FE Mechanical · knowledge area 11 · 10–15 questions
FE thermodynamics review, tables, laws, cycles
Thermodynamics is 10–15 questions and the area where the property tables decide most answers. Find the state, pick the right table, then apply an energy balance to the right kind of system. Engineering thermodynamics for the whole site is explained here.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Mechanical
questions on the exam10–15
worked problems here18
of 14 knowledge areas11
Checked
What the area covers
The specification lists eleven topics. Grouped:
Properties: ideal gases and pure substances, steam and refrigerant tables, quality in the two-phase region.
Energy transfers and the laws: work, heat, the first law for closed and open systems, entropy and the second law.
Processes and components: isothermal, isentropic, constant-pressure and constant-volume processes; turbines, compressors, pumps, nozzles, valves and heat exchangers.
Cycles: power cycles (Rankine, Brayton, Otto, Diesel) and refrigeration and heat pump cycles.
Mixtures and air: nonreacting gas mixtures, psychrometrics and HVAC processes.
Combustion: fuel and air, theoretical and excess air, products of combustion.
18 thermodynamics problems, every option explained
0 / 18 answered · 0 correct
Question 1 of 18 · Thermodynamics · 10–15 items
Refrigerant-134a enters a condenser at 1.0 MPa as a saturated mixture with a quality of 0.90. It is fully condensed and leaves as saturated liquid at the same pressure. The mass flow rate is 0.050 kg/s. With saturation properties at 1.0 MPa of hf = 107.32 kJ/kg and hg = 270.99 kJ/kg, what is the rate of heat rejection from the condenser?
FE Reference Handbook, Thermodynamics: steady-flow energy equation (heat exchangers).
ṁ·hg uses an absolute enthalpy; heat rejection is the enthalpy difference h1 − hf.
ṁ·h1 is the inlet enthalpy flow, not the change from inlet to exit.
This condenses the full mass flow, as if the inlet were saturated vapor (x = 1).
Only the vapor fraction condenses: Q̇ = ṁ·x·hfg = 0.050 × 0.90 × 163.67 ≈ 7.37 kW.
This uses (1 − x), the liquid already present at the inlet; it is the vapor fraction that releases latent heat.
Question 2 of 18 · Thermodynamics · 10–15 items
In a throttling process, which property remains constant?
Solution
q − w = Δh + ΔKE + ΔPE with q = w = 0 → h2 = h1
The purpose of a throttle is a large pressure drop.
Temperature stays constant only for an ideal gas; a real fluid such as a refrigerant cools sharply when throttled.
Throttling is irreversible, so entropy increases; it is not an isentropic process.
A throttling valve has no work, negligible heat transfer and negligible kinetic energy change, so h1 = h2.
Question 3 of 18 · Thermodynamics · 10–15 items
What is the coefficient of performance (COP) of a refrigerator operating between −5 °C and 25 °C, assuming it operates on an ideal reversed Carnot cycle?
FE Reference Handbook, Thermodynamics: Carnot cycle (refrigeration).
Plugging Celsius values into TL/(TH − TL) (5/30, sign dropped) gives a meaningless ratio; absolute temperatures are required.
TL/TH = 0.90 is not a COP; a Carnot refrigerator COP divides TL by the temperature difference.
COPR = TL/(TH − TL) = 268.15/30 = 8.94, with temperatures in kelvins.
TH/(TH − TL) is the Carnot heat pump COP, which is the refrigerator COP plus one.
Question 4 of 18 · Thermodynamics · 10–15 items
A Carnot engine operates between a high-temperature reservoir and a low-temperature reservoir. If the engine has an efficiency of 40% when the low-temperature reservoir is maintained at 300 K, determine the percentage increase in the high-temperature reservoir temperature required to achieve an efficiency of 50%.
FE Reference Handbook, Thermodynamics: Carnot cycle.
600/500 = 1.20 is the ratio of the temperatures, reported as a percentage; the increase is 20.0%.
100/300 compares the change with the cold reservoir temperature instead of the original hot one.
This is the relative increase in efficiency (0.5/0.4), not in the reservoir temperature.
TH = TL/(1 − η): 300/0.6 = 500 K at 40% and 300/0.5 = 600 K at 50%, a 20.0% increase.
