Q: Specific Heat Ratio of Air – Typical Value Near Room Conditions For dry air near standard conditions, the specific heat ratio gamma = Cp/Cv is approximately 1.4.

Introduction / Context: The specific heat ratio gamma affects speed of sound, isentropic relations, nozzle flows, and cycle efficiencies. Knowing its typical value for air is crucial in many thermofluid calculations. Given Data / Assumptions: Dry air modeled as a diatomic ideal gas near room temperature. Calorically perfect approximation over a modest temperature range. Cp ≈ 1.005 kJ/kg-K, Cv ≈ 0.718 kJ/kg-K. Concept / Approach: Gamma is defined as Cp/Cv. For diatomic gases like nitrogen and oxygen (major constituents of air), rotational degrees of freedom lead to Cp and Cv values that yield gamma ≈ 1.4 at ambient conditions. Gamma decreases slightly with temperature as vibrational modes activate. Step-by-Step Solution: Recall Cp and Cv for air: Cp ≈ 1.005, Cv ≈ 0.718 kJ/kg-K.Compute gamma = 1.005 / 0.718 ≈ 1.40.State that this value is standard in compressible-flow and acoustic calculations.Note mild temperature dependence for precision work. Verification / Alternative check: Speed of sound a = sqrt(gamma * R * T) using gamma = 1.4 matches measured acoustic velocities in air near 20°C, supporting the typical value. Why Other Options Are Wrong: 1 implies monatomic-like ratio at very high temperatures (not typical); 1.8 is unrealistic for air; 23 is nonsensical; 1.2 is more typical of high-temperature effective gamma in some mixtures, not room conditions. Common Pitfalls: Using gamma = 1.4 indiscriminately at very high temperatures or with humid air where deviations occur; confusing Cp/Cv with Cv/Cp. Final Answer: 1.4

Q: Stirling Cycle – Process Composition Check A Stirling cycle consists of two isothermal processes and two constant-volume (isochoric) regenerative processes.

Introduction / Context: The Stirling cycle is a classic external-combustion, closed-cycle model employing regeneration. Understanding its process makeup clarifies how it can, in principle, approach Carnot efficiency with perfect regeneration and isothermal heat exchange. Given Data / Assumptions: Ideal gas working fluid. Two isothermal processes at T_H and T_L. Two isochoric regenerative processes that shuttle heat internally. Concept / Approach: The Stirling sequence: isothermal compression at T_L with heat rejection to a sink, constant-volume regenerative heating, isothermal expansion at T_H with heat addition from a source, and constant-volume regenerative cooling back to T_L. The constant-volume steps store/recover heat in the regenerator matrix, minimizing external heat transfer at intermediate temperatures. Step-by-Step Solution: Identify the two isotherms: expansion at T_H and compression at T_L.Recognize the remaining two legs are at constant volume (regeneration steps).Confirm no isentropic (adiabatic and reversible) legs appear in the ideal Stirling cycle.Thus, the correct completion is “two constant volume”. Verification / Alternative check: T–s and p–v diagrams of the ideal Stirling show horizontal isotherms and vertical isochors; the regenerator provides internal heat interchange between those isochors. Why Other Options Are Wrong: Constant-pressure and isentropic legs characterize Brayton/Ericsson or Otto/Diesel cycles, not Stirling. Mixed CP/CV is not Stirling either. Common Pitfalls: Confusing Stirling with Ericsson (which has constant-pressure, not constant-volume, regeneration) or Brayton (adiabatic compression/expansion). Final Answer: two constant volume

Q: Carbonisation of Coal – Temperature Range for Hard (Metallurgical) Coke Hard coke is obtained by high-temperature carbonisation of coal, typically in the range of about 900° to 1100°C.

Introduction / Context: Coking converts coal into a carbon-rich, porous solid used in metallurgical furnaces. The temperature regime of carbonisation determines the properties of the product, including strength, porosity, and volatile content. Given Data / Assumptions: Carbonisation in the absence of air (destructive distillation). Controlled heating with by-product recovery in modern coke ovens. Distinct low- and high-temperature regimes produce different coke qualities. Concept / Approach: Low-temperature carbonisation (about 500°–700°C) yields softer, smokeless fuels with higher volatile content. High-temperature carbonisation (about 900°–1100°C) drives off volatiles more completely and sinters the carbon matrix, forming hard, strong metallurgical coke required for blast furnaces. Step-by-Step Solution: Classify regimes: low-temperature vs. high-temperature carbonisation.Associate product properties: soft vs. hard coke.Select the temperature band that delivers hard coke: 900°–1100°C.Conclude this is the correct operating range for metallurgical coke. Verification / Alternative check: Industrial practice and materials handbooks consistently cite ~1000°C as a representative temperature for strong coke production. Why Other Options Are Wrong: Lower ranges leave too much volatile matter and inadequate structure; significantly higher ranges are uncommon and risk excessive energy input and structural degradation. Common Pitfalls: Confusing soft (domestic) coke with metallurgical coke; overlooking the role of heating rate and coal blend on final properties. Final Answer: 900° to 1100°C

