Question 1

Desirable ignition characteristics of a good fuel A good industrial fuel should ideally have what kind of ignition point (ignition temperature) to balance safety and ease of lighting?

Accepted Answer

Introduction / Context: Ignition point (ignition temperature) is the minimum temperature at which a fuel begins self-sustained combustion under specified conditions. For practical handling and use, fuels should neither ignite too easily nor be too difficult to light. Given Data / Assumptions: Fuel used in industrial burners, boilers, or furnaces. Safety (storage/transport) and operability (start-up) are both important. Ambient conditions vary; standard definitions apply. Concept / Approach: A moderate ignition point ensures the fuel does not present undue fire hazard at normal temperatures (unlike very low ignition point) while still permitting reliable lighting and stable flame without excessive preheating (unlike very high ignition point). Step-by-Step Solution: Consider safety: fuels that ignite near ambient temperature are hazardous in storage.Consider usability: fuels requiring extremely high preheat complicate start-up and control.Balance dictates a moderate ignition temperature as the desirable property.Therefore select “moderate.” Verification / Alternative check: Design specifications for burner systems often include auto-ignition temperatures in a manageable range; diesel vs. gasoline comparisons illustrate trade-offs between ignition quality and safety. Why Other Options Are Wrong: Low or near ambient increases accidental ignition risk. High or extremely high hinders start-up and may require auxiliary fuels or intensive preheating. Common Pitfalls: Confusing ignition temperature with flash point; while related, they are distinct parameters. The qualitative answer remains “moderate.” Final Answer: moderate

Question 2

Second law of thermodynamics – Clausius statement (direction of heat flow) Which of the following correctly reflects the Clausius statement of the second law? Select the statement that captures the impossibility without external aid.

Accepted Answer

Introduction / Context: The Clausius statement of the second law formalizes the natural direction of spontaneous heat flow and the requirement of external work to move heat against this direction. Given Data / Assumptions: Macroscopic, classical thermodynamics. External source means an external agency supplying work or equivalent high-grade energy. Refrigerators and heat pumps are practical embodiments. Concept / Approach: Clausius statement: No process is possible whose sole result is the transfer of heat from a cooler body to a hotter body. Equivalently, cold-to-hot transfer requires external aid (work input). The question asks for the statement that emphasizes the impossibility without aid. Step-by-Step Solution: Identify the spontaneous direction: hot → cold.Recognize that reversing this requires external work.Choose the option that forbids cold → hot without aid.Hence, select the impossibility statement (option b). Verification / Alternative check: Refrigeration cycles explicitly consume work to move heat from low to high temperature, satisfying the Clausius statement while achieving the transfer with external aid. Why Other Options Are Wrong: (a) Omits the “external aid” requirement and is false as a general spontaneous statement. (c) True in practice but not the Clausius prohibition; it describes an application, not the principle. (d) Incorrect because (b) is correct. (e) Misstates spontaneity; spontaneous flow is hot → cold without work. Common Pitfalls: Confusing the descriptive statement (refrigeration is possible with work) with the normative prohibition (impossible without work). Final Answer: It is impossible to transfer heat from a lower-temperature body to a higher-temperature body without the aid of an external source.

Question 3

Definition check – higher vs. lower calorific value of a fuel Fill in the blank: The heat obtained by the complete combustion of 1 kg of a fuel when the products are cooled to the temperature of the supplied air (so that the steam formed is condense…

Accepted Answer

Introduction / Context: Calorific values come in two standard forms: higher (gross) and lower (net). The distinction lies in whether the latent heat of water formed during combustion is recovered. Given Data / Assumptions: Complete combustion of 1 kg of fuel. Products cooled to intake-air temperature, below the dew point so water condenses. Latent heat of vaporisation is included in the measured heat. Concept / Approach: When products are cooled enough to condense water, the latent heat released is captured. This condition defines the higher calorific value (HCV), also called the gross heating value. The lower calorific value (LCV) excludes this latent heat by keeping water as vapour in the products. Step-by-Step Solution: Recognize condition: products cooled to initial air temperature → steam condensed.Recovered latent heat is counted.Therefore the appropriate term is higher calorific value. Verification / Alternative check: Fuel datasheets may list both HCV and LCV; HCV ≥ LCV, with the difference approximately proportional to the fuel hydrogen content. Why Other Options Are Wrong: Lower: excludes latent heat, contrary to the stated cooling to condense steam. Other labels (net/apparent/gross for gases only) misuse terminology. Common Pitfalls: Forgetting that “gross” = “higher” and “net” = “lower.” Always check whether product water is condensed. Final Answer: higher

Question 4

Ideal-gas specific heats – identities connecting cp, cv, γ, and R Which of the following relations correctly connect specific heats at constant pressure and volume with the gas constant and γ = cp/cv?