Question 5 of 18 · Thermodynamics · 10–15 items
Temp (°C)
Sat. Pressure (kPa)
Spec. Vol. Liquid, vf (m3/kg)
Spec. Vol. Vapor, vg (m3/kg)
150
476.16
0.001090
0.39248
160
618.23
0.001102
0.30680
170
792.18
0.001114
0.24260
Water is held in a closed system at 160 °C with a specific volume of 0.4500 m3/kg. Based on the saturation data provided, which phase region describes the state of the water, and which property table should be used to find its remaining intensive properties?
FE Reference Handbook, Thermodynamics: second law (entropy).
Both have positive or zero total entropy change; heat absorbed from the surroundings lowers Ssurr only by Q/T.
Process 3 gives Sgen = 1.0 − 400/400 = 0, a reversible process, which is allowed.
Process 1 has ΔSsys < 0, but the surroundings gain 800/300 = 2.67 kJ/K, so Sgen = +0.67 kJ/K.
Process 3 gives Sgen = 1.0 − 400/400 = 0, a reversible process; it is allowed, not impossible.
Sgen = ΔSsys + ΔSsurr: process 1 +0.67, process 2 −3.0 + 600/300 = −1.0, process 3 0 (reversible), process 4 +0.5 kJ/K; only process 2 is negative.
Question 7 of 18 · Thermodynamics · 10–15 items
A gas turbine receives air at 1,000 K and 60 m/s and discharges it at 700 K and 150 m/s. The mass flow rate of the air is 4.0 kg/s, and the turbine loses 35 kW of heat to the surroundings. Assuming a constant specific heat Cp = 1.1 kJ/kg·K and neglecting potential energy changes, what is the power output of the turbine?
This subtracts the heat loss but ignores the kinetic energy gain of the exhaust, which also takes energy from the shaft.
Adding the kinetic energy term reverses its sign; the air leaves faster, so that energy is not available as work.
This reverses the kinetic energy sign and also ignores the 35 kW heat loss.
This adds the heat loss and the kinetic energy term instead of subtracting them; both reduce the shaft output.
Question 8 of 18 · Thermodynamics · 10–15 items
A rigid, adiabatic mixing chamber receives two streams of water (Cp = 4.18 kJ/kg·K). Stream 1 enters at 3.0 kg/s and 360 K, while Stream 2 enters at 2.0 kg/s and 300 K. The streams mix completely and exit as a single stream at uniform temperature. What is the rate of entropy generation for this process?
Adiabatic does not mean isentropic: mixing streams at different temperatures is irreversible.
Question 9 of 18 · Thermodynamics · 10–15 items
A piston-cylinder device contains 0.2 kg of air at 100 kPa and 300 K. The air is compressed isentropically to 600 kPa. Assuming air is an ideal gas with k = 1.4 and Cv = 0.718 kJ/kg·K, what is the work input required for this process?
FE Reference Handbook, Thermodynamics: ideal gas isentropic process; closed-system first law.
m·R·ΔT is the boundary work of a constant-pressure process, not of an adiabatic compression.
T2 = 300 × 60.286 = 500.6 K; for a closed system W = m·cv(T2 − T1) = 0.2 × 0.718 × 200.6 ≈ 29 kJ.
m·R·T·ln(6) is the work of an isothermal compression; an isentropic process follows a different path.
m·cp·ΔT is the work of a steady-flow compressor; a piston-cylinder (closed) system uses cv.
Using exponent k − 1 on the pressure ratio overstates T2; with pressures the exponent is (k − 1)/k.
Question 10 of 18 · Thermodynamics · 10–15 items
An adiabatic compressor takes in superheated vapor at 100 kPa and compresses it to 500 kPa with an isentropic efficiency of 80%. Using the property table for the real fluid below, what is the actual work required per unit mass of fluid?
State
Pressure (kPa)
Temperature (°C)
Enthalpy (kJ/kg)
Entropy (kJ/(kg·K))
Inlet
100
150
2100
7.20
Exit A
500
280
2860
7.20
Exit B
500
350
3050
7.55
Exit C
500
420
3240
7.90
Given
Inlet: 100 kPa, 150 °C, h = 2100 kJ/kg, s = 7.20 kJ/(kg·K)
Exit A: 500 kPa, h = 2860 kJ/kg, s = 7.20 kJ/(kg·K)
Exit B: 500 kPa, h = 3050 kJ/kg, s = 7.55 kJ/(kg·K)
Exit C: 500 kPa, h = 3240 kJ/kg, s = 7.90 kJ/(kg·K)
FE Reference Handbook, Thermodynamics: isentropic efficiency of compressors.
