving at a final answer that suggests different intermediate values. Answer B makes a critical sign error in the potential energy term, using (50−30)(50 - 30) instead of (30−50)(30 - 50), adding 196.2 J/kg instead of subtracting it. This gives 7,446 J/kg instead of the correct 6,054 J/kg. Answer D reverses the kinetic energy calculation to (1002−1502)(100^2 - 150^2), making kinetic energy negative when the gas is actually accelerating. Remember that energy changes always follow the convention of "final minus initial" conditions. The gas gains kinetic energy (accelerating) but loses potential energy (descending), so be meticulous with your signs and velocity terms.

Question 5

A system's internal energy is measured relative to a reference state where U = 0. If the system undergoes three sequential processes with internal energy changes of +50 kJ, -30 kJ, and +20 kJ respectively, and then returns to the reference state via a fourth process, what is the internal energy change in the fourth process?

  1. -40 kJ because internal energy must return to zero for a complete cycle (correct answer)
  2. +40 kJ because the system must overcome the previous energy changes
  3. 0 kJ because the fourth process simply maintains the existing energy state
  4. -140 kJ because all previous energy changes must be reversed
  5. +90 kJ because the total positive changes must be reinforced
Explanation: When you encounter problems involving internal energy changes through multiple processes, remember that internal energy is a state function - it depends only on the current state of the system, not on how the system got there. This means that for any complete cycle returning to the initial state, the total change in internal energy must be zero. Let's trace through this step-by-step. Starting from the reference state (U = 0), the system undergoes three processes with changes of +50 kJ, -30 kJ, and +20 kJ. After these three processes, the total change is: +50+(−30)+(+20)=+40 kJ+50 + (-30) + (+20) = +40 \text{ kJ} So the system now has an internal energy of +40 kJ relative to the reference state. Since the fourth process returns the system to the reference state (U = 0), the change must be -40 kJ to complete the cycle. Answer A correctly identifies this: the internal energy change is -40 kJ because the system must return to zero for a complete cycle. Answer B (+40 kJ) would actually double the energy above the reference state instead of returning to it. Answer C (0 kJ) incorrectly assumes no change is needed, ignoring that the system is currently 40 kJ above the reference. Answer D (-140 kJ) mistakenly suggests reversing all individual changes rather than just achieving the net correction needed. Study tip: For any cyclic process in thermodynamics, remember that state functions (internal energy, enthalpy, entropy) must have zero net change. Only calculate what's needed to return to the starting state.

Question 6

For a system where internal energy depends only on temperature U = nCᵥT, and pressure and volume are related by PV = nRT, which expression correctly represents enthalpy in terms of temperature alone?

  1. H = nCᵥT because enthalpy has the same temperature dependence as internal energy
  2. H = nCᵥT + nRT = nT(Cᵥ + R) because enthalpy includes the pressure-volume term (correct answer)
  3. H = nCᵥT + PV = nCᵥT + nRT only at constant pressure conditions
  4. H = nCᵥT - nRT because the pressure-volume work opposes internal energy
  5. H = n(Cᵥ + R)T only for reversible processes involving this ideal system
Explanation: When you encounter problems involving enthalpy and internal energy relationships, remember that enthalpy is defined as H=U+PVH = U + PV. This fundamental relationship is key to solving temperature-dependent expressions for enthalpy. Starting with the given information: internal energy U=nCvTU = nC_vT and the ideal gas law PV=nRTPV = nRT, you can substitute directly into the enthalpy definition. This gives you H=nCvT+nRT=nT(Cv+R)H = nC_vT + nRT = nT(C_v + R). This expression is valid for any ideal gas system, regardless of the process conditions, because it stems from the fundamental definition of enthalpy. Choice A is incorrect because it ignores the PVPV term entirely. Enthalpy and internal energy are fundamentally different quantities—enthalpy always includes the pressure-volume contribution, which for an ideal gas equals nRTnRT. Choice C contains a critical misconception. While the expression H=nCvT+nRTH = nC_vT + nRT is mathematically correct, the claim that it only applies "at constant pressure conditions" is wrong. This temperature dependence of enthalpy holds for ideal gases under any conditions, not just constant pressure processes. Choice D incorrectly uses subtraction instead of addition. The enthalpy definition requires adding the PVPV term to internal energy, not subtracting it. The PVPV term represents the energy associated with the system's ability to do expansion work, which adds to the total enthalpy. Study tip: Always remember that enthalpy equals internal energy plus PVPV, never minus. For ideal gases, this PVPV term always equals nRTnRT.