Accepted Answer

Introduction / Context: For ideal gases with temperature-independent specific heats over a range, cp, cv, γ, and the gas constant R are linked by simple identities used throughout thermodynamics and gas dynamics. Given Data / Assumptions: Ideal-gas behaviour. Specific heats cp and cv are taken as constants over the temperature interval. γ is defined as cp/cv. Concept / Approach: From the ideal-gas relations and definitions, two cornerstone identities follow: cp − cv = R and cp/cv = γ. Combining them yields convenient forms for cp and cv in terms of R and γ: cp = γR/(γ − 1) and cv = R/(γ − 1). Step-by-Step Solution: Start with cp − cv = R.Use γ = cp/cv ⇒ cp = γcv.Substitute into cp − cv = R → γcv − cv = R → cv(γ − 1) = R.Solve: cv = R/(γ − 1) and cp = γR/(γ − 1). Verification / Alternative check: Dimensional check: cp, cv, and R share units J/(kg·K) (for specific values); formulas are dimensionally consistent. Plugging back reproduces cp − cv = R and γ = cp/cv. Why Other Options Are Wrong: (a) is true by definition but incomplete alone; the question asks for relations connecting to R as well. (e) cp − cv = γR is incorrect; it would imply cp/cv varies improperly with R. Common Pitfalls: Confusing molar vs. specific (per mass) quantities; the same relations hold for both as long as R is the corresponding gas constant (Rmolar or Rspecific). Final Answer: Both (b) and (c)

Question 5

Foundational Chemistry – Nature of Elements as Atomic Particles In basic chemistry, an element is defined as a pure substance made of extremely small, discrete, and chemically indivisible particles called atoms. An atom is the smallest unit of an el…

Accepted Answer

Introduction / Context: Understanding what constitutes an element is the first step in chemistry and materials science. The statement asserts that elements are composed of minute, chemically indivisible particles known as atoms. This directly ties to the atomic theory and the definition of chemical identity. Given Data / Assumptions: Chemical reactions do not alter the nuclei of atoms; they rearrange electrons and bonds. Atoms are the smallest units that preserve the chemical properties of an element in ordinary chemical processes. Subatomic particles exist, but splitting atoms requires nuclear, not chemical, processes. Concept / Approach: Dalton’s atomic theory and modern quantum chemistry agree that elements consist of atoms. While some elements exist as molecules (e.g., O2), each molecule is formed from atoms of the same element. Chemical indivisibility means no chemical method can break an atom into smaller entities that retain the original element’s chemical identity. Step-by-Step Solution: Identify what “element” means in chemistry: a substance with atoms of the same atomic number.Clarify “chemically indivisible”: breaking atoms changes the element; only nuclear reactions do this.Conclude: the statement aligns with atomic theory, hence it is true. Verification / Alternative check: Electrolysis decomposes compounds into elements but does not split atoms. Nuclear fission/fusion alter atomic nuclei and transmute elements, which are not chemical processes. Therefore, chemical indivisibility holds for atoms. Why Other Options Are Wrong: Limiting the claim to metals is arbitrary; all elements are atomic. Saying elements are “molecules only” ignores monatomic elements (e.g., noble gases). Atoms are not chemically divisible; division requires nuclear changes. Common Pitfalls: Confusing diatomic elemental molecules (H2, O2) with the definition of element; assuming the existence of subatomic particles refutes chemical indivisibility. Final Answer: True

Question 6

Steady-Flow Processes in Thermodynamics – Identify the Invariant Quantity For a steady-flow process through a control volume (e.g., nozzle, turbine, heat exchanger), properties at fixed locations do not change with time. In particular, the mass flow…

