Free HVAC Excellence Practice Test
55 original questions on electrical, refrigeration, airflow, heat pumps, gas heat and safety, with every answer explained. This free HVAC Excellence practice test covers selected Employment Ready fundamentals and supplemental refrigerant safety — unofficial practice, not real exam questions or a full simulation of any one subject exam.
Electrical and controls (Questions 1–15)
Question 1 of 55 · Electrical and controls
In a 24-volt DC circuit, the supply voltage stays the same while a resistor changes from 48 ohms to 96 ohms. What happens to the current?
- A. It rises from 0.50 A to 1.00 A
- B. It stays at 0.50 A
- C. It falls from 0.50 A to 0.25 A
- D. It falls from 0.50 A to 0.05 A
Show answer and explanation
Answer: C. It falls from 0.50 A to 0.25 A
Current is voltage divided by resistance (I = V/R). Before: 24 ÷ 48 = 0.50 A. After: 24 ÷ 96 = 0.25 A. With voltage held steady, doubling the resistance halves the current. A has the relationship backwards. B ignores the change in resistance. D drops the current to one-tenth instead of one-half.
Source: OpenStax College Physics 2e, 20.2 Ohm's Law (Section 20.2).
Question 2 of 55 · Electrical and controls
An ideal control transformer has 240 volts on its primary and 24 volts on its secondary. What is its primary-to-secondary turns ratio?
- A. 1:10
- B. 10:1
- C. 24:1
- D. 240:1
Show answer and explanation
Answer: B. 10:1
An ideal transformer's voltage ratio equals its turns ratio (Vs/Vp = Ns/Np). 240 ÷ 24 = 10, so the primary has 10 turns for every 1 turn on the secondary. That's a step-down transformer. A is a step-up ratio. C and D confuse the voltages themselves with the ratio between them.
Source: OpenStax College Physics 2e, 23.7 Transformers (Equation 23.26; negligible coil-resistance assumption).
Question 3 of 55 · Electrical and controls
A 40 VA control transformer supplies a 24-volt circuit. About how much current can the secondary deliver at its full rating?
- A. 0.6 A
- B. 16.7 A
- C. 960 A
- D. 1.67 A
Show answer and explanation
Answer: D. 1.67 A
VA means volts × amps, so amps = VA ÷ volts = 40 ÷ 24 ≈ 1.67 A. A divides the wrong way (24 ÷ 40). B slips a decimal. C multiplies instead of dividing.
Sources: OpenStax College Physics 2e, 20.4 Electric Power and Energy (Power in Electric Circuits; The Cost of Electricity) · OpenStax College Physics 2e, 23.7 Transformers (Equation 23.26; negligible coil-resistance assumption).
Question 4 of 55 · Electrical and controls
A 10 kW electric heat strip runs on 240 volts. About how much current does it draw?
- A. 41.7 A
- B. 4.2 A
- C. 24 A
- D. 83.3 A
Show answer and explanation
Answer: A. 41.7 A
I = P ÷ V = 10,000 W ÷ 240 V ≈ 41.7 A. Convert kilowatts to watts first. B divides by 2,400 instead of 240. C doesn't come from any valid setup of these numbers. D assumes the heater still produces 10 kW at 120 volts. With the same resistance, halving voltage halves current to about 20.8 A and reduces power to about 2.5 kW.
Source: OpenStax College Physics 2e, 20.4 Electric Power and Energy (Power in Electric Circuits; The Cost of Electricity).
Question 5 of 55 · Electrical and controls
A heating element is rated 5 kW at 240 volts. What is its resistance?
- A. 20.8 Ω
- B. 11.5 Ω
- C. 48 Ω
- D. 0.048 Ω
Show answer and explanation
Answer: B. 11.5 Ω
Use P = V²/R, rearranged: R = V² ÷ P = 240² ÷ 5,000 = 57,600 ÷ 5,000 = 11.52 Ω. Check it: 240 ÷ 11.52 ≈ 20.8 A, and 20.8 A × 240 V ≈ 5,000 W. That check shows why A is a trap — 20.8 is the element's current in amps, not its resistance. C divides 240 by 5 without converting kilowatts to watts. D is 240 ÷ 5,000.