Question 7

A steam turbine receives steam with specific enthalpy 3000 kJ/kg and velocity 100 m/s, and discharges steam with specific enthalpy 2500 kJ/kg and velocity 200 m/s. If elevation changes are negligible, what is the change in specific total energy?

  1. -485 kJ/kg because total energy includes both enthalpy and kinetic energy changes (correct answer)
  2. -500 kJ/kg because enthalpy change dominates over kinetic energy effects in turbines
  3. -515 kJ/kg because kinetic energy increase opposes the enthalpy decrease
  4. -300 kJ/kg because turbine operation modifies the relationship between enthalpy and total energy
  5. -700 kJ/kg because both enthalpy and kinetic energy changes contribute negatively
Explanation: When analyzing steam turbines, you need to consider specific total energy, which includes both the thermal energy (represented by specific enthalpy) and the mechanical kinetic energy of the flowing steam. The change in specific total energy is calculated as: Δetotal=Δh+Δke\Delta e_{total} = \Delta h + \Delta ke where Δke=V22−V122\Delta ke = \frac{V_2^2 - V_1^2}{2} Let's calculate each component:
  • Enthalpy change: Δh=2500−3000=−500 kJ/kg\Delta h = 2500 - 3000 = -500 \text{ kJ/kg}
  • Kinetic energy change: Δke=2002−10022=40000−100002=15000 J/kg=15 kJ/kg\Delta ke = \frac{200^2 - 100^2}{2} = \frac{40000 - 10000}{2} = 15000 \text{ J/kg} = 15 \text{ kJ/kg}
Therefore: Δetotal=−500+15=−485 kJ/kg\Delta e_{total} = -500 + 15 = -485 \text{ kJ/kg} Answer A is correct because it properly accounts for both enthalpy and kinetic energy changes. Answer B (-500 kJ/kg) ignores the kinetic energy contribution entirely, considering only the enthalpy change. This is a common error when students forget that total energy includes all energy forms. Answer C incorrectly describes the kinetic energy as "opposing" the enthalpy decrease, leading to a wrong calculation. Answer D (-300 kJ/kg) appears to result from a fundamental misunderstanding of energy conservation principles in turbines. Study tip: Always remember that specific total energy in flowing systems equals specific enthalpy plus specific kinetic energy. Don't neglect kinetic energy changes, especially when velocities change significantly, even though enthalpy changes are typically much larger in magnitude.

Question 8

A piston-cylinder system contains gas at initial state (P₁, V₁, T₁). The gas undergoes a polytropic process PVⁿ = constant to final state (P₂, V₂, T₂). If the internal energy change is ΔU and work done by the gas is W, which expression represents the enthalpy change?

