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CUET-UG for B.Tech Agricultural Engineering Engineering Thermodynamics: practice questions

CUET-UG (ICAR 20% All-India Quota) for B.Tech Agricultural Engineering · India · National Testing Agency

Original practice questions written to the published syllabus · not past papers
10worked here, free
16written to this exam in this subject
6syllabus units

About CUET-UG for B.Tech Agricultural Engineering

CUET-UG (ICAR 20% All-India Quota) for B.Tech Agricultural Engineering is a university entrance exam run by National Testing Agency, for candidates in India. CUET-UG domain tests; the exact subject combination for B.Tech Agricultural Engineering was not verified.

Official site: exams.nta.ac.in.

We have not been able to confirm these details against the exam body's own specification, so fees, dates and pass marks are left out on purpose. Check the current format, eligibility, fees and dates on the official site.

Engineering Thermodynamics in CUET-UG for B.Tech Agricultural Engineering

Energy, work, heat and the cycles behind engines, refrigerators and dryers.

Our 16 CUET-UG for B.Tech Agricultural Engineering questions in this subject: 16 easy, 0 medium, 0 hard, across 6 units.

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

These 10 were written to CUET-UG for B.Tech Agricultural Engineering and are not shown on any other exam's page.

Question 1 · easy

Which of the following is an intensive property?

  1. Total volume of the gas
  2. Total internal energy
  3. Specific volume
  4. Mass of the charge
Answer and working

C. Specific volume

Specific volume (volume per unit mass) does not depend on how much matter is present, so it is intensive. Total volume, total internal energy and mass all scale with the size of the system, so they are extensive.

Question 2 · easy

A cold store for frozen produce is held at −18 °C. On the Fahrenheit scale this is

  1. -0.4 °F
  2. -32.4 °F
  3. 25.2 °F
  4. 22.0 °F
Answer and working

A. -0.4 °F

t(°F) = 1.8 t(°C) + 32 = 1.8 × (−18) + 32 = -0.4 °F. −32.4 °F forgets to add 32, 25.2 °F adds 32 before multiplying, and 22.0 °F divides by 1.8 instead of multiplying.

Question 3 · easy

A 2.5 m³ rigid tank holds 3 kg of air. The specific volume of the air is

  1. 1.20 m³/kg
  2. 0.833 m³/kg
  3. 7.50 m³/kg
  4. 0.120 m³/kg
Answer and working

B. 0.833 m³/kg

v = V/m = 2.5/3 = 0.833 m³/kg. 1.20 is the density (m/V) given the wrong unit, 7.50 multiplies volume by mass, and 0.120 is a misplaced decimal of the density.

Question 4 · easy

The internal energy of an ideal gas depends on

  1. temperature only
  2. pressure only
  3. pressure and volume together
  4. volume only
Answer and working

A. temperature only

Joule's free-expansion experiment showed no temperature change when a low-pressure gas expanded into a vacuum, so u of an ideal gas is a function of temperature alone. Pressure and volume can change at constant u provided T is unchanged.

Question 5 · easy

Air in a cylinder expands at a constant pressure of 200 kPa from 0.1 m³ to 0.4 m³. The work done by the air is

  1. 80 kJ
  2. 60 kJ
  3. 20 kJ
  4. 0.06 kJ
Answer and working

B. 60 kJ

W = P(V2 − V1) = 200 × (0.4 − 0.1) = 60 kJ. 80 kJ uses the final volume instead of the change, 20 kJ uses the initial volume, and 0.06 kJ is a unit slip (kPa × m³ already gives kJ).

Question 6 · easy

On a P–v diagram, the area under the curve of a quasi-static process represents

  1. the heat transferred per unit mass during the process
  2. the change of internal energy per unit mass in the process
  3. the change of entropy per unit mass during the process
  4. the boundary work per unit mass done during the process
Answer and working

D. the boundary work per unit mass done during the process

Boundary work is ∫P dv, the area under the P–v curve. Heat for a reversible process is the area under a T–s curve, not a P–v curve. Internal energy and entropy changes are not areas on a P–v diagram.

Question 7 · easy

The characteristic gas constant of carbon dioxide (molar mass 44 kg/kmol; universal gas constant 8.314 kJ/kmol·K) is

  1. 8.314 kJ/kg·K
  2. 365.8 kJ/kg·K
  3. 0.189 kJ/kg·K
  4. 5.29 kJ/kg·K
Answer and working

C. 0.189 kJ/kg·K

R = R̄/M = 8.314/44 = 0.189 kJ/kg·K. 8.314 is the universal constant per kmol, not per kg; 365.8 multiplies instead of dividing; 5.29 inverts the ratio.

Question 8 · easy

A reversible engine works between a source at 527 °C and a sink at 27 °C. Its efficiency is

  1. 62.5 %
  2. 94.9 %
  3. 37.5 %
  4. 5.1 %
Answer and working

A. 62.5 %

η = 1 − TL/TH with kelvin: 1 − 300/800 = 62.5 %. 94.9 % uses Celsius temperatures, 37.5 % is TL/TH (the fraction of heat rejected), and 5.1 % is the Celsius ratio itself.

Question 9 · easy

At the triple point of water,

  1. ice, liquid water and vapour coexist in equilibrium
  2. the liquid and vapour phases become indistinguishable
  3. the latent heat of vaporisation is exactly zero
  4. water boils at 100 °C under one standard atmosphere
Answer and working

A. ice, liquid water and vapour coexist in equilibrium

The triple point (0.01 °C, 0.6117 kPa) is where all three phases coexist. Indistinguishable liquid and vapour, and zero latent heat, describe the critical point; 100 °C at one atmosphere is the normal boiling point.

Question 10 · easy

When air is heated sensibly in a dryer's air heater, its

  1. humidity ratio rises and relative humidity stays constant
  2. humidity ratio falls and relative humidity rises sharply
  3. humidity ratio stays constant and relative humidity falls
  4. dew point rises in step with the dry-bulb temperature
Answer and working

C. humidity ratio stays constant and relative humidity falls

No water is added or removed, so W and the dew point stay constant; the saturation pressure rises with temperature, so RH falls — which is why heated air can dry grain. The other options would need moisture to be added or removed.

A study plan for Engineering Thermodynamics in CUET-UG for B.Tech Agricultural Engineering

This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.

  1. First, the heavy units (69% of the 16 questions): First law of thermodynamics (6, 38%); Systems, properties and the zeroth law (5, 31%).

    Read the manual chapters for these first and do every worked question you can find on them.

  2. Then the middle band: Second law, entropy and the Carnot cycle (2, 13%); Ideal gas processes (1, 6%).

    Aim for one sitting per unit, then mixed questions across them.

  3. Then the lighter units: Introduction to psychrometry (1, 6%); Properties of steam, steam tables and the Mollier chart (1, 6%).

    Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.

  4. Bank the easy marks: First law of thermodynamics (6 easy of 6); Systems, properties and the zeroth law (5 easy of 5).

    Do these early in revision so they are fast on the day.

  5. Finish timed. Sit the subject as a timed set without notes, mark it, and send each wrong answer back to its unit above.

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