FLORIDA BOARD OF PROFESSIONAL
ENGINEERS PRINCIPLES AND PRACTICE
MECHANICAL EXAM WITH ACTUAL
QUESTIONS AND VERIFIED ANSWERS,
PLUS EXPLAINED RATIONALES/EXPERT
VERIFIED FOR GUARANTEED 100% PASS
2026/LATEST UPDATE/INSTANT
DOWNLOAD PDF
1. Thermodynamics — First Law
A closed system contains 2 kg of an ideal gas. During a process, the gas
receives 150 kJ of heat and performs 90 kJ of boundary work on its
surroundings. Neglect changes in kinetic and potential energy. What is
the change in internal energy of the gas?
A. −240 kJ
B. −60 kJ
C. +60 kJ
D. +240 kJ
Answer: C. +60 kJ
Rationale: For a closed system, the first law is ΔU = Q − W when Q
is positive into the system and W is positive when work is done by
the system. Therefore, ΔU = 150 − 90 = 60 kJ. The gas has gained
internal energy because the heat input exceeds the work output. The
mass of the gas is not needed because the total heat and work are
already given.*
2. Ideal Gas — Specific Volume
1
,Air at 100 kPa and 300 K is treated as an ideal gas with R = 0.287
kJ/(kg·K). What is its specific volume?
A. 0.287 m³/kg
B. 0.861 m³/kg
C. 1.23 m³/kg
D. 2.87 m³/kg
Answer: B. 0.861 m³/kg
Rationale: For an ideal gas, Pv = RT. Therefore, v = RT/P =
(0.287)(300)/100 = 0.861 m³/kg. The important examination skill is
maintaining consistent units: kPa·m³/kg equals kJ/kg.*
3. Entropy — Second Law
A heat engine operates between a high-temperature reservoir at 800 K
and a low-temperature reservoir at 400 K. If the engine operates
reversibly, what is its maximum possible thermal efficiency?
A. 25%
B. 40%
C. 50%
D. 75%
Answer: C. 50%
Rationale: For a reversible heat engine, the maximum efficiency is
the Carnot efficiency: η = 1 − TL/TH. Thus η = 1 − 400/800 = 0.50,
or 50%. Any real heat engine operating between these same
temperatures must have an efficiency less than 50% because
irreversibilities generate entropy.*
4. Brayton Cycle
2
,An ideal gas-turbine Brayton cycle has a compressor inlet temperature
of 300 K and a turbine inlet temperature of 1,200 K. The pressure ratio
is 10, and γ = 1.4. What is the approximate compressor exit temperature
assuming isentropic compression?
A. 420 K
B. 579 K
C. 720 K
D. 930 K
Answer: B. 579 K
Rationale: For isentropic compression, T₂/T₁ = (P₂/P₁)^[(γ−1)/γ].
Therefore, T₂ = 300(10)^[(0.4)/(1.4)]. The exponent is approximately
0.286, giving 10^0.286 ≈ 1.93. Thus T₂ ≈ 300(1.93) ≈ 579 K.*
5. Rankine Cycle
In a steam power plant, increasing the boiler pressure while maintaining
the same turbine inlet temperature generally has what effect on ideal
Rankine-cycle thermal efficiency?
A. It always decreases efficiency
B. It generally increases efficiency
C. It has no effect
D. It reduces turbine work to zero
Answer: B. It generally increases efficiency
Rationale: Increasing boiler pressure generally raises the average
temperature at which heat is added, improving the ideal Rankine-
cycle thermal efficiency. However, excessive boiler pressure can
increase moisture at the turbine exhaust depending on the turbine
inlet condition, which can create blade-erosion concerns. In
practical design, the pressure increase must therefore be evaluated
together with turbine exhaust quality and equipment constraints.*
3
, 6. Heat Transfer — Conduction
A plane wall is 0.20 m thick and has a thermal conductivity of 1.5
W/(m·K). The two surfaces are maintained at 200°C and 50°C. What is
the steady one-dimensional heat flux through the wall?
A. 375 W/m²
B. 750 W/m²
C. 1,125 W/m²
D. 2,250 W/m²
Answer: C. 1,125 W/m²
Rationale: Fourier's law for steady one-dimensional conduction
through a plane wall is q″ = k(T₁ − T₂)/L. Therefore q″ = 1.5(200 −
50)/0.20 = 1,125 W/m². The heat flows from the higher-temperature
surface toward the lower-temperature surface.*
7. Thermal Resistance
A wall consists of two layers in series. Layer 1 has thickness 0.10 m and
k = 0.5 W/(m·K). Layer 2 has thickness 0.05 m and k = 0.25 W/(m·K).
What is the total conduction resistance per unit area?
A. 0.20 m²·K/W
B. 0.40 m²·K/W
C. 0.50 m²·K/W
D. 0.60 m²·K/W
Answer: D. 0.40 + 0.20 = 0.60 m²·K/W
Rationale: For layers in series, conduction resistances are additive.
R₁″ = L₁/k₁ = 0.10/0.5 = 0.20 m²·K/W. R₂″ = 0.05/0.25 = 0.20
m²·K/W. Therefore the total resistance is 0.40 m²·K/W, not 0.60.
The correct answer is therefore B.
4
ENGINEERS PRINCIPLES AND PRACTICE
MECHANICAL EXAM WITH ACTUAL
QUESTIONS AND VERIFIED ANSWERS,
PLUS EXPLAINED RATIONALES/EXPERT
VERIFIED FOR GUARANTEED 100% PASS
2026/LATEST UPDATE/INSTANT
DOWNLOAD PDF
1. Thermodynamics — First Law
A closed system contains 2 kg of an ideal gas. During a process, the gas
receives 150 kJ of heat and performs 90 kJ of boundary work on its
surroundings. Neglect changes in kinetic and potential energy. What is
the change in internal energy of the gas?
A. −240 kJ
B. −60 kJ
C. +60 kJ
D. +240 kJ
Answer: C. +60 kJ
Rationale: For a closed system, the first law is ΔU = Q − W when Q
is positive into the system and W is positive when work is done by
the system. Therefore, ΔU = 150 − 90 = 60 kJ. The gas has gained
internal energy because the heat input exceeds the work output. The
mass of the gas is not needed because the total heat and work are
already given.*
2. Ideal Gas — Specific Volume
1
,Air at 100 kPa and 300 K is treated as an ideal gas with R = 0.287
kJ/(kg·K). What is its specific volume?
A. 0.287 m³/kg
B. 0.861 m³/kg
C. 1.23 m³/kg
D. 2.87 m³/kg
Answer: B. 0.861 m³/kg
Rationale: For an ideal gas, Pv = RT. Therefore, v = RT/P =
(0.287)(300)/100 = 0.861 m³/kg. The important examination skill is
maintaining consistent units: kPa·m³/kg equals kJ/kg.*
3. Entropy — Second Law
A heat engine operates between a high-temperature reservoir at 800 K
and a low-temperature reservoir at 400 K. If the engine operates
reversibly, what is its maximum possible thermal efficiency?
A. 25%
B. 40%
C. 50%
D. 75%
Answer: C. 50%
Rationale: For a reversible heat engine, the maximum efficiency is
the Carnot efficiency: η = 1 − TL/TH. Thus η = 1 − 400/800 = 0.50,
or 50%. Any real heat engine operating between these same
temperatures must have an efficiency less than 50% because
irreversibilities generate entropy.*
4. Brayton Cycle
2
,An ideal gas-turbine Brayton cycle has a compressor inlet temperature
of 300 K and a turbine inlet temperature of 1,200 K. The pressure ratio
is 10, and γ = 1.4. What is the approximate compressor exit temperature
assuming isentropic compression?
A. 420 K
B. 579 K
C. 720 K
D. 930 K
Answer: B. 579 K
Rationale: For isentropic compression, T₂/T₁ = (P₂/P₁)^[(γ−1)/γ].
Therefore, T₂ = 300(10)^[(0.4)/(1.4)]. The exponent is approximately
0.286, giving 10^0.286 ≈ 1.93. Thus T₂ ≈ 300(1.93) ≈ 579 K.*
5. Rankine Cycle
In a steam power plant, increasing the boiler pressure while maintaining
the same turbine inlet temperature generally has what effect on ideal
Rankine-cycle thermal efficiency?
A. It always decreases efficiency
B. It generally increases efficiency
C. It has no effect
D. It reduces turbine work to zero
Answer: B. It generally increases efficiency
Rationale: Increasing boiler pressure generally raises the average
temperature at which heat is added, improving the ideal Rankine-
cycle thermal efficiency. However, excessive boiler pressure can
increase moisture at the turbine exhaust depending on the turbine
inlet condition, which can create blade-erosion concerns. In
practical design, the pressure increase must therefore be evaluated
together with turbine exhaust quality and equipment constraints.*
3
, 6. Heat Transfer — Conduction
A plane wall is 0.20 m thick and has a thermal conductivity of 1.5
W/(m·K). The two surfaces are maintained at 200°C and 50°C. What is
the steady one-dimensional heat flux through the wall?
A. 375 W/m²
B. 750 W/m²
C. 1,125 W/m²
D. 2,250 W/m²
Answer: C. 1,125 W/m²
Rationale: Fourier's law for steady one-dimensional conduction
through a plane wall is q″ = k(T₁ − T₂)/L. Therefore q″ = 1.5(200 −
50)/0.20 = 1,125 W/m². The heat flows from the higher-temperature
surface toward the lower-temperature surface.*
7. Thermal Resistance
A wall consists of two layers in series. Layer 1 has thickness 0.10 m and
k = 0.5 W/(m·K). Layer 2 has thickness 0.05 m and k = 0.25 W/(m·K).
What is the total conduction resistance per unit area?
A. 0.20 m²·K/W
B. 0.40 m²·K/W
C. 0.50 m²·K/W
D. 0.60 m²·K/W
Answer: D. 0.40 + 0.20 = 0.60 m²·K/W
Rationale: For layers in series, conduction resistances are additive.
R₁″ = L₁/k₁ = 0.10/0.5 = 0.20 m²·K/W. R₂″ = 0.05/0.25 = 0.20
m²·K/W. Therefore the total resistance is 0.40 m²·K/W, not 0.60.
The correct answer is therefore B.
4