  1. ΔH = ΔU + W because enthalpy includes the work done by the system
  2. ΔH = ΔU + P₂V₂ - P₁V₁ because enthalpy change includes pressure-volume term changes (correct answer)
  3. ΔH = ΔU - W + P₂V₂ - P₁V₁ because work opposes internal energy in enthalpy calculation
  4. ΔH = ΔU + (P₁V₁ + P₂V₂) because enthalpy accounts for both initial and final PV terms
  5. ΔH = ΔU + nR(T₂ - T₁) because enthalpy depends on temperature change for polytropic processes
Explanation: When you encounter thermodynamics problems involving enthalpy changes, remember that enthalpy H is defined as H=U+PVH = U + PV, where U is internal energy. This definition is crucial for understanding how enthalpy changes during any process. For any process, the change in enthalpy is ΔH=ΔU+Δ(PV)\Delta H = \Delta U + \Delta(PV). Since we're dealing with a finite process from state 1 to state 2, this becomes ΔH=ΔU+(P2V2−P1V1)\Delta H = \Delta U + (P_2V_2 - P_1V_1). This relationship holds regardless of the specific process path - whether it's isothermal, adiabatic, or polytropic as in this problem. Choice A incorrectly suggests that enthalpy change equals internal energy change plus work done by the gas (ΔH=ΔU+W\Delta H = \Delta U + W). This confuses the definition of enthalpy with the first law of thermodynamics. Work W relates to energy transfer, but enthalpy specifically accounts for the PV term changes. Choice C adds an extra work term that doesn't belong in the enthalpy definition. This represents a fundamental misunderstanding of what enthalpy measures. Choice D incorrectly adds the initial and final PV terms (P1V1+P2V2P_1V_1 + P_2V_2) rather than taking their difference. Remember, we need the change in the PV term, which is the final value minus the initial value. Choice B correctly applies the fundamental enthalpy relationship: ΔH=ΔU+P2V2−P1V1\Delta H = \Delta U + P_2V_2 - P_1V_1. Study tip: Always start with fundamental definitions in thermodynamics. When you see enthalpy problems, immediately recall H=U+PVH = U + PV and work from there.

Question 9

A gas turbine operates with gas entering at high velocity and exiting at low velocity. If the specific internal energy decreases by 200 kJ/kg, kinetic energy decreases by 50 kJ/kg, and potential energy changes are negligible, what is the change in specific total energy?

  1. -200 kJ/kg because total energy change equals internal energy change when other effects are small
  2. -150 kJ/kg because total energy includes both internal and kinetic energy changes
  3. -250 kJ/kg because all forms of energy decrease and their effects are additive (correct answer)
  4. 50 kJ/kg because kinetic energy change opposes internal energy change in turbines
  5. 200 kJ/kg because the kinetic energy decrease compensates for internal energy decrease
Explanation: When analyzing energy changes in turbomachinery, you need to consider specific total energy, which encompasses all forms of energy per unit mass: internal energy, kinetic energy, and potential energy. The specific total energy change is simply the algebraic sum of all individual energy changes. Here, the internal energy decreases by 200 kJ/kg (change = -200 kJ/kg), kinetic energy decreases by 50 kJ/kg (change = -50 kJ/kg), and potential energy change is negligible (≈ 0). Therefore: Δetotal=Δu+Δke+Δpe=(−200)+(−50)+(0)=−250 kJ/kg\Delta e_{total} = \Delta u + \Delta ke + \Delta pe = (-200) + (-50) + (0) = -250 \text{ kJ/kg} This confirms answer C is correct. Answer A incorrectly assumes you can ignore kinetic energy changes when they're "small." While 50 kJ/kg might seem small compared to 200 kJ/kg, it's still significant and must be included in the total energy calculation. Answer B makes a sign error or conceptual mistake. It appears to subtract the kinetic energy decrease from the internal energy decrease (-200 - (-50) = -150), which doesn't represent any meaningful physical quantity in this context. Answer D completely misunderstands the problem by suggesting the changes "oppose" each other. Both internal and kinetic energy are decreasing (both negative changes), so they add together in the same direction, not oppose each other. Study tip: For turbomachinery problems, always account for all forms of energy explicitly. Total energy is the straightforward sum of internal, kinetic, and potential energy changes—no special interactions or cancellations occur between different energy forms.

Question 10

An ideal gas expands isothermally from volume V to 3V. If the initial internal energy is U₀, what is the relationship between the final enthalpy H₂ and initial enthalpy H₁?