Question 1

Phase terminology – state after complete evaporation from the liquid The state of a substance whose evaporation from its liquid state is complete is commonly referred to as:

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Introduction / Context: Thermodynamics distinguishes between phases and special names used for specific substances. After a liquid has completely evaporated, the substance is in the gaseous phase; general terminology is important for clarity. Given Data / Assumptions: No liquid remains; the state is entirely gaseous at the prevailing pressure and temperature. Terminology is generic (not restricted to water). Equilibrium state is assumed. Concept / Approach: For any substance, the gaseous state coexisting with or derived from the liquid near saturation is called vapour. For water specifically, the term steam is used; when perfectly dry at saturation it is called dry saturated steam. In generic multiple-choice contexts, “vapour” is the broad, substance-agnostic term. Step-by-Step Solution: Identify that the liquid has fully evaporated → the phase is gaseous.Choose the general term for post-evaporation state: vapour.Note: for water, “steam” is common; however, the question does not limit to water. Verification / Alternative check: Steam tables are specific to water; refrigerant tables label the gaseous phase as refrigerant vapour, underscoring the general applicability of the term “vapour.” Why Other Options Are Wrong: Perfect gas: an idealization with specific equation-of-state assumptions; not a phase label. Air: a particular gas mixture, not a phase of a single substance. Steam: correct for water specifically but not general to all substances; the question is generic. Liquid–vapour mixture: implies partial, not complete, evaporation. Common Pitfalls: Using “steam” as a universal term; it properly refers only to water in engineering usage. Final Answer: vapour

Question 2

Specific-heat ratio – monoatomic ideal gas Select the correct value of the ratio of molar specific heats gamma = cp/cv for a monoatomic ideal gas:

Accepted Answer

Introduction / Context: The specific-heat ratio gamma influences wave speeds, nozzle flows, and thermodynamic relations. Its value depends on molecular degrees of freedom. Given Data / Assumptions: Monoatomic ideal gas (e.g., He, Ne, Ar). Equipartition theorem applies at ordinary temperatures. Molar specific heats are constant over the range considered. Concept / Approach: For a monoatomic ideal gas, only three translational degrees of freedom contribute at room temperature. Therefore cv = (3/2) * R and cp = cv + R = (5/2) * R, giving gamma = cp/cv = (5/2)R / [(3/2)R] = 5/3 ≈ 1.67. Step-by-Step Solution: Write cv = 3R/2.Compute cp = cv + R = 5R/2.Form the ratio gamma = cp/cv = (5/2)/(3/2) = 5/3.Convert to decimal: 5/3 ≈ 1.67. Verification / Alternative check: Measured gamma for noble gases at room temperature matches ~1.66–1.67, validating the ideal-gas, constant-heat-capacity assumption. Why Other Options Are Wrong: 1.4 is typical for diatomic gases (e.g., air) at moderate temperatures. 1 or 1.3 contradict monoatomic degrees of freedom. 1.87 exceeds typical values even for restricted rotational modes. Common Pitfalls: Using air's gamma for all gases; gamma depends strongly on molecular structure. Final Answer: 1.67

Question 3

Phase terminology – partial evaporation and the definition of steam Evaluate the statement: “The state of a substance whose evaporation from its liquid state is partial is known as steam.”

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Introduction / Context: Correct naming of phase states prevents misinterpretation of property tables. Partial evaporation produces a two-phase mixture; we must distinguish this from a pure vapor state and from the special term “steam.” Given Data / Assumptions: Substance has both liquid and vapor present (wet condition). Equilibrium at a given pressure–temperature pair on the saturation dome. Engineering usage of the word “steam” is considered. Concept / Approach: When evaporation is partial, the state is a saturated liquid–vapor mixture (wet state). For water, the term “wet steam” is used, but “steam” by itself generally signifies water vapor, often dry saturated or superheated. The statement claiming “steam” as the generic name for any partial evaporation is therefore inaccurate. Step-by-Step Solution: Identify the state: partial evaporation → two-phase mixture.Name the state correctly: saturated mixture (wet region) with quality x between 0 and 1.Clarify terminology: “steam” commonly refers to the vapor phase of water; “wet steam” specifies a mixture, not a pure state.Conclude the given statement is false as written. Verification / Alternative check: Steam tables differentiate saturated liquid, saturated vapor (dry steam), and wet mixture with defined quality x; the terminology supports the distinction. Why Other Options Are Wrong: Conditional agreements do not fix the categorical misuse; the correct generic term is “saturated mixture,” not simply “steam.” Common Pitfalls: Using “steam” ambiguously; always specify wet, dry saturated, or superheated to avoid confusion. Final Answer: Disagree

Question 4

Sources of irreversibility – identifying the primary causes Select the main causes of thermodynamic irreversibility in real processes:

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Introduction / Context: Irreversibility reduces the maximum useful work obtainable and increases entropy. Recognizing its sources helps engineers design more efficient systems. Given Data / Assumptions: Macroscopic, classical thermodynamics framework. Real processes involving fluids and machines. Neglect of microscopic fluctuations; focus on engineering causes. Concept / Approach: Common irreversibilities include friction (mechanical, viscous), free expansion and unrestrained mixing, and heat transfer across finite temperature differences. Each produces entropy and prevents recovery of energy as useful work. Step-by-Step Solution: Mechanical/fluid friction dissipates mechanical energy into internal energy, increasing entropy.Unrestricted expansion or mixing proceeds without a restoring force, creating entropy directly.Heat transfer with finite ΔT causes entropy generation because Q flows spontaneously from hot to cold with σ = Q * (1/T_cold − 1/T_hot) > 0.Therefore, all listed mechanisms contribute to irreversibility. Verification / Alternative check: Entropy balance for control volumes confirms positive entropy generation (S_gen ≥ 0) in each of these cases, aligning with the second law. Why Other Options Are Wrong: Any single cause alone is incomplete; real systems typically have several irreversibilities simultaneously. “None of the above” contradicts basic thermodynamics and observation. Common Pitfalls: Assuming that eliminating one source (e.g., mechanical friction) makes a process reversible; finite ΔT heat transfer or fluid friction still generate entropy. Final Answer: all of the above

Question 5

Kinetic Theory of Gases – Translational Kinetic Energy vs. Absolute Temperature According to the kinetic theory, the average translational kinetic energy of a gas molecule is directly proportional to the absolute temperature (on the Kelvin scale).

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Introduction / Context: The kinetic theory of gases links microscopic molecular motion with macroscopic thermodynamic properties. A cornerstone result is that the average translational kinetic energy of molecules scales with absolute temperature, providing a physical meaning to temperature as a measure of molecular agitation. Given Data / Assumptions: Ideal gas with negligible intermolecular potential energy during collisions. Random, isotropic molecular motion (no preferred direction). Equilibrium state so that averages are meaningful. Concept / Approach: For an ideal monatomic gas, the mean translational kinetic energy per molecule is (3/2) * k_B * T, where k_B is Boltzmann’s constant and T is absolute temperature. This shows a linear dependence on T. While the numeric prefactor can change with degrees of freedom in more complex molecules, the proportionality to T for translational modes remains fundamental. Step-by-Step Solution: Interpret temperature microscopically: higher T means faster molecules on average.Relate molecular speed to energy: KE_avg ∝ m * (speed_rms)^2.Use kinetic-theory result: KE_avg = (3/2) * k_B * T → linear in T.Conclude the proportionality: translational kinetic energy ∝ T. Verification / Alternative check: Root-mean-square speed u_rms satisfies u_rms ∝ sqrt(T). Since KE ∝ u_rms^2, it follows directly that KE ∝ T. Calorimetric measurements of heat capacity also align with this relationship for gases near ideal behavior. Why Other Options Are Wrong: 1/T and constant independence contradict the kinetic-theory equations. T^2 is too strong a dependence; the correct law is first power of T. Pressure-only dependence ignores the equation of state and microscopic basis. Common Pitfalls: Confusing energy per mole with energy per molecule; mixing total internal energy (which includes all modes) with purely translational kinetic energy; using Celsius instead of Kelvin. Final Answer: proportional to T (absolute temperature)

Question 6

Irreversible Process – What Must Be True? In any natural (irreversible) thermodynamic process, the entropy generation is positive; equivalently, the entropy of the universe increases during the process.

Accepted Answer

Introduction / Context: Reversibility classifies processes by whether they can be undone without leaving changes in the surroundings. The second law states that real, natural processes are irreversible and produce entropy. Recognizing the correct hallmark of irreversibility is fundamental to thermal system analysis and efficiency estimation. Given Data / Assumptions: Closed or open systems may be considered; surroundings are large reservoirs. Irreversibilities include friction, unrestrained expansion, heat transfer across a finite temperature difference, mixing, and inelastic deformation. Standard sign conventions for heat and work. Concept / Approach: The combined system (system + surroundings) obeys ΔS_universe = S_gen ≥ 0. For an internally reversible process, S_gen = 0; for an irreversible one, S_gen > 0. The presence or absence of heat loss is not a defining feature, because a process can be adiabatic yet still irreversible (e.g., throttling, rapid expansion). Step-by-Step Solution: Identify the second-law criterion: ΔS_universe = S_gen.For irreversible processes, set S_gen > 0 regardless of Q or W specifics.Note that individual system heat or work terms can be positive, negative, or zero; irreversibility depends on entropy generation.Conclude the correct invariant statement: S_gen is positive in irreversibility. Verification / Alternative check: Examples: Throttling valve (Joule–Thomson) has Q ≈ 0 and W ≈ 0 but S_gen > 0 due to internal friction and mixing. Heat transfer across a finite ΔT yields S_gen = Q * (1/T_cold − 1/T_hot) > 0. Why Other Options Are Wrong: Net heat loss is not necessary; adiabatic yet irreversible paths exist. Zero work is not required; turbines produce work irreversibly. Spontaneous reversal violates the second law. Constant system entropy implies reversibility only when the environment change is appropriately accounted for. Common Pitfalls: Equating “irreversible” with “adiabatic” or “no heat”. Focusing on system-only entropy without considering surroundings. Assuming irreversibility always reduces temperature or pressure (it need not). Final Answer: Entropy generation is positive (S_gen > 0).