Source: OpenStax College Physics 2e, 20.4 Electric Power and Energy (Power in Electric Circuits; The Cost of Electricity).
Question 6 of 55 · Electrical and controls
An electric heater rated for 240 volts is installed on a 208-volt supply. If its resistance stays the same, its heat output will be about what percentage of its rating?
- A. 87%
- B. 100%
- C. 75%
- D. 115%
Show answer and explanation
Answer: C. 75%
With fixed resistance, power follows the square of the voltage (P = V²/R). (208 ÷ 240)² = 0.867² ≈ 0.75, so about 75%. A uses the voltage ratio without squaring it. B ignores the lower voltage. D would take a higher voltage, not a lower one.
Source: OpenStax College Physics 2e, 20.4 Electric Power and Energy (Power in Electric Circuits; The Cost of Electricity).
Question 7 of 55 · Electrical and controls
Three resistive loads are connected in parallel across 120 volts. They draw 2 A, 3 A and 5 A. How much current does the supply conductor carry?
- A. 5 A
- B. 3.3 A
- C. 30 A
- D. 10 A
Show answer and explanation
Answer: D. 10 A
Each parallel branch gets the full 120 volts and draws its own current. The supply carries the sum: 2 + 3 + 5 = 10 A. A picks only the largest branch. B averages them. C multiplies them — parallel currents add.
Source: OpenStax College Physics 2e, 21.1 Resistors in Series and Parallel (Section 21.1).
Question 8 of 55 · Electrical and controls
A 24-volt circuit has a 4 Ω resistance and an 8 Ω resistance in series. What voltage appears across the 8 Ω resistance?
- A. 16 V
- B. 8 V
- C. 12 V
- D. 24 V
Show answer and explanation
Answer: A. 16 V
Series resistances add: 4 + 8 = 12 Ω. Current is 24 ÷ 12 = 2 A, and the same current flows through both. The drop across the 8 Ω part is 2 A × 8 Ω = 16 V; the 4 Ω part drops the other 8 V (16 + 8 = 24). The bigger resistance takes the bigger share. B is the drop across the 4 Ω part. C splits the voltage evenly, which only works for equal resistances. D is the full source voltage.
Sources: OpenStax College Physics 2e, 20.2 Ohm's Law (Section 20.2) · OpenStax College Physics 2e, 21.1 Resistors in Series and Parallel (Section 21.1).
Question 9 of 55 · Electrical and controls
A condensing unit draws a constant 3.5 kW of electrical power while running 8 hours a day for 30 days. Electricity costs $0.15 per kWh. What does it cost to run for the month?
- A. $4.20
- B. $126
- C. $15.75
- D. $1,260
Show answer and explanation
Answer: B. $126
Energy is power × time: 3.5 kW × 8 h/day × 30 days = 840 kWh. Cost: 840 × $0.15 = $126. A is one day. C leaves out the 8 hours a day. D slips a decimal.
Source: OpenStax College Physics 2e, 20.4 Electric Power and Energy (Power in Electric Circuits; The Cost of Electricity).
Question 10 of 55 · Electrical and controls
About how much heat does a 5 kW electric heat strip put out, in BTU per hour?
- A. 1,465 BTU/h
- B. 3,412 BTU/h
- C. 17,060 BTU/h
- D. 170,600 BTU/h
Show answer and explanation
Answer: C. 17,060 BTU/h
One kilowatt-hour is about 3,412 Btu, so each kilowatt of resistance heat gives 3,412 BTU/h. 5 × 3,412 = 17,060 BTU/h. A divides instead of multiplying. B is only 1 kW. D slips a decimal.
Source: EIA, British thermal units (Btu) (Definition; Sample Btu conversion factors).
Question 11 of 55 · Electrical and controls
A relay's coil runs on 24 volts AC. Its contacts switch a separate 240-volt compressor circuit. What makes this arrangement work?