Accepted Answer

Introduction / Context: Steady-flow analysis underpins the design of turbines, compressors, nozzles, and heat exchangers. The concept of “steady” means that at any fixed point in the device, intensive properties and flow variables do not change with time, enabling averaged energy and mass balances. Given Data / Assumptions: Control volume with steady operation (no net accumulation of mass over time). Single or multiple inlets/outlets possible. Transient start-up/shutdown excluded from the definition of steady state. Concept / Approach: For steady flow, the mass conservation statement is m_dot_in = m_dot_out when the control-volume mass remains constant. Heat and work rates can be time-dependent in practice (e.g., changing load), yet the process can still be steady if properties at a point do not vary with time. Therefore, the safest invariant to pick from the options is mass flow balance, not necessarily constant heat or work rates. Step-by-Step Solution: Define steady flow: ∂()/∂t at fixed locations equals zero for properties like p, T, v, h.Apply continuity: accumulation term is zero → m_dot_in − m_dot_out = 0.Conclude mass flow in equals mass flow out for the control volume at steady state.Note that Q_dot and W_dot may vary yet still satisfy steady operation if property fields are time-invariant. Verification / Alternative check: Plant data logs often show small fluctuations in heat or shaft power during “steady” operation, while flow and state variables remain effectively constant—confirming that strict constancy of Q_dot or W_dot is not required by the definition. Why Other Options Are Wrong: Heat/work transfer rates need not be constant to qualify as steady; enthalpy can change between inlet and outlet (e.g., across heaters or turbines), so it is not invariant either. Common Pitfalls: Equating “steady” with “constant everything”; forgetting that steady is about time invariance at a location, not equality of inlet and outlet properties except for conserved totals like mass. Final Answer: mass flow at inlet and outlet is same

Question 7

Carbonisation of Coal – Product from 500° to 700°C (Low-Temperature Range) When coal is carbonised in the lower temperature range of about 500° to 700°C in limited air, the product formed is commonly called soft coke (low-temperature coke).

Accepted Answer

Introduction / Context: Coal carbonisation temperature directly controls the nature of the coke produced. Industrial practice distinguishes low-temperature carbonisation (domestic or soft coke) from high-temperature carbonisation (metallurgical or hard coke), each suited to different applications. Given Data / Assumptions: Carbonisation range: about 500°–700°C. Oxygen-limited (destructive distillation) environment. Objective: identify the principal solid product name. Concept / Approach: At 500°–700°C, coal releases volatiles but not as completely as at higher temperatures. The resulting coke retains more volatile matter, is more reactive, and is mechanically weaker, hence called soft coke. By contrast, hard (metallurgical) coke is produced at roughly 900°–1100°C and has higher strength and lower volatile content, appropriate for blast furnace use. Step-by-Step Solution: Associate temperature band with carbonisation regime: 500°–700°C → low-temperature carbonisation.Recall the typical product names: low temp → soft coke; high temp → hard coke.Eliminate distractors: pulverised coal is a preparation method; bituminous and anthracite are coal ranks, not carbonisation products.Select “soft coke” as the correct product. Verification / Alternative check: Materials and fuel handbooks consistently classify 500°–700°C carbonisation yields as soft coke with higher volatiles, suitable for domestic heating or gas production by-products. Why Other Options Are Wrong: Hard coke requires ~900°–1100°C. Pulverised coal is not a product of carbonisation. Bituminous and anthracite are starting coal ranks, not outcomes of this process. Common Pitfalls: Confusing coal rank (geological) with coke type (process outcome); assuming higher temperature always better without considering end-use requirements. Final Answer: soft coke

Question 8

Thermodynamic Properties – Definition Check for Extensive Properties An extensive property of a system is one whose value for the whole system equals the sum of its values for all non-overlapping parts (e.g., mass, volume, total internal energy). Is…

Accepted Answer

Introduction / Context: Classifying properties as intensive or extensive is fundamental to formulating balances and scaling processes. The question asks whether the additivity definition correctly captures the essence of extensive properties. Given Data / Assumptions: System can be subdivided into non-overlapping parts. Properties considered include mass, volume, total energy, and entropy (with care for interactions at interfaces). No chemical reaction across the imagined partitions changing totals. Concept / Approach: An extensive property scales with system size and is additive over disjoint subsystems: X_total = Σ X_i. Examples: mass, volume, total internal energy, total enthalpy. In contrast, intensive properties (pressure, temperature, density) are not additive; they do not scale simply with size and are defined locally or as ratios of extensive properties (e.g., specific volume v = V/m). Step-by-Step Solution: State the additivity criterion: for property X, if system splits into parts i, then X_total = Σ X_i.Match to examples: mass_total and volume_total clearly add; energy adds for non-interacting partitions.Confirm definition meets the standard of “extensive”.Therefore, the statement is correct. Verification / Alternative check: Dimensional analysis: properties proportional to amount of substance (moles, mass) are typically extensive. Taking specific or molar forms (dividing by an extensive amount) yields intensive counterparts. Why Other Options Are Wrong: Homogeneity is not required; additivity holds as long as subsystems are disjoint. Saying properties “never add” contradicts conservation laws. Temperature constraints are irrelevant to extensivity. Common Pitfalls: Confusing entropy’s subtlety; despite path dependence, total entropy is extensive for non-interacting aggregates. Also, mixing up “sum” vs. “average” when combining subsystems. Final Answer: Yes