Question 7

Heating a Gas at Constant Volume – What Changes? When a gas is heated in a rigid, constant-volume container, its temperature rises and, for an ideal gas, the pressure increases proportionally to the absolute temperature.

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Introduction / Context: Understanding which properties change during constrained heating is essential for safe vessel design and control. The classic case of a rigid tank emphasizes the role of the ideal gas law and constant-volume specific heat in predicting state changes. Given Data / Assumptions: Rigid, sealed container (V = constant). Gas behaves ideally over the temperature range considered. No mass enters or leaves; neglect changes in kinetic and potential energy. Concept / Approach: Energy balance for a closed, rigid system gives Q = ΔU = m * Cv * (T2 − T1). Thus, adding heat raises temperature. The ideal gas law pV = mRT then implies p ∝ T when V and m are fixed, so pressure also increases proportionally to absolute temperature. Step-by-Step Solution: Write Q = ΔU since W = ∫ p dV = 0 for constant volume.Relate ΔU to temperature: ΔU = m * Cv * (T2 − T1) → T2 > T1.Use ideal gas law: p2/p1 = T2/T1 at constant m and V.Conclude: both T and p increase upon heating at constant volume. Verification / Alternative check: Manometer readings in a heated rigid cylinder rise with temperature; tabulated compressibility factors near unity corroborate the proportionality for many gases at moderate pressures. Why Other Options Are Wrong: Temperature or pressure increasing alone contradicts the coupled relationship via pV = mRT. No change in either violates the first law for nonzero Q. A decrease in temperature upon heating is unphysical for a simple gas in a rigid container. Common Pitfalls: Confusing constant-volume with constant-pressure heating; forgetting that even small temperature rises can cause significant pressure increases in tightly sealed vessels. Final Answer: both temperature and pressure will increase

Question 8

Ericsson Cycle vs. Carnot – Ideal Thermal Efficiency with Perfect Regeneration For an ideal Ericsson cycle (two isothermal and two constant-pressure processes) operating with perfect regeneration between the isobars, the thermal efficiency is equal …

Accepted Answer

Introduction / Context: The Ericsson and Stirling cycles are regenerative cycles composed of isothermal heat transfer processes. With an ideal (perfect) regenerator, both can theoretically achieve the Carnot efficiency when operating between the same high and low temperatures, despite having different process combinations than Carnot. Given Data / Assumptions: Ideal gas as working fluid. Two isothermal and two constant-pressure processes (Ericsson). Perfect regenerator shifting heat internally between the two isobars; no pressure losses; fully reversible processes. Concept / Approach: Carnot efficiency depends only on the two reservoir temperatures: eta_Carnot = 1 − T_L/T_H. An ideal Ericsson cycle, made fully reversible via perfect regeneration, has no net internal temperature differences during heat transfer other than at the reservoirs, making its efficiency depend solely on T_H and T_L, the same as Carnot. Step-by-Step Solution: Identify Ericsson sequence: isothermal expansion at T_H, isobaric regeneration cooling, isothermal compression at T_L, isobaric regeneration heating.Assume perfect regenerator so that external heat is added only at T_H and rejected only at T_L.Conclude that the cycle is internally reversible and thus achieves eta = 1 − T_L/T_H.Therefore Ericsson efficiency equals Carnot under these idealizations. Verification / Alternative check: Derivations using T–s diagrams show equal net area (work) relative to heat in/out purely at isotherms when regeneration is perfect. Any regenerator inefficiency or pressure drop reduces efficiency below Carnot in practice. Why Other Options Are Wrong: Greater than Carnot is impossible by the second law. Less than applies when regeneration is imperfect, but the ideal statement specifies perfect regeneration. “Undefined” ignores the specified idealization. Common Pitfalls: Confusing Ericsson with Brayton (which has adiabatic legs, not isothermal) and overlooking the crucial role of perfect regeneration. Final Answer: equal to

Question 9

Fuels – Coal Rank vs. Calorific Value Among the listed types of coal, anthracite has the highest calorific value (highest fixed carbon and lowest volatile matter).

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Introduction / Context: Coal is categorized by rank based on fixed carbon, volatile matter, moisture, and mineral content. Calorific value typically increases with rank. Recognizing which coal yields more energy per unit mass is important for combustion system design and fuel selection. Given Data / Assumptions: Standard coal ranks: peat, lignite, sub-bituminous, bituminous, anthracite. Higher fixed carbon and lower moisture/volatiles raise heating value. Comparison at similar reference conditions (as-received may vary). Concept / Approach: Anthracite, the highest rank coal, has the greatest fixed carbon content and the least volatile matter, leading to the highest gross calorific value among common coals. Peat and lignite are low-rank with high moisture and low heating value; bituminous is intermediate. Step-by-Step Solution: Order ranks by maturity: peat < lignite < sub-bituminous < bituminous < anthracite.Associate calorific value with rank: increases along the list.Select the highest-rank option: anthracite.Conclude it has the highest calorific value among the choices. Verification / Alternative check: Typical higher heating values: lignite ~10–20 MJ/kg, bituminous ~24–35 MJ/kg, anthracite ~30–35+ MJ/kg, with lower moisture making anthracite most energy-dense. Why Other Options Are Wrong: Peat/lignite are moisture-rich and energy-poor; bituminous is lower than anthracite; sub-bituminous is intermediate, not maximal. Common Pitfalls: Confusing “best flame stability” or “ease of ignition” with highest calorific value; considering as-received vs. dry basis inconsistently. Final Answer: Anthracite coal

Question 10

Specific Heat Ratio of Air – Typical Value Near Room Conditions For dry air near standard conditions, the specific heat ratio gamma = Cp/Cv is approximately 1.4.