Multiplying by the efficiency is the turbine form; a compressor needs more work than the isentropic ideal.
760 kJ/kg is the isentropic work; the 80% efficiency raises the actual work.
The isentropic exit is state A (same s): ws = 2860 − 2100 = 760 kJ/kg, and wactual = ws/η = 760/0.80 = 950 kJ/kg.
This reads state C as the exit; the actual exit enthalpy is 2100 + 950 = 3050 kJ/kg, which is state B.
Question 11 of 18 · Thermodynamics · 10–15 items
Stream
Fluid state
Specific enthalpy
Mass flow rate
1
Superheated steam
h1
ṁ1
2
Cooling water
h2
ṁ2
3
Saturated steam
h3
ṁ3
A direct-contact desuperheater is used in an industrial facility to reduce the temperature of a steam supply. Superheated steam is mixed with cooling water to produce saturated steam at the exit. The device operates at steady state and is well-insulated. Based on the variable definitions in the provided table, which system of equations correctly represents the complete mass and energy conservation balances for this device?
The device is well insulated, so there is no heat-loss term.
This treats the cooling water as an outlet; both streams 1 and 2 enter.
The cooling water's energy flow enters the device, so it adds rather than subtracts.
Specific enthalpies do not add; each must be weighted by its mass flow.
An insulated mixing device with no work conserves mass and enthalpy flow: what comes in leaves in the single exit stream.
Question 12 of 18 · Thermodynamics · 10–15 items
For one mole of an ideal gas undergoing a reversible isothermal expansion at 350 K, the pressure decreases from 5 atm to 1 atm. Calculate the change in entropy of the system.
FE Reference Handbook, Thermodynamics: ideal gas entropy change.
Using ln(p2/p1) flips the sign; entropy rises when a gas expands.
This uses log10 5 instead of the natural logarithm.
This is R alone; the pressure ratio enters through ln(p1/p2).
For an isothermal ideal gas, ΔS = nR ln(p1/p2) = 1 × 8.314 × ln 5 = 13.4 J/K.
Question 13 of 18 · Thermodynamics · 10–15 items
An ideal diatomic gas (k = 1.4) undergoes a reversible adiabatic expansion in which its volume increases by a factor of 4. What happens to its absolute temperature?
Given
Specific heat ratio k = 1.4
Volume ratio V2/V1 = 4
Find
Temperature ratio T2/T1
Solution
T2/T1 = (V1/V2)k − 1 = (1/4)0.4 = 0.574
FE Reference Handbook, Thermodynamics: ideal gas isentropic process.
T2/T1 = (V1/V2)k − 1 = (1/4)0.4 = 0.574.
43% is the drop, 1 − 0.574, not the remaining fraction.
Temperature stays constant only in an isothermal expansion; an adiabatic expansion does work at the expense of internal energy.
(1/4)0.286 uses the pressure-ratio exponent (k − 1)/k with a volume ratio.
Question 14 of 18 · Thermodynamics · 10–15 items
On a T-s diagram for a pure substance, what does the area under the curve of a reversible process represent?
Solution
qrev = ∫T ds (area under the path on the T-s diagram)
Work is the area under the path on a p-v diagram, not a T-s diagram.
For a reversible process δq = T ds, so the area under the curve on T-s axes is the heat transferred per unit mass.
Internal energy is a property change set by the end states; the area under the path depends on the path.
Enthalpy change equals the area only in a reversible constant-pressure process; in general it does not.
Question 15 of 18 · Thermodynamics · 10–15 items
For a pure substance, at what condition do the liquid and vapor phases have the same density?
At the critical point the saturated liquid and vapor lines meet, and the two phases become indistinguishable with equal density.
At the triple point three phases coexist, each with its own distinct density.
At the freezing point solid and liquid coexist; liquid and vapor densities are far apart.
At ordinary boiling, vapor density is far below liquid density, about 1/1600 for water at 1 atm.
Question 16 of 18 · Thermodynamics · 10–15 items
A Carnot engine operates between reservoirs at 650 K (hot) and 300 K (cold). If the engine absorbs 500 kJ of heat from the hot reservoir, what is the amount of heat rejected to the cold reservoir?