- A. The coil and the contacts are electrically isolated, so the control and load circuits can run at different voltages
- B. The relay steps 24 volts up to 240 volts, like a transformer
- C. The compressor current flows through the relay coil
- D. The coil must be rated 240 volts to switch a 240-volt load
Show answer and explanation
Answer: A. The coil and the contacts are electrically isolated, so the control and load circuits can run at different voltages
A relay's coil and its contacts are separate circuits. The coil's magnetic field moves the contacts; no current passes from one circuit to the other. That's how a low-voltage thermostat circuit can switch a line-voltage load. B describes a transformer, not a relay. C is wrong — load current flows through the contacts, never the coil. D is wrong: the coil is selected for the control circuit, while the contacts must be rated for the load they switch.
Source: All About Circuits, Relay Construction (Relay contacts; electrical isolation).
Question 12 of 55 · Electrical and controls
With the circuit turned off and verified dead, you check a heating element in place. Its expected resistance is about 20 Ω, but your meter reads 12 Ω. Other components are connected in parallel with it. What should you conclude?
- A. The element is shorted and must be replaced
- B. Other paths in parallel can pull the reading down, so isolate the element and measure again before judging it
- C. The element is fine, because the reading is within 10 Ω
- D. The element should be retested with power on
Show answer and explanation
Answer: B. Other paths in parallel can pull the reading down, so isolate the element and measure again before judging it
When a component stays connected, your ohmmeter reads every path in parallel with it, and those extra paths usually lower the reading. Disconnect the element (or at least one of its leads) and measure it alone. A jumps to a verdict the reading can't support. C invents a tolerance. D is unsafe and wrong — resistance is measured with the circuit de-energized.
Sources: Fluke, How to measure resistance (Parallel paths and component isolation) · OpenStax College Physics 2e, 21.1 Resistors in Series and Parallel (Section 21.1).
Question 13 of 55 · Electrical and controls
A run capacitor is labeled 7.5 µF ±6%. With the circuit de-energized and verified dead, and the capacitor safely discharged and disconnected, it measures 6.8 µF. Ignoring meter accuracy, is it within its printed tolerance?
- A. Yes, because it's within 1 µF of the rating
- B. Yes, because it's within 10% of the rating
- C. No — the label allows 7.05 to 7.95 µF
- D. You can't tell without the motor's amp draw
Show answer and explanation
Answer: C. No — the label allows 7.05 to 7.95 µF
Work out the band from the label: 6% of 7.5 = 0.45, so the acceptable range is 7.5 − 0.45 = 7.05 µF to 7.5 + 0.45 = 7.95 µF. At 6.8 µF, it's below the range. A and B use a tolerance that isn't on the label; always use the one printed on the part. D is wrong — the capacitor's own reading answers the question. Note the setup: capacitance is measured with power off, the capacitor discharged, and the part detached from the circuit.
Source: Fluke, How to measure capacitance (Power off, discharge, detach from circuit).
Question 14 of 55 · Electrical and controls
On a conventional thermostat, which terminal typically energizes the compressor contactor for first-stage cooling?
- A. G
- B. W
- C. C
- D. Y
Show answer and explanation
Answer: D. Y
In common manufacturer designations, Y controls the compressor contactor (stage 1), G the indoor fan, W stage-1 heat, and C is the 24 VAC common. Always confirm against the equipment's wiring diagram, since designations can vary.
Source: Resideo T6 Pro installation instructions (33-00181EFS) (Printed pp. 3 and 5: terminal designations and heat-pump wiring; O/B footnote 7).
Question 15 of 55 · Electrical and controls
A smart thermostat in a conventional, single-transformer system needs steady 24 VAC power. Between which two terminals is that voltage normally supplied?
- A. Y and G
- B. R and C
- C. W and Y
- D. G and C
Show answer and explanation
Answer: B. R and C
R (or Rc) carries 24 VAC from the transformer, and C is the 24 VAC common. Together they power the thermostat without switching any equipment. Y, G and W are outputs that switch the compressor contactor, fan and heat, so they aren't a steady supply.