Accepted Answer

Introduction / Context: The specific heat ratio gamma affects speed of sound, isentropic relations, nozzle flows, and cycle efficiencies. Knowing its typical value for air is crucial in many thermofluid calculations. Given Data / Assumptions: Dry air modeled as a diatomic ideal gas near room temperature. Calorically perfect approximation over a modest temperature range. Cp ≈ 1.005 kJ/kg-K, Cv ≈ 0.718 kJ/kg-K. Concept / Approach: Gamma is defined as Cp/Cv. For diatomic gases like nitrogen and oxygen (major constituents of air), rotational degrees of freedom lead to Cp and Cv values that yield gamma ≈ 1.4 at ambient conditions. Gamma decreases slightly with temperature as vibrational modes activate. Step-by-Step Solution: Recall Cp and Cv for air: Cp ≈ 1.005, Cv ≈ 0.718 kJ/kg-K.Compute gamma = 1.005 / 0.718 ≈ 1.40.State that this value is standard in compressible-flow and acoustic calculations.Note mild temperature dependence for precision work. Verification / Alternative check: Speed of sound a = sqrt(gamma * R * T) using gamma = 1.4 matches measured acoustic velocities in air near 20°C, supporting the typical value. Why Other Options Are Wrong: 1 implies monatomic-like ratio at very high temperatures (not typical); 1.8 is unrealistic for air; 23 is nonsensical; 1.2 is more typical of high-temperature effective gamma in some mixtures, not room conditions. Common Pitfalls: Using gamma = 1.4 indiscriminately at very high temperatures or with humid air where deviations occur; confusing Cp/Cv with Cv/Cp. Final Answer: 1.4

Question 11

Stirling Cycle – Process Composition Check A Stirling cycle consists of two isothermal processes and two constant-volume (isochoric) regenerative processes.

Accepted Answer

Introduction / Context: The Stirling cycle is a classic external-combustion, closed-cycle model employing regeneration. Understanding its process makeup clarifies how it can, in principle, approach Carnot efficiency with perfect regeneration and isothermal heat exchange. Given Data / Assumptions: Ideal gas working fluid. Two isothermal processes at T_H and T_L. Two isochoric regenerative processes that shuttle heat internally. Concept / Approach: The Stirling sequence: isothermal compression at T_L with heat rejection to a sink, constant-volume regenerative heating, isothermal expansion at T_H with heat addition from a source, and constant-volume regenerative cooling back to T_L. The constant-volume steps store/recover heat in the regenerator matrix, minimizing external heat transfer at intermediate temperatures. Step-by-Step Solution: Identify the two isotherms: expansion at T_H and compression at T_L.Recognize the remaining two legs are at constant volume (regeneration steps).Confirm no isentropic (adiabatic and reversible) legs appear in the ideal Stirling cycle.Thus, the correct completion is “two constant volume”. Verification / Alternative check: T–s and p–v diagrams of the ideal Stirling show horizontal isotherms and vertical isochors; the regenerator provides internal heat interchange between those isochors. Why Other Options Are Wrong: Constant-pressure and isentropic legs characterize Brayton/Ericsson or Otto/Diesel cycles, not Stirling. Mixed CP/CV is not Stirling either. Common Pitfalls: Confusing Stirling with Ericsson (which has constant-pressure, not constant-volume, regeneration) or Brayton (adiabatic compression/expansion). Final Answer: two constant volume

Question 12

Carbonisation of Coal – Temperature Range for Hard (Metallurgical) Coke Hard coke is obtained by high-temperature carbonisation of coal, typically in the range of about 900° to 1100°C.

Accepted Answer

Introduction / Context: Coking converts coal into a carbon-rich, porous solid used in metallurgical furnaces. The temperature regime of carbonisation determines the properties of the product, including strength, porosity, and volatile content. Given Data / Assumptions: Carbonisation in the absence of air (destructive distillation). Controlled heating with by-product recovery in modern coke ovens. Distinct low- and high-temperature regimes produce different coke qualities. Concept / Approach: Low-temperature carbonisation (about 500°–700°C) yields softer, smokeless fuels with higher volatile content. High-temperature carbonisation (about 900°–1100°C) drives off volatiles more completely and sinters the carbon matrix, forming hard, strong metallurgical coke required for blast furnaces. Step-by-Step Solution: Classify regimes: low-temperature vs. high-temperature carbonisation.Associate product properties: soft vs. hard coke.Select the temperature band that delivers hard coke: 900°–1100°C.Conclude this is the correct operating range for metallurgical coke. Verification / Alternative check: Industrial practice and materials handbooks consistently cite ~1000°C as a representative temperature for strong coke production. Why Other Options Are Wrong: Lower ranges leave too much volatile matter and inadequate structure; significantly higher ranges are uncommon and risk excessive energy input and structural degradation. Common Pitfalls: Confusing soft (domestic) coke with metallurgical coke; overlooking the role of heating rate and coal blend on final properties. Final Answer: 900° to 1100°C

Question 13

Open vs. Closed Cycle Gas Turbines – Identify the Wrong Statement Select the incorrect statement regarding open- and closed-cycle gas turbine plants.