Question 9

Ideal Brayton (Gas Turbine) Efficiency in Terms of Pressure Ratio For an ideal simple gas turbine cycle with isentropic compressor and turbine and no regeneration, the thermal efficiency in terms of pressure ratio r and specific heat ratio gamma is …

Accepted Answer

Introduction / Context: Expressing Brayton-cycle efficiency in terms of pressure ratio r and gamma provides a rapid sizing and performance estimate for gas turbines in preliminary design. It is derived from isentropic relations and the definition of thermal efficiency as 1 − Q_out/Q_in. Given Data / Assumptions: Ideal gas with constant specific heats. Two isentropic and two isobaric processes; no regeneration or intercooling. Compressor pressure ratio r = p2/p1 = p3/p4. Concept / Approach: Isentropic relations give T2/T1 = r^((gamma−1)/gamma) and T3/T4 = r^((gamma−1)/gamma). Writing Q_in = mCp(T3 − T2) and Q_out = mCp(T4 − T1) and substituting the temperature ratios yields the compact formula eta_th = 1 − 1 / r^((gamma − 1)/gamma). Step-by-Step Solution: Relate compressor temperature rise to r using isentropic equation.Relate turbine temperature drop similarly.Form eta_th = 1 − Q_out/Q_in and cancel Cp and m.Algebraically reduce to 1 − 1 / r^((gamma − 1)/gamma). Verification / Alternative check: Temperature-form efficiency eta_th = 1 − (T4 − T1)/(T3 − T2) reduces to the r–gamma form under isentropic assumptions, confirming consistency. Why Other Options Are Wrong: r^(gamma − 1) and 1 − r^(gamma − 1) lack correct dimensions/limits. r/(gamma − 1) is unrelated. 1 − T1/T3 is Carnot-like and not Brayton’s general result. Common Pitfalls: Applying this ideal expression to real engines without accounting for component efficiencies and pressure drops; mixing absolute and gauge pressures when computing r. Final Answer: eta_th = 1 − 1 / r^((gamma − 1)/gamma)

Question 10

Reversible Non-Flow Processes – Which Types Can Be Reversible? Identify which of the listed non-flow processes can, in principle, be carried out reversibly for a closed system if executed quasi-statically with no friction, finite temperature differe…

Accepted Answer

Introduction / Context: Reversibility is an idealization used to define limits and to derive reference formulas. Many named processes (constant volume, constant pressure, or pv = constant) can be reversible if executed infinitely slowly with perfect insulation or heat transfer at infinitesimal temperature differences and without friction. Given Data / Assumptions: Closed system undergoing a path in p–v–T space. Quasi-static evolution with no dissipative effects (no friction, turbulence, electrical resistance, or mixing). Heat transfer, if present, occurs with negligible temperature difference. Concept / Approach: Reversibility depends on how a process is executed, not on its label. Isochoric (v = constant), isobaric (p = constant), and hyperbolic (pv = constant) paths can all be traced reversibly if constraints remove sources of entropy generation. Any of these processes may also be irreversible if conducted rapidly or with finite gradients. Step-by-Step Solution: Define reversibility: process and its reverse leave system and surroundings unchanged.Show feasibility: for isochoric, keep volume fixed and exchange heat with an infinite sequence of reservoirs at dT steps.For isobaric, adjust piston load and heat to maintain p constant with infinitesimal gradients.For hyperbolic pv = constant, guide the system along that locus quasi-statically with negligible gradients. Verification / Alternative check: T–s diagrams for reversible paths show ds integrals that match q_rev/T with no excess entropy generation. Any deviation producing positive S_gen marks irreversibility. Why Other Options Are Wrong: “None of these” ignores the central idea that reversibility is about execution conditions, not process names. Picking only one would be arbitrarily restrictive. Common Pitfalls: Equating “adiabatic” with “reversible” automatically; running processes quickly and still labeling them reversible; forgetting surface friction at pistons causes S_gen. Final Answer: all of these

Question 11

Fuel Testing – What Calorific Value Does a Bomb Calorimeter Measure? A bomb calorimeter determines the higher (gross) calorific value of solid and liquid fuels, since combustion occurs at constant volume with water formed condensed to liquid.