  1. H₂ = H₁ because both internal energy and temperature remain constant during isothermal expansion (correct answer)
  2. H₂ = 3H₁ because enthalpy is proportional to volume for an ideal gas at constant temperature
  3. H₂ = H₁/3 because pressure decreases by factor of 3 while volume increases by factor of 3
  4. H₂ = H₁ + 2PV because the enthalpy includes the additional pressure-volume work term
  5. H₂ < H₁ because the gas does work during expansion, reducing its enthalpy content
Explanation: When analyzing isothermal processes for ideal gases, focus on what "isothermal" means and how it affects the fundamental thermodynamic properties. An isothermal process occurs at constant temperature, which has profound implications for ideal gas behavior. For an ideal gas, both internal energy and enthalpy depend only on temperature. Since U=nCVTU = nC_VT and H=nCPTH = nC_PT for an ideal gas, when temperature remains constant during an isothermal process, both internal energy and enthalpy remain constant regardless of volume or pressure changes. Therefore, H2=H1H_2 = H_1 because the temperature hasn't changed. Let's examine why the other options are incorrect: Option B incorrectly suggests enthalpy is proportional to volume at constant temperature. While volume triples, enthalpy depends only on temperature for ideal gases, not volume. Option C makes the error of trying to relate enthalpy changes to pressure and volume changes independently. Though pressure does decrease by a factor of 3 (from PV=nRTPV = nRT), this doesn't create an inverse relationship with enthalpy. Option D confuses enthalpy with work done. While the gas does perform work during expansion (W=nRTln⁡(Vf/Vi)W = nRT\ln(V_f/V_i)), this work doesn't add to the enthalpy. The enthalpy already accounts for both internal energy and the PVPV term. Study tip: Remember that for ideal gases, internal energy and enthalpy are functions of temperature only. In any isothermal process involving an ideal gas, these state functions remain constant even when pressure, volume, and other properties change dramatically.

Question 11

A gas in a piston-cylinder assembly undergoes a process where its internal energy increases by 80 kJ while 120 kJ of heat is added. If the gas also experiences an increase in kinetic energy of 15 kJ and a decrease in potential energy of 25 kJ, what is the work done by the gas?

  1. 40 kJ because work equals heat added minus internal energy increase (correct answer)
  2. 50 kJ because the net energy change must account for kinetic and potential energy changes
  3. 30 kJ because total energy considerations modify the basic first law calculation
  4. 65 kJ because work includes both thermodynamic work and mechanical energy changes
  5. 200 kJ because work represents the total energy transfer in the process
Explanation: When you encounter a thermodynamics problem involving energy changes in a gas system, you need to apply the first law of thermodynamics, but be careful about which energies are relevant to the thermodynamic analysis. The first law of thermodynamics states: Q=ΔU+WQ = \Delta U + W, where Q is heat added, ΔU is the change in internal energy, and W is work done by the system. Rearranging: W=Q−ΔU=120 kJ−80 kJ=40 kJW = Q - \Delta U = 120 \text{ kJ} - 80 \text{ kJ} = 40 \text{ kJ} Answer A is correct because it properly applies the fundamental first law relationship. The work done by the gas is simply the difference between heat added and internal energy increase. Answer B incorrectly assumes you must include kinetic and potential energy changes in the work calculation. While these are real energy changes, they don't affect the thermodynamic work done by the gas on its surroundings through volume changes. Answer C falls into the same trap as B, attempting to modify the first law calculation with the mechanical energy changes (40−15+25=5040 - 15 + 25 = 50, then somehow getting 30 kJ). The basic first law calculation doesn't need modification for these external energy changes. Answer D incorrectly adds mechanical energy changes to the thermodynamic work (40+15+25=8040 + 15 + 25 = 80, close to 65). This confuses thermodynamic work with total energy accounting. Study tip: In first law problems, distinguish between thermodynamic properties (internal energy, heat, work) and mechanical energies (kinetic, potential). The classic Q=ΔU+WQ = \Delta U + W relationship only involves thermodynamic quantities—don't let additional energy information distract you from this core principle.

Question 12

A system's total energy is defined as E = U + KE + PE. During a process, internal energy increases by 100 J, kinetic energy decreases by 30 J, potential energy increases by 20 J, and 150 J of heat is added. What work was done by the system?