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Introduction / Context: Gas turbine plants are implemented as open (Brayton with atmospheric intake and exhaust) or closed cycles (recirculated working fluid through external heat addition and rejection). Distinguishing their characteristics is vital for plant design and performance analysis. Given Data / Assumptions: Open cycle exchanges mass with atmosphere; closed cycle does not. Closed cycle uses external heat exchangers (heater, cooler, regenerator). Pressures in a closed cycle are set by plant design, not by ambient pressure. Concept / Approach: The incorrect statement is the one tying closed-cycle pressure range to atmospheric pressure. In closed cycles, absolute pressure levels and ratios are chosen independently (within mechanical limits) since the working fluid is contained. In contrast, open cycles usually have compressor inlet near atmospheric pressure, so their absolute levels depend on ambient conditions. Step-by-Step Solution: Evaluate (a): true—heat is added externally to the working fluid in heaters.Evaluate (b): false—closed-cycle pressures are not constrained by atmosphere.Evaluate (c): acceptable phrasing—combustion is external with associated heat-exchanger surfaces.Evaluate (d): true—open cycle exhausts to atmosphere acting as the sink.Include (e) to clarify true closed-cycle feature. Verification / Alternative check: Textbook schematics show a closed Brayton loop pressurized to several bars for compactness and efficiency, independent of ambient pressure. Why Other Options Are Wrong: Options (a), (d), and (e) reflect standard definitions; (c) is a qualitative statement about external surfaces and does not claim false dependencies. Common Pitfalls: Assuming all gas turbines must inhale/exhaust to atmosphere; forgetting that pressurizing a closed loop improves heat-exchanger effectiveness and specific power. Final Answer: In the closed cycle gas turbine, the pressure range depends upon the atmospheric pressure.

Question 14

First Law for a Closed System – Choose the Correct Statement Which of the following correctly expresses the First Law of Thermodynamics for a closed system over a process from state 1 to 2?

Accepted Answer

Introduction / Context: The First Law of Thermodynamics is the statement of energy conservation for thermodynamic systems. For closed systems, it relates heat transfer, work, and the change in total energy (internal, kinetic, potential). Selecting the correct form avoids many conceptual mistakes in problem solving. Given Data / Assumptions: Closed system (fixed mass) undergoes a process from state 1 to state 2. Sign convention: heat added to the system positive; work done by the system positive. Total energy E includes U + KE + PE unless stated negligible. Concept / Approach: The general First-Law statement for a closed system is ΔE = Q − W. In words: the increase in system energy equals the net heat added minus the work done by the system. Special cases (adiabatic, isochoric, cyclic) are all contained within this master equation by setting appropriate terms to zero. Step-by-Step Solution: Write ΔE = E2 − E1 = (U2 − U1) + ΔKE + ΔPE.Apply the First Law: ΔE = Q − W for the chosen sign convention.Recover common results: adiabatic (Q = 0) → ΔE = −W; isochoric (W = 0) → ΔE = Q.Recognize that equality Q = W holds only when ΔE = 0 (e.g., steady cyclic), not in general. Verification / Alternative check: Control-mass piston–cylinder analyses and calorimetry experiments consistently validate ΔE = Q − W across diverse processes when all energy contributions are accounted for. Why Other Options Are Wrong: (a) Internal energy is not constant in general; (c) is only true for ΔE = 0 cases; (d) incorrectly groups properties—entropy is not conserved in irreversible processes and enthalpy is path-dependent except in special cases; (e) is false because heat and work are path functions, not state properties. Common Pitfalls: Mixing sign conventions, neglecting kinetic/potential energies when they matter, and assuming cyclic identities apply to noncyclic processes. Final Answer: The total energy change equals heat added minus work done: ΔE = Q − W.