Accepted Answer

Introduction / Context: Calorific value indicates energy released during complete combustion. Laboratory calorimeters come in different types, with the bomb calorimeter being the standard for solids and liquids. Knowing whether it yields higher (gross) or lower (net) values is essential for comparing fuels and designing boilers or engines. Given Data / Assumptions: Bomb calorimeter operates at constant volume in a sealed “bomb”. Products, including water from hydrogen combustion, are cooled back to initial temperature and water is condensed. Complete combustion with sufficient oxygen is ensured. Concept / Approach: The higher (gross) calorific value includes the latent heat of condensation of the water produced. Because the bomb calorimeter cools the products to the initial temperature and water is liquid at the end, the measured heat releases include this latent heat, yielding the higher value. The lower (net) value subtracts the latent heat associated with water remaining as vapor, typical of gas-turbine or internal-combustion contexts. Step-by-Step Solution: Perform combustion at constant volume in the bomb.Capture released heat in a known water jacket; compute heat using mC(ΔT) with corrections.Note condensed water at final state → latent heat captured → higher CV measured.Report higher calorific value after standard acid and ignition corrections. Verification / Alternative check: Standards (e.g., ASTM) define bomb calorimetry procedures that return gross heating value, with formulas to convert to lower value by subtracting latent heat of water and minor corrections. Why Other Options Are Wrong: “Lower” applies to processes where water leaves as vapor. “Net only at constant pressure” and “neither” misunderstand the apparatus. “Apparent without correction” ignores standard correction steps but still yields higher value fundamentally. Common Pitfalls: Confusing constant-volume vs. constant-pressure calorimetry; neglecting nitric/sulfuric acid corrections for fuels containing N/S. Final Answer: higher

Question 12

Improving Carnot Efficiency – Parameter Sensitivity For a Carnot heat engine operating between T_H (highest temperature) and T_L (lowest temperature), efficiency increases by raising T_H and/or lowering T_L. Which listed action increases efficiency?

Accepted Answer

Introduction / Context: The Carnot cycle provides the upper theoretical efficiency bound for heat engines. Its efficiency depends only on the two reservoir temperatures, not on the working fluid or specific mechanism. Understanding sensitivity to T_H and T_L guides practical design goals for turbines and boilers. Given Data / Assumptions: Carnot efficiency formula: eta_C = 1 − T_L/T_H (temperatures in Kelvin). Ideal reversible operation; no internal losses. Other variables (mass flow, pressures) are secondary to reservoir temperatures. Concept / Approach: From eta_C = 1 − T_L/T_H, increasing T_H (with T_L fixed) makes the ratio T_L/T_H smaller, thus increasing efficiency. Conversely, decreasing T_H reduces efficiency, and increasing T_L also reduces efficiency. Holding T_L “constant” does not by itself increase efficiency; it simply states no change. Raising both T_H and T_L by the same absolute amount does not necessarily increase efficiency; what matters is their ratio. Step-by-Step Solution: Write eta_C = 1 − T_L/T_H.Consider differential: d(eta_C) ≈ (T_L/T_H^2) dT_H − (1/T_H) dT_L.Set dT_L = 0: positive dT_H → eta_C increases.Conclude that increasing T_H is the listed action that increases efficiency. Verification / Alternative check: Example: T_L = 300 K. If T_H rises from 900 K to 1200 K, eta_C increases from 1 − 300/900 = 0.667 to 1 − 300/1200 = 0.75. Why Other Options Are Wrong: Decreasing T_H or increasing T_L worsens efficiency; keeping T_L constant is not an action that increases it; increasing both temperatures equally may or may not change T_L/T_H beneficially. Common Pitfalls: Using Celsius instead of Kelvin; focusing on absolute temperature difference ΔT rather than the ratio T_L/T_H that appears in the Carnot expression. Final Answer: increasing the highest temperature

Question 13

Which Engine Achieves Carnot (Maximum Reversible) Efficiency? Among engine types, only a fully reversible engine operating between two reservoirs can, in principle, achieve Carnot’s maximum efficiency limit.