  1. 50 J because work equals heat minus internal energy change according to the first law (correct answer)
  2. 60 J because work must account for all energy changes in the system
  3. 40 J because kinetic and potential energy changes modify the work calculation
  4. 150 J because work done equals heat added when total energy increases
  5. 90 J because work includes both thermodynamic and mechanical energy effects
Explanation: When you encounter a problem involving both the first law of thermodynamics and mechanical energy changes, you need to carefully apply the first law to the system's internal energy, not its total energy. The first law states that ΔU=Q−W\Delta U = Q - W, where ΔU\Delta U is the change in internal energy, QQ is heat added, and WW is work done by the system. Given that internal energy increases by 100 J and 150 J of heat is added, you can solve directly: 100=150−W100 = 150 - W, so W=50W = 50 J. The kinetic and potential energy changes represent mechanical energy transfers that don't affect the thermodynamic work calculation between the system and surroundings. Looking at the incorrect answers: Answer B (60 J) mistakenly tries to incorporate all energy changes, but the first law specifically relates heat, work, and internal energy only. Answer C (40 J) falls into the same trap by attempting to include kinetic and potential energy changes in the work calculation, when these represent separate mechanical effects. Answer D (150 J) incorrectly assumes work equals heat input, ignoring that some heat went into increasing internal energy. Answer A correctly identifies that work equals heat minus internal energy change according to the first law, giving 50 J. Study tip: Always remember that thermodynamic work in the first law refers specifically to energy transfer between system and surroundings through volume changes or other boundary work—not mechanical energy changes within the system. Keep ΔU=Q−W\Delta U = Q - W separate from kinetic and potential energy accounting.

Question 13

A rigid container holds 2 kg of air initially at rest. The container is then placed on a moving platform that accelerates upward at 3 m/s23 \text{ m/s}^2 while simultaneously being heated until the air temperature increases by 50 K. The platform reaches a final velocity of 10 m/s at height 15 m above its starting position. For air, cv=0.718c_v = 0.718 kJ/kg·K. Which energy component contributes most significantly to the total energy change?

  1. Internal energy change of 71.8 kJ dominates due to the substantial temperature rise in the system (correct answer)
  2. Kinetic energy change of 100 kJ is the primary contributor from the platform's acceleration to final velocity
  3. Potential energy change of 294 kJ overshadows other contributions due to gravitational effects at height
  4. All energy components contribute nearly equally, with no single term exceeding 40% of total change
Explanation: Let's calculate each energy component: Internal energy change: ΔU = mc_v ΔT = 2 kg × 0.718 kJ/kg·K × 50 K = 71.8 kJ. Kinetic energy change: ΔKE = ½m(v₂² - v₁²) = ½ × 2 × (10² - 0²) = 100 J = 0.1 kJ. Potential energy change: ΔPE = mgh = 2 × 9.81 × 15 = 294.3 J = 0.294 kJ. The internal energy change (71.8 kJ) is by far the largest component, being over 200 times larger than the kinetic and potential energy changes. Choice B incorrectly states kinetic energy in kJ rather than J. Choice C makes the same unit error for potential energy. Choice D incorrectly suggests comparable magnitudes.

Question 14

A piston-cylinder device contains gas that undergoes an isothermal expansion while work is extracted. During this process, the internal energy remains constant, but the system's elevation changes as the entire apparatus is lifted 25 m vertically. The cylinder mass is 50 kg, and 800 J of heat is added to maintain constant temperature. What is the relationship between the different energy quantities for this system?

  1. Work output and heat input are independent of potential energy change since that affects only mechanical energy
  2. Work output exceeds heat input by the amount of potential energy gained by the apparatus during lifting
  3. Heat input exceeds work output by exactly the potential energy increase needed to lift the apparatus
  4. Work output equals heat input since internal energy change is zero, while total energy increases by gravitational potential (correct answer)
Explanation: For the gas system alone (applying first law to the working fluid): Q = ΔU + W. Since the process is isothermal and ΔU = 0, we have Q = W, so work output equals heat input (800 J). However, considering the total system including the apparatus, the potential energy increases by mgh = 50 × 9.81 × 25 = 12,262.5 J. The total energy of the complete system increases by this potential energy amount while the thermal portion (gas) maintains energy balance between heat and work. Choice B incorrectly suggests work exceeds heat input. Choice C incorrectly reverses the relationship. Choice D incorrectly suggests independence between thermal and mechanical energy changes.

Question 15

An engineer must determine the total energy per unit mass of fluid in a pipeline carrying liquid water at 80°C, 500 kPa, flowing at 8 m/s through a pipe located 45 m above a reference datum. Given that specific internal energy u=335u = 335 kJ/kg and specific volume v=0.001029v = 0.001029 m³/kg at these conditions, which component contributes least to the total specific energy?