Question 15

Thermodynamics — Definition of an Isolated System In an isolated thermodynamic system, neither heat nor work crosses the system boundary, and no mass transfer occurs. Evaluate the statement: “In an isolated system, neither heat and work nor the mass…

Accepted Answer

Introduction / Context: Thermodynamic systems are categorized by what can cross their boundaries. The strictest case is an isolated system, a cornerstone concept for applying conservation laws and the second law to the “universe” comprising system plus surroundings. Given Data / Assumptions: System boundary is perfectly adiabatic (no heat transfer). Boundary is perfectly rigid and non-working (no shaft/boundary work or any other work mode). No mass can enter or leave the system control mass. Concept / Approach: By definition, an isolated system exchanges neither energy nor matter with its surroundings. Energy exchanges include heat (Q) and work (W) of any kind (shaft, electrical, boundary). Therefore, for an isolated system over any process: Q = 0, W = 0, and mass_in = mass_out = 0. Step-by-Step Solution: Identify what the statement claims: no heat, no work, and no mass across the boundary.Recall the formal definition of an isolated system: no interaction at all with surroundings in terms of energy or matter.Match the two: the statement precisely restates the definition of an isolated system. Verification / Alternative check: Considering the “system + surroundings” as an isolated whole (the universe) helps verify that internal exchanges may occur within subsystems, but nothing crosses the overarching isolated boundary. Classic examples include an idealized, perfectly insulated, sealed rigid tank undergoing internal mixing with no external connections. Why Other Options Are Wrong: “False” would allow some combination of heat, work, or mass transfer, contradicting the strict definition of an isolated system. Common Pitfalls: Confusing “closed” systems (no mass but possible heat/work) with “isolated” systems. Another confusion is assuming only heat is blocked; in isolation, both heat and all forms of work are prevented. Final Answer: True

Question 16

Air-Standard Cycles — Diesel vs. Otto at the Same Compression Ratio For the same compression ratio r, how does the air-standard efficiency of the Diesel cycle compare with that of the Otto cycle?

Accepted Answer

Introduction / Context: Otto and Diesel cycles are idealized models for spark-ignition and compression-ignition engines. Comparing their efficiencies at equal compression ratio illuminates the effect of heat-addition mode (constant volume vs. constant pressure) on cycle performance. Given Data / Assumptions: Air-standard analysis with ideal gas, constant specific heats. Same compression ratio r = V1/V2 for both cycles. Reversible, internally idealized processes (no friction, no real losses). Concept / Approach: For the Otto cycle, heat is added at constant volume, producing a higher peak temperature for the same r. For the Diesel cycle, a portion of heat is added at constant pressure (finite cutoff ratio), lowering the average temperature at which heat is added and reducing thermal efficiency relative to Otto when r is the same. Hence, eta_Diesel < eta_Otto for equal r (for any cutoff ratio > 1). Step-by-Step Solution: Start from standard efficiency relations: eta_Otto = 1 - 1/r^(k-1).Diesel efficiency depends on both r and cutoff ratio rc: eta_Diesel = 1 - (1/r^(k-1)) * ((rc^k - 1)/(k*(rc - 1))).For rc > 1, the multiplicative factor reduces eta_Diesel below eta_Otto at the same r. Verification / Alternative check: Plotting efficiencies vs. r for typical k = 1.4 shows the Otto curve above the Diesel curve for the same r, converging only as rc → 1 (which would approach Otto behavior). Why Other Options Are Wrong: “Greater than” or “equal to” would require zero cutoff (rc = 1) or different r; those conditions violate the stated comparison with typical Diesel heat addition. Common Pitfalls: Confusing real-engine compression ratios (Diesel engines often use higher r than SI engines) with the theoretical comparison at equal r. In practice, higher Diesel r can yield higher efficiency, but that is a different comparison. Final Answer: less than

Question 17

Reversible Cycles — Necessary Conditions for Reversibility Which condition(s) must be satisfied for a thermodynamic cycle to be reversible in practice and principle?

Accepted Answer

Introduction / Context: Reversibility is an idealization central to defining the upper limit of efficiency. Real cycles can only approach, never reach, perfect reversibility. Understanding the conditions that must hold clarifies why all real devices generate entropy and fall short of reversible bounds. Given Data / Assumptions: No heat transfer across finite temperature differences. No friction or viscous dissipation in any moving part or fluid flow. Quasi-static (infinitesimal gradients) evolution at every stage. Concept / Approach: Reversible processes require an infinite sequence of equilibrium states (quasi-static) and the absence of irreversibilities: mechanical (friction), thermal (finite temperature gradients), and non-equilibrium mixing or throttling. Consequently, slow operation, near-equality with surroundings, and frictionless components are all necessary conditions in the ideal model. Step-by-Step Solution: Assess option A: small or negligible driving forces ensure near-equilibrium heat/work exchange.Assess option B: extremely slow change is the practical route to maintain quasi-equilibrium.Assess option C: frictionless components prevent mechanical irreversibilities.Collectively, all must hold; therefore “all of the above” is correct. Verification / Alternative check: Any finite difference in temperature or pressure creates entropy generation. Likewise, friction converts mechanical work into internal energy, making the reverse path non-recoverable, violating reversibility. Why Other Options Are Wrong: Selecting only one or two conditions ignores other irreversible effects that still preclude reversibility; all are necessary in the idealization. Common Pitfalls: Thinking reversibility is achieved merely by slow operation; without eliminating friction and temperature differences, slow alone is insufficient. Final Answer: all of the above

Question 18

Heat Capacity vs. Water Equivalent — Are Their Numerical Values the Same? Judge the statement: “The numerical value of the thermal capacity (heat capacity) of a body and its water equivalent are the same.”