Accepted Answer

Introduction / Context: The Carnot theorem states that no heat engine operating between two thermal reservoirs can be more efficient than a reversible engine working between the same reservoirs—and all reversible engines between the same temperatures have the same efficiency. This question asks which engine type fits that idealization. Given Data / Assumptions: Two thermal reservoirs at fixed temperatures T_H and T_L. Engine cycle can be idealized as reversible or irreversible. No restrictions on working fluid; reversibility is the key criterion. Concept / Approach: Real petrol (Otto) and diesel (Diesel) engines are inherently irreversible due to friction, finite-rate combustion, pressure losses, and heat transfer across finite differences. Therefore they cannot achieve the Carnot limit. Only a hypothetical reversible engine—one that operates quasi-statically with no entropy generation—matches Carnot efficiency. Step-by-Step Solution: Recall Carnot efficiency: eta_C = 1 − T_L/T_H.Apply Carnot theorem: eta_real ≤ eta_C, with equality only for reversible engines.Classify petrol/diesel engines as irreversible due to real losses.Conclude: “reversible engine” is the only correct selection. Verification / Alternative check: Exergy analysis of actual engines shows positive entropy generation and associated exergy destruction, proving eta < eta_C. Laboratory approximations can approach but never equal the reversible limit. Why Other Options Are Wrong: Petrol and diesel engines are specific real cycles (Otto/Diesel) with irreversibilities; irreversible engines by definition cannot reach Carnot’s maximum; “any real engine” contradicts the second law. Common Pitfalls: Assuming higher compression ratio or better fuel alone allows surpassing fundamental reversible limits; confusing high component efficiencies with cycle reversibility. Final Answer: reversible engine

Question 14

Regnault’s Law (Perfect Gas Approximation) – Behavior of Specific Heats According to Regnault’s law for ideal gases in an approximate range, the specific heats at constant pressure (cp) and at constant volume (cv) are taken as constants—i.e., they d…

Accepted Answer

Introduction / Context: Regnault’s law is a classical engineering approximation used to simplify calculations with ideal gases: take cp and cv as constants over a moderate temperature range. Although modern data show mild variation, the assumption remains practical in many problems and underpins closed-form formulas in basic thermodynamics. Given Data / Assumptions: Gas behaves ideally in the range of interest. Moderate temperatures and pressures where property variation is small. Engineering accuracy acceptable for introductory analysis. Concept / Approach: Perfect-gas models often assume cp and cv are constants (calorically perfect gas). This leads to useful relations like h = cpT and u = cvT and supports isentropic formulas with a constant gamma = cp/cv. Regnault’s empirical observations motivated treating specific heats as essentially constant over limited ranges, even though precision work may require temperature-dependent property data. Step-by-Step Solution: State the approximation: cp and cv ≈ constants for ideal gases.Use it to get h = cpT and u = cvT with cp, cv independent of T and p.Then gamma remains roughly constant, enabling Brayton/Otto closed-form expressions.Acknowledge that outside the range, real-gas tables show mild temperature dependence. Verification / Alternative check: Comparing ideal-gas predictions using constant cp to NASA polynomials shows small errors near room temperature but growing deviations at high T, confirming the approximation’s limited yet useful domain. Why Other Options Are Wrong: Monotonic pressure dependencies are negligible for ideal gases; “become zero” is unphysical; complex opposing trends are not part of the ideal-gas model. Common Pitfalls: Applying constant specific heats at very high temperatures (e.g., combustion) where vibrational modes raise cp noticeably; confusing ideal-gas assumptions with real-gas behavior at high pressure. Final Answer: do not change

Question 15

Foundational thermodynamics — identifying the law: “The heat and work are mutually convertible.” This statement refers to which law of thermodynamics? Choose the most appropriate law by name.

Accepted Answer

Introduction / Context: Thermodynamics organizes our understanding of energy interactions into four numbered laws. The statement “heat and work are mutually convertible” captures the essence of energy conservation when energy crosses a boundary as heat or as work. Correctly mapping such a statement to the appropriate law is essential for engineering analysis. Given Data / Assumptions: Closed or open systems exchanging energy as heat and/or work. No chemical reactions are required to interpret the statement. Classical (macroscopic) thermodynamics viewpoint. Concept / Approach: The First Law of Thermodynamics is an energy balance: the change in total energy of a system equals energy added by heat minus energy leaving as work (plus flow terms for open systems). It formalizes that heat and work are forms of energy transfer and, under proper accounting, one can be converted into the other while conserving total energy. The Zeroth law addresses thermal equilibrium and temperature; the Second law addresses directionality and limits to conversion (entropy); the Third law concerns absolute zero properties. Step-by-Step Solution: Identify the key claim: “heat and work are mutually convertible.”Recall the First Law energy balance: ΔE_system = Q_in − W_out (closed system form).Conclude the statement aligns with the First Law’s conservation principle, not with the Zeroth/Second/Third laws. Verification / Alternative check: Practical examples—electrical heaters convert work (electric) to heat; steam engines convert heat to work—demonstrate mutual convertibility within energy conservation limits prescribed by the First Law. Why Other Options Are Wrong: Zeroth law defines temperature via thermal equilibrium; Second law limits conversion efficiency and introduces entropy; Third law describes entropy behavior near 0 K. None of these state energy conservation between heat and work. Common Pitfalls: Confusing “possible to convert” (First Law) with “how much can be converted” (Second Law). The First Law does not guarantee 100% conversion from heat to work; the Second Law limits that. Final Answer: First law of thermodynamics