  1. Flow energy (pressure-volume work) contributes only 0.515 kJ/kg, making it the smallest component
  2. Kinetic energy contributes only 0.032 kJ/kg, representing the minimal component of total energy (correct answer)
  3. Potential energy contributes only 0.441 kJ/kg, making it smaller than other energy forms
  4. Internal energy at 335 kJ/kg is actually the smallest when compared to other specific energy forms
Explanation: Let's calculate each specific energy component: Internal energy: u = 335 kJ/kg (given). Flow energy (Pv): Pv = 500 × 0.001029 = 0.5145 kJ/kg. Kinetic energy: ½v² = ½(8²)/1000 = 0.032 kJ/kg. Potential energy: gz = 9.81 × 45/1000 = 0.441 kJ/kg. Comparing all components: internal energy (335 kJ/kg) >> flow energy (0.515 kJ/kg) > potential energy (0.441 kJ/kg) > kinetic energy (0.032 kJ/kg). Kinetic energy is clearly the smallest component. Choice A incorrectly identifies flow energy as smallest. Choice C incorrectly identifies potential energy as smallest. Choice D absurdly suggests internal energy is smallest when it's actually dominant.

Question 16

Two identical blocks of copper, each with mass 5 kg and initial temperature 300 K, undergo different processes. Block A is heated at constant volume until its temperature reaches 400 K. Block B is heated at constant pressure until it also reaches 400 K, but during this process it expands and performs work against the atmosphere. If cv=0.385c_v = 0.385 kJ/kg·K and cp=0.395c_p = 0.395 kJ/kg·K for copper, what is the fundamental difference in internal energy changes between the two blocks?

  1. Block A has greater internal energy change by 50 kJ due to constant volume heating being more efficient
  2. Block B has greater internal energy change by 5 kJ because constant pressure processes add more thermal energy
  3. Both blocks have identical internal energy changes of 192.5 kJ since internal energy depends only on temperature for solids (correct answer)
  4. Block A has internal energy change of 192.5 kJ while Block B has 197.5 kJ due to pressure-volume work effects
Explanation: Internal energy is a state function that depends only on temperature for solids (and ideal gases). Since both blocks start at 300 K and end at 400 K, they undergo the same change in internal energy: ΔU = mc_v ΔT = 5 × 0.385 × (400-300) = 192.5 kJ for both blocks. The difference between the processes lies in the heat transfer: Block A requires Q = ΔU = 192.5 kJ, while Block B requires Q = mc_p ΔT = 5 × 0.395 × 100 = 197.5 kJ. The extra 5 kJ for Block B goes into expansion work, not internal energy change. Choices A, B, and D incorrectly suggest different internal energy changes based on process path rather than recognizing internal energy as a state function.

Question 17

A closed system consists of 3 kg of an ideal gas initially at 25°C. The system undergoes a process where 15 kJ of work is done on the gas while 8 kJ of heat is rejected to the surroundings. Simultaneously, the entire system (container included) is accelerated from rest to 20 m/s horizontally. If the container mass is 12 kg, what is the change in internal energy of the gas alone?

  1. Internal energy increases by 23 kJ due to work addition and heat rejection effects on the gas
  2. Internal energy increases by 7 kJ from the net effect of work input minus heat rejection (correct answer)
  3. Internal energy increases by 25.4 kJ when including the kinetic energy gained by the gas motion
  4. Internal energy decreases by 7 kJ since heat rejection exceeds work input for the gas system
Explanation: For the gas alone as a closed system, the first law states: ΔU = Q - W. Here, Q = -8 kJ (heat rejected, negative sign) and W = -15 kJ (work done on the system, negative sign in our convention where positive work is done by the system). Therefore: ΔU = (-8) - (-15) = -8 + 15 = 7 kJ. The acceleration of the container affects the kinetic energy of the system but not the internal energy of the gas, which depends only on molecular motion and temperature. Choice A incorrectly adds work and heat. Choice C incorrectly includes macroscopic kinetic energy in internal energy. Choice D uses incorrect sign convention for work.