Accepted Answer

Introduction / Context: Heat capacity (thermal capacity) and water equivalent are related but distinct concepts used in calorimetry. Confusing them leads to unit mistakes and wrong interpretations of experimental results. Given Data / Assumptions: Heat capacity C_body = m_body * c_body with units J/K. Specific heat of water c_water ≈ 4186 J/(kg·K) in SI. Water equivalent W_eq is defined as mass of water that has the same heat capacity as the body. Concept / Approach: By definition, W_eq (kg) satisfies C_body = W_eq * c_water, so W_eq = C_body / c_water. Therefore, numerical equality holds only in unit systems where c_water is numerically 1 (e.g., 1 cal/(g·K) in CGS-calorie units). In SI, c_water ≠ 1, so the numerical values differ unless coincidentally scaled. Step-by-Step Solution: Write C_body = m_body * c_body.Set C_body = W_eq * c_water ⇒ W_eq = C_body / c_water.Since c_water ≈ 4186 in SI, W_eq and C_body cannot have the same numerical value in SI units. Verification / Alternative check: Example: if C_body = 2093 J/K, then W_eq = 2093/4186 ≈ 0.5 kg of water. The numbers (2093 vs. 0.5) are not equal, proving the general statement false in SI. Why Other Options Are Wrong: Agreeing would only be valid in a special unit system (calorie-gram) and is not generally true; the question asks for general correctness. Common Pitfalls: Dropping units and comparing “numbers” alone; mixing unit systems; assuming c_water = 1 in SI. Final Answer: Disagree

Question 19

IC Engines — Definition of Compression Ratio In internal combustion engines, the compression ratio r is defined as which of the following volume ratios?

Accepted Answer

Introduction / Context: Compression ratio is a critical geometric parameter influencing efficiency, knock tendency, and emissions. A precise definition avoids confusion among swept, clearance, and total volumes in reciprocating engines. Given Data / Assumptions: Total (maximum) cylinder volume V1 = swept volume Vs + clearance volume Vc. Minimum volume at top dead center (TDC) is V2 = Vc. Air-standard viewpoint for definition only. Concept / Approach: The compression ratio r is defined as r = V1 / V2. Since V2 = Vc and V1 = Vs + Vc, r = (Vs + Vc)/Vc = 1 + Vs/Vc. This connects to thermodynamic efficiency relations in air-standard cycles such as Otto and Diesel. Step-by-Step Solution: Identify total volume: V1 = Vs + Vc.Identify minimum volume: V2 = Vc.Compute r = V1 / V2 = (Vs + Vc) / Vc = 1 + Vs/Vc.Therefore, the correct descriptive ratio is “total volume to clearance volume.” Verification / Alternative check: Textbook definitions and manufacturer specifications express r this way; example: r = 10:1 indicates V1 is ten times V2 (the clearance volume). Why Other Options Are Wrong: Options A and B mix swept with total in the wrong order; option C (Vs/Vc) is a related quantity, but the formal definition requires V1/V2, not Vs/Vc alone. Common Pitfalls: Equating Vs/Vc with r directly; ignoring that r must be > 1 by definition; confusing geometric with effective compression in forced-induction engines. Final Answer: total volume to clearance volume

Question 20

Fuel Classification — Identify a Natural Solid Fuel Which of the following is classified as a natural solid fuel (as opposed to a manufactured fuel)?

Accepted Answer

Introduction / Context: Engineering fuels are broadly categorized as natural (occurring in nature with minimal processing) and manufactured (produced by industrial treatment). Recognizing the category helps anticipate ash content, volatility, and combustion behavior. Given Data / Assumptions: “Natural” implies not produced by carbonization or extensive industrial processing. “Manufactured” implies deliberately processed to alter properties (e.g., coking). Common examples: wood, coal ranks, coke, briquettes. Concept / Approach: Wood is a naturally occurring biomass fuel used with minimal processing (cutting, drying). Coke is a manufactured product from destructive distillation of coal. Pulverised coal refers to a prepared/processed state of coal (manufactured form for firing), not a fuel that occurs as such in nature. Anthracite is a natural coal rank; however, many basic classifications in introductory courses emphasize “wood” as the archetypal natural solid fuel used directly without mining/beneficiation, while coke and pulverised forms are manufactured. To keep a single correct choice among the given distractors, “wood” is selected here as the unambiguously natural, minimally processed fuel. Step-by-Step Solution: Identify which options require industrial processing: coke (yes), pulverised coal (yes).Wood is harvested and dried; no industrial transformation is required for combustion.Select “wood” as the natural solid fuel in this context. Verification / Alternative check: Introductory fuel-technology texts list wood and raw coal under natural fuels; since “pulverised coal” denotes a processed form and “coke” is manufactured, the clearest natural, minimally processed choice is wood. Why Other Options Are Wrong: Coke is manufactured via carbonization; pulverised coal is a processed size-fraction for burners; anthracite coal is natural but leads to ambiguity in one-right-answer format, so wood is the intended unambiguous selection. Common Pitfalls: Confusing “coal rank” (natural) with “fuel form” (manufactured); overlooking that test questions often use wood as the canonical natural fuel. Final Answer: wood

Thermodynamics Questions