Question 16

Units and dimensions — selecting the S.I. unit of energy: Which of the following is the correct S.I. (International System) unit for energy?

Accepted Answer

Introduction / Context: Dimensional consistency and correct units are crucial in engineering calculations. Energy appears across mechanics, heat transfer, and electricity; selecting the correct S.I. unit prevents cascading errors in analyses and designs. Given Data / Assumptions: S.I. base units and derived units are used. Energy may be mechanical (work), thermal (heat), or electrical in form. Power and force units are related but distinct from energy. Concept / Approach: In S.I., the unit of energy (and work) is the joule, defined as 1 J = 1 N·m. The watt (W) is power, equal to J/s. Compound variants such as J/m or J·m represent line energy density or energy times length and are not standard units for energy alone. Step-by-Step Solution: Recall: energy (work) W = force * displacement ⇒ unit = N·m.By definition, 1 N = 1 kg·m/s^2 ⇒ N·m has dimensions kg·m^2/s^2, named the joule (J).Therefore, the correct S.I. unit is Joule (J). Verification / Alternative check: Electrical energy from power P integrated over time t yields E = P * t with units W·s = J, confirming consistency across domains. Why Other Options Are Wrong: J·m is energy multiplied by length (not energy). W is power (rate of energy). J/m is energy per length (a density), not energy itself. Common Pitfalls: Interchanging power and energy; forgetting that torque also uses N·m but is not energy (axial moment), despite sharing units. Final Answer: Joule (J)

Question 17

Gas laws — identifying Charles’ law: Select the statement that correctly represents Charles’ law for a fixed mass of gas.

Accepted Answer

Introduction / Context: Classical gas laws provide limiting relationships of the ideal-gas equation when one property is held fixed. Recognizing which statement corresponds to Charles’ law is vital for quick problem setup in thermodynamics and fluid mechanics. Given Data / Assumptions: Ideal-gas behavior with fixed mass. One property (pressure, volume, or temperature) held constant as specified. Absolute (Kelvin) temperature scale is used. Concept / Approach: Charles’ law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature. Mathematically, v ∝ T at constant p, or v/T = constant. Boyle’s law applies to p and v at constant T (p * v = constant), while Gay–Lussac’s law applies to p and T at constant v (p/T = constant). Step-by-Step Solution: From p * v = m * R * T, hold p constant.Then v is proportional to T ⇒ v/T = constant.Hence the correct statement is option B. Verification / Alternative check: Linear v–T plots at constant p extrapolate to zero volume at T = 0 K, consistent with the absolute scale. Why Other Options Are Wrong: Option A is Boyle’s law (constant temperature). Option C is Gay–Lussac’s law. Option D is a reciprocal form that is not the conventional statement at constant v. Common Pitfalls: Using Celsius instead of Kelvin in proportional relationships; confusing which property is held constant. Final Answer: v / T = constant, if p is kept constant

Question 18

Fuel technology — briquetted coal production: “Briquetted coal is produced from finely ground coal by moulding under pressure, with or without a binding material.” State whether this industrial statement is correct.

Accepted Answer

Introduction / Context: Briquetting improves the handling, combustion, and transportation characteristics of fine coal (slurry, fines, dust) by agglomerating particles into larger, stronger shapes. Understanding how briquettes are produced is useful for fuel preparation and emissions control topics. Given Data / Assumptions: Finely divided coal feed (fines or slack). Application of pressure in moulds or roll presses. Presence of a binder is optional depending on coal rank and process. Concept / Approach: Briquetting uses mechanical compaction, sometimes with thermal assistance, to form dense solids. Binders such as molasses, pitch, tar, or starch may be added to improve green strength and durability. Some high-rank coals can briquette without binder under suitable pressure/temperature due to plasticity of inherent resins/tars. Step-by-Step Solution: Start with coal fines, screen and dry as needed.Optionally blend with binder and water to a target moisture/consistency.Compact in presses or moulds at high pressure; optionally cure/dry to develop strength.Use or store briquettes, benefiting from reduced dusting and improved combustion control. Verification / Alternative check: Industry practice includes pillow briquettes and roller-pressed ovoids produced exactly by this description, confirming correctness. Why Other Options Are Wrong: “False” would deny a standard, widely deployed coal preparation method. Common Pitfalls: Assuming a binder is always mandatory; many processes achieve binderless briquetting depending on coal properties and process temperature. Final Answer: True