Source: Resideo T6 Pro installation instructions (33-00181EFS) (Printed pp. 3 and 5: terminal designations and heat-pump wiring; O/B footnote 7).
Refrigeration and air conditioning (Questions 16–25)
Question 16 of 55 · Refrigeration and air conditioning
A British thermal unit (Btu) is the amount of heat needed to:
- A. Raise 1 gram of water 1°C
- B. Raise 1 pound of water 1°F
- C. Melt 1 pound of ice
- D. Boil 1 pound of water
Show answer and explanation
Answer: B. Raise 1 pound of water 1°F
EIA defines one Btu as the heat needed to raise one pound of water by 1°F. A describes a calorie. C and D are phase changes, which take far more heat than one Btu.
Source: EIA, British thermal units (Btu) (Definition; Sample Btu conversion factors).
Question 17 of 55 · Refrigeration and air conditioning
The indoor blower pushes warm room air across a cold evaporator coil. Heat moving from the moving air to the coil is mainly transferred by:
- A. Convection
- B. Radiation
- C. Evaporation of the air
- D. Conduction only
Show answer and explanation
Answer: A. Convection
Convection is heat carried by a moving fluid — here, air driven by the blower. Conduction then moves the heat through the fins and tubing, but the air-to-coil transfer is convective. B is electromagnetic energy like sunlight, which doesn't depend on moving air. C isn't a heat-transfer method for the air side. D leaves out the moving air entirely.
Source: OpenStax College Physics 2e, 14.4 Heat Transfer Methods (Section 14.4).
Question 18 of 55 · Refrigeration and air conditioning
Heat a pure refrigerant absorbs while it boils in the evaporator at constant pressure, without its temperature changing, is called:
- A. Sensible heat
- B. Specific heat
- C. Superheat
- D. Latent heat
Show answer and explanation
Answer: D. Latent heat
For a pure refrigerant boiling at constant pressure, energy goes in with no change in temperature. That energy is latent heat, not a temperature rise. A is heat that does change temperature. B is a material property (heat per pound per degree). C is the temperature rise of vapor above its boiling point, after the phase change is complete.
Sources: OpenStax College Physics 2e, 14.3 Phase Change and Latent Heat (Section 14.3) · National Refrigerants, How to Use a Two-Column Pressure-Temperature Chart (PDF p. 1: single-component versus blend charts; dew/bubble values).
Question 19 of 55 · Refrigeration and air conditioning
In a basic vapor-compression cycle, where does refrigerant go right after it leaves the compressor?
- A. Evaporator
- B. Metering device
- C. Condenser
- D. Suction line
Show answer and explanation
Answer: C. Condenser
The loop runs compressor → condenser → metering device → evaporator → back to the compressor. The compressor raises the pressure and temperature so the refrigerant can give off heat in the condenser. A and B come later in the loop. D is the line returning to the compressor.
Sources: OpenStax College Physics 2e, 15.5 Heat Pumps and Refrigerators (Heat-pump cycle and heating-mode coil roles; COPhp = Qh/W) · Parker Sporlan, Thermostatic Expansion Valves: Theory of Operation (Printed p. 3: refrigeration system; fixed-area restrictor; TEV control).
Question 20 of 55 · Refrigeration and air conditioning
Refrigerant in the discharge line, just after the compressor, is normally a:
- A. High-pressure, high-temperature vapor
- B. Low-pressure, low-temperature vapor
- C. Low-pressure, low-temperature liquid
- D. High-pressure, subcooled liquid
Show answer and explanation
Answer: A. High-pressure, high-temperature vapor
The compressor raises the vapor's pressure and temperature so it's hotter than the air or water it will reject heat to. B and C describe the low side of the system. D describes the liquid line after the condenser.
Source: Parker Sporlan, Thermostatic Expansion Valves: Theory of Operation (Printed p. 3: refrigeration system; fixed-area restrictor; TEV control).