Given
Hot reservoir TH = 650 K
Cold reservoir TC = 300 K
Heat absorbed QH = 500 kJ
Find
Heat rejected QL (kJ)
Solution
QL = QH·TL/TH = 500 × 300/650 = 230.77 kJ
FE Reference Handbook, Thermodynamics: Carnot cycle.
For a Carnot engine QL/QH = TL/TH, so QL = 500 × 300/650 = 230.77 kJ.
This is the net work QH − QL, not the heat rejected.
Rejecting all the heat absorbed would mean zero work output; a Carnot engine converts 54% of it.
Using TH/TL instead of TL/TH (500 × 650/300) makes the heat rejected exceed the heat absorbed.
Question 17 of 18 · Thermodynamics · 10–15 items
For an isentropic steam turbine with a known inlet state, which exit-state property must be found from the steam tables to compute the turbine work?
Solution
wT = h1 − h2s, h2s = h(p2, s2 = s1)
FE Reference Handbook, Thermodynamics: steady-flow devices (turbines).
Pressure alone does not fix the exit state; a second property (s2 = s1) is needed, and the work is the enthalpy difference.
Specific volume enters boundary or flow work for liquids, but turbine work comes from the enthalpy drop.
Turbine work per unit mass is h1 − h2; for an isentropic turbine the exit state is fixed by the exit pressure and s2 = s1.
Temperature alone does not give the work, and in the wet region it does not even fix the state.
Question 18 of 18 · Thermodynamics · 10–15 items
A simple Brayton cycle operates with air flowing at steady state. Air enters the compressor at 300 K and exits at 550 K. Following heat addition, the air enters the turbine at 1300 K and exits at 720 K. Assuming constant specific heats with cp = 1.005 kJ/(kg · K), determine the net specific work output of the cycle.
This is the turbine work alone; part of it drives the compressor.
cp(1300 − 550) is the heat added in the combustor, not the net work.
Adding the compressor work to the turbine work gets its sign wrong; the compressor consumes work.
Answer every question to see your score
Common traps
Checked against the Thermodynamics chapter of the FE Reference Handbook and Çengel and Boles, Thermodynamics: An Engineering Approach:
Absolute temperature. Ideal-gas relations, entropy changes and every Carnot limit use kelvin or rankine. A ratio of Celsius temperatures is meaningless.
Closed or open system. A piston-cylinder is a closed system and its energy balance uses internal energy u. A turbine, compressor or nozzle with steady flow uses enthalpy h. Choosing the wrong one drops the flow work.
Isentropic efficiency runs both ways. For a turbine, actual work is smaller than isentropic work; for a compressor or pump, actual work is larger. Divide or multiply accordingly.
Locate the state before reading a table. Compare the given pressure or specific volume with saturation values first. Interpolating inside the two-phase table for a superheated state is a common, confident mistake.
R must match the units. The specific gas constant goes with kg or lbm, the universal constant with moles; pressure in pv = RT is absolute (gauge versus absolute).
Sign of work. Texts differ on whether work done by the system is positive. Pick one convention and check the sign of the answer against the physical picture.
Psychrometrics: the basics here, the systems on the PE
The outline includes psychrometrics and HVAC processes, so expect a question or two on reading moist-air properties at sea level and on an ideal vapor-compression cycle. Coil processes, mixing of outdoor and return air and loads are covered in depth on PE HVAC&R loads and psychrometrics, real refrigeration equipment on PE HVAC&R equipment and components.
Thermodynamics beyond this page
The steady-flow energy equation is shared with fluid mechanics, and heat exchangers analyzed here as black boxes are sized in heat transfer. Combustion uses the Handbook's combustion-in-air relations for the air-fuel ratio.
Other exams on this site point to this page for engineering thermodynamics. Chemical equilibrium, heats of reaction and vapor-liquid equilibrium belong to FE Chemical thermodynamics.
Yes. Steam and refrigerant property tables and charts are part of the on-screen FE Reference Handbook. Practice finding a state with the Handbook's search box, because Ctrl+F does not work on exam day.
How many thermodynamics questions are on the FE Mechanical exam?
10–15, one of the four largest ranges in the outline alongside dynamics, fluid mechanics and mechanical design.
More thermodynamics problems on your phone
FE Mechanical questions, answers and worked solutions in a study app.