Question 19

Entropy concepts — choose the correct statement: Which of the following best aligns with the thermodynamic meaning of entropy and its relation to heat–to–work conversion?

Accepted Answer

Introduction / Context: Entropy connects heat transfer with temperature and sets fundamental limits on converting heat into work. Identifying a correct qualitative statement about entropy strengthens intuition for the Second Law and availability (exergy) concepts used in real plant analysis. Given Data / Assumptions: Reversible reference for entropy change: dS = dQ_rev / T. Temperature measured on the Kelvin scale. Availability (exergy) represents the useful work potential relative to an environment. Concept / Approach: Entropy growth quantifies degradation of energy quality. For a given heat quantity, adding it at low temperature increases entropy more (dS = dQ/T), which reduces the portion potentially convertible to work. Thus, entropy change is tightly linked to the unavailability of energy for work, not to the “maximum obtainable work per degree” (that belongs to exergy analysis and Carnot factors). Step-by-Step Solution: Interpret option B: it states entropy change gauges unavailability of heat for work. This aligns with Second-Law reasoning and exergy loss being T0 * ΔS_gen.Consider option A: while it is true that dS is larger at lower T for a given dQ, it is incomplete as the “best” statement of entropy’s role in availability.Evaluate option C: claims entropy directly equals maximum work per degree drop—this is incorrect; maximum work relates to temperature levels via Carnot efficiency and exergy, not entropy alone.Hence, the most appropriate and correct statement is B. Verification / Alternative check: Exergy destruction equals T0 * S_gen. Larger entropy generation directly implies greater lost work, reinforcing option B’s framing. Why Other Options Are Wrong: A is true but narrower and less definitive about entropy’s meaning; C is incorrect in definition; D cannot be right because C is wrong. Common Pitfalls: Equating entropy with “disorder” only; ignoring that engineering relevance is in work potential relative to an environment at T0. Final Answer: The change in entropy may be regarded as a measure of the unavailability of heat for transformation into work.

Question 20

Polytropic processes — defining the polytropic index (n): For a polytropic process of a perfect gas described by p * v^n = constant between states 1 and 2, which expression correctly gives the polytropic index n?

Accepted Answer

Introduction / Context: Many real compression and expansion processes do not follow purely isothermal (n = 1) or adiabatic (n = c_p/c_v = k) paths. The polytropic model p * v^n = constant fits intermediate behavior. Determining the polytropic index n from end states is a frequent task in cycle analysis and compressor/turbine testing. Given Data / Assumptions: States 1 and 2 of a closed-system gas process are known. Polytropic relation holds: p * v^n = constant. Ideal-gas behavior (for interpretive clarity). Concept / Approach: From p1 * v1^n = p2 * v2^n, rearrange to isolate n. Taking natural logarithms yields n = ln(p2/p1) / ln(v1/v2). This relation uses only end-state pressure and volume ratios and is independent of path length, provided the process is truly polytropic. Step-by-Step Solution: Start: p1 * v1^n = p2 * v2^n.Divide: (p2/p1) = (v1/v2)^n.Take ln: ln(p2/p1) = n * ln(v1/v2).Solve for n: n = ln(p2/p1) / ln(v1/v2). Verification / Alternative check: Special cases: if n = 1, then p * v = constant (isothermal); if n = k = c_p/c_v, the path is adiabatic for an ideal gas—confirming the formula’s consistency when applied to appropriate p, v ratios. Why Other Options Are Wrong: Option B flips the volume ratio, giving a sign error. Option C mixes temperature and volume and is not the general definition. Option D equals k (adiabatic index), which applies only to reversible adiabatic processes, not general polytropes. Common Pitfalls: Using base-10 vs. natural logs inconsistently; any logarithm base works as long as the same base is used in numerator and denominator. Final Answer: n = ln(p2/p1) / ln(v1/v2)

Thermodynamics Questions