Question 21 of 55 · Refrigeration and air conditioning
Just after the metering device, refrigerant entering the evaporator is normally a:
- A. High-pressure vapor
- B. High-pressure subcooled liquid
- C. Low-pressure mixture of liquid and vapor
- D. Superheated vapor
Show answer and explanation
Answer: C. Low-pressure mixture of liquid and vapor
The metering device drops the pressure, which drops the boiling temperature. Some of the liquid flashes to vapor right away, and the rest boils in the evaporator as it picks up heat. A and D are vapor states found at the compressor or the evaporator outlet. B is the condition going into the metering device, not coming out.
Sources: Parker Sporlan, Thermostatic Expansion Valves: Theory of Operation (Printed p. 3: refrigeration system; fixed-area restrictor; TEV control) · OpenStax College Physics 2e, 14.3 Phase Change and Latent Heat (Section 14.3).
Question 22 of 55 · Refrigeration and air conditioning
A heat pump delivers 36,000 BTU/h of heat while drawing 3,000 watts. What is its coefficient of performance (COP)?
- A. 12
- B. 0.28
- C. About 3.5
- D. 1.0
Show answer and explanation
Answer: C. About 3.5
COP is heat delivered ÷ energy input, in the same units. Convert the input first: 3 kW × 3,412 = 10,236 BTU/h. Then 36,000 ÷ 10,236 ≈ 3.5. A divides BTU/h by watts, mixing units. B flips the ratio. D is plain resistance heat, where every watt becomes one watt of heat.
Sources: OpenStax College Physics 2e, 15.5 Heat Pumps and Refrigerators (Heat-pump cycle and heating-mode coil roles; COPhp = Qh/W) · EIA, British thermal units (Btu) (Definition; Sample Btu conversion factors).
Question 23 of 55 · Refrigeration and air conditioning
A worksheet lists a suction-line pressure and temperature but not which refrigerant is in the system. What do you need before you can find the saturation temperature for that pressure?
- A. The system's tonnage
- B. The refrigerant's identity, so you can use its own pressure-temperature data
- C. The age of the equipment
- D. The size of the suction line
Show answer and explanation
Answer: B. The refrigerant's identity, so you can use its own pressure-temperature data
Every refrigerant has its own pressure-temperature relationship. For a blend with temperature glide, use the vapor/dew-point or liquid/bubble-point value appropriate to the calculation; not every chart displays two columns. Reading the wrong refrigerant's chart gives the wrong saturation temperature and every calculation after it. A, C and D don't change how pressure relates to saturation temperature.
Source: National Refrigerants, How to Use a Two-Column Pressure-Temperature Chart (PDF p. 1: single-component versus blend charts; dew/bubble values).
Question 24 of 55 · Refrigeration and air conditioning
In ENERGY STAR's Single-Family New Homes National HVAC Commissioning Checklist, Rev. 14, which quantity is compared with the manufacturer's target in the standard cooling-mode charge check for a thermostatic expansion valve (TXV) system?
- A. Superheat
- B. Suction pressure alone
- C. Subcooling
- D. Compressor amperage alone
Show answer and explanation
Answer: C. Subcooling
The checklist uses subcooling for TXV systems and superheat for fixed-orifice systems, each compared with the manufacturer's target. A TXV regulates evaporator superheat; this checklist uses liquid-line subcooling to check charge. The checklist also permits an OEM-specified alternative procedure. B and D alone don't verify charge. The equipment manufacturer's charging instructions always govern a specific unit.
Sources: ENERGY STAR Single-Family New Homes National HVAC Commissioning Checklist, Rev. 14 (Printed p. 1, §2: 2.7–2.10 (TXV), 2.11–2.14 (fixed orifice), 2.16 (OEM alternative)) · Parker Sporlan, Thermostatic Expansion Valves: Theory of Operation (Printed p. 3: refrigeration system; fixed-area restrictor; TEV control).
Question 25 of 55 · Refrigeration and air conditioning
In ENERGY STAR's Single-Family New Homes National HVAC Commissioning Checklist, Rev. 14, which quantity is compared with its target in the standard cooling-mode charge check for a fixed-orifice system?
- A. Superheat
- B. Subcooling
- C. Discharge temperature alone
- D. Condensate flow
Show answer and explanation
Answer: A. Superheat
A fixed orifice has a fixed opening, not a constant flow rate: flow still changes with pressure conditions. The checklist compares superheat against a target for the test conditions. B is the TXV method. C and D don't verify charge.
Sources: ENERGY STAR Single-Family New Homes National HVAC Commissioning Checklist, Rev. 14 (Printed p. 1, §2: 2.7–2.10 (TXV), 2.11–2.14 (fixed orifice), 2.16 (OEM alternative)) · Parker Sporlan, Thermostatic Expansion Valves: Theory of Operation (Printed p. 3: refrigeration system; fixed-area restrictor; TEV control).
Airflow and moisture (Questions 26–31)
Question 26 of 55 · Airflow and moisture
A rectangular duct measures 18 by 12 inches inside. A traverse gives an average velocity of 700 feet per minute. What is the airflow?
- A. 700 CFM
- B. 1,050 CFM
- C. 8,400 CFM
- D. 151,200 CFM
Show answer and explanation
Answer: B. 1,050 CFM
Airflow (CFM) = velocity (FPM) × area (square feet). Area: 18 × 12 = 216 square inches ÷ 144 = 1.5 square feet. Then 700 × 1.5 = 1,050 CFM. A skips the area. C multiplies by 12 inches instead of the area in square feet. D multiplies by square inches without converting.
Source: Fluke, Measuring air velocity with the velocity probe (Q = V x A; measurements at grilles, registers and diffusers).
Question 27 of 55 · Airflow and moisture
You get one velocity reading of 900 feet per minute at the center of a duct. What's the better way to find the duct's total airflow?
- A. Multiply the center reading by the duct area
- B. Double the center reading
- C. Use the single highest reading you can find
- D. Take readings across the duct in a traverse, average them, and multiply by the area
Show answer and explanation
Answer: D. Take readings across the duct in a traverse, average them, and multiply by the area
A straight duct commonly has higher velocity away from the walls, but elbows and other disturbances can distort the profile. One center reading does not establish the average. A traverse takes readings across the whole cross-section and averages them. A assumes the center speed is the average. B and C have no basis.
Sources: Fluke, Duct traversal airflow measurement (Conducting a duct traverse; velocity pressure) · Fluke, Measuring air velocity with the velocity probe (Q = V x A; measurements at grilles, registers and diffusers).
Question 28 of 55 · Airflow and moisture
At one point in a duct, total pressure is +0.20 in. w.c. and static pressure is −0.05 in. w.c. (in. w.c. means inches of water column). What is the velocity pressure?
- A. 0.15 in. w.c.
- B. −0.25 in. w.c.
- C. 0.05 in. w.c.
- D. 0.25 in. w.c.
Show answer and explanation
Answer: D. 0.25 in. w.c.
Velocity pressure = total pressure − static pressure = 0.20 − (−0.05) = 0.25 in. w.c. Subtracting a negative adds it. A treats −0.05 as if it were +0.05. B flips the sign. C has no valid path from these readings.
Source: Fluke, Duct traversal airflow measurement (Conducting a duct traverse; velocity pressure).
Question 29 of 55 · Airflow and moisture
A supply grille has a nominal face of 2.0 square feet, and its manufacturer lists an effective area of 0.60 square feet. Measured using the manufacturer’s specified method for that effective area, the average face velocity is 500 feet per minute. What is the airflow?
- A. 300 CFM
- B. 500 CFM
- C. 833 CFM
- D. 1,000 CFM
Show answer and explanation
Answer: A. 300 CFM
At grilles and registers, use the manufacturer's effective area, not the face size: 500 × 0.60 = 300 CFM. D is the trap — it uses the 2.0-square-foot face size. B skips the area. C divides by the effective area instead of multiplying.
Source: Fluke, Measuring air velocity with the velocity probe (Q = V x A; measurements at grilles, registers and diffusers).
Question 30 of 55 · Airflow and moisture
Room air has a dew point of 57°F. Which of these surfaces is cold enough for moisture from that air to condense on it?
- A. 85°F
- B. 75°F
- C. 65°F
- D. 50°F
Show answer and explanation
Answer: D. 50°F
The dew point is the temperature air has to be cooled to before it's saturated. Air touching a 50°F surface cools below 57°F, so water vapor condenses — the same reason an evaporator coil drips and cold ducts sweat. The other surfaces are all above the dew point, so air touching them never reaches saturation.
Source: National Weather Service Louisville, Humidity (Relative Humidity; Dewpoint).
Question 31 of 55 · Airflow and moisture
Air is heated at essentially constant pressure with no moisture added or removed. What happens to its relative humidity?
- A. It goes down
- B. It goes up
- C. It stays the same
- D. It rises to 100%
Show answer and explanation
Answer: A. It goes down
At a higher temperature, more water vapor is needed for saturation. With the same amount of moisture present, warming the air lowers its relative humidity. B and D describe what happens when air cools toward its dew point. C ignores the temperature change. A lower RH here doesn't mean any water was removed.
Source: National Weather Service Louisville, Humidity (Relative Humidity; Dewpoint).
Heat pumps (Questions 32–40)
Question 32 of 55 · Heat pumps
In heating mode, the outdoor coil of an air-source heat pump acts as the:
- A. Condenser
- B. Metering device
- C. Evaporator
- D. Compressor
Show answer and explanation
Answer: C. Evaporator
In heating, the heat pump picks up heat from outdoor air at the outdoor coil (the evaporator) and releases it indoors at the indoor coil (the condenser). A is the outdoor coil's role in cooling. B and D are separate components, not coil roles.
Source: OpenStax College Physics 2e, 15.5 Heat Pumps and Refrigerators (Heat-pump cycle and heating-mode coil roles; COPhp = Qh/W).
Question 33 of 55 · Heat pumps
Which component lets a heat pump switch between heating and cooling?
- A. The reversing (changeover) valve
- B. The accumulator
- C. The crankcase heater
- D. The TXV
Show answer and explanation
Answer: A. The reversing (changeover) valve
The reversing valve sends the compressor's hot discharge gas to either the indoor or the outdoor coil, which swaps which coil condenses and which evaporates. The thermostat controls it through the O or B terminal. B and C do not switch refrigerant flow between the coils. D meters refrigerant into the evaporator.
Sources: Resideo T6 Pro installation instructions (33-00181EFS) (Printed pp. 3 and 5: terminal designations and heat-pump wiring; O/B footnote 7) · OpenStax College Physics 2e, 15.5 Heat Pumps and Refrigerators (Heat-pump cycle and heating-mode coil roles; COPhp = Qh/W) · Parker Sporlan, Thermostatic Expansion Valves: Theory of Operation (Printed p. 3: refrigeration system; fixed-area restrictor; TEV control).
Question 34 of 55 · Heat pumps
A heat pump thermostat is set up for "O." In which mode is the changeover valve energized?
- A. Heating
- B. Defrost only
- C. Emergency heat
- D. Cooling
Show answer and explanation
Answer: D. Cooling
Set up for O, the changeover valve is energized in cooling; set up for B, it's energized in heating. The outdoor unit's design decides which one is correct, so the thermostat setting has to match the equipment. A describes a B setup.
Source: Resideo T6 Pro installation instructions (33-00181EFS) (Printed pp. 3 and 5: terminal designations and heat-pump wiring; O/B footnote 7).
Question 35 of 55 · Heat pumps
An air-source heat pump is heating. Outdoor air is 28°F, and refrigerant in the outdoor coil is boiling at 14°F. Which way does heat flow at the outdoor coil?
- A. From the air into the refrigerant
- B. No heat flows, because the air is below freezing
- C. From the refrigerant into the air
- D. Heat flows only during defrost
Show answer and explanation
Answer: A. From the air into the refrigerant
Heat flows from warmer to colder. At 28°F the outdoor air is still warmer than 14°F refrigerant, so the refrigerant picks up heat from it. Below-freezing air still contains thermal energy, so it can supply heat to colder refrigerant. B confuses freezing with having no heat. C reverses the direction. D describes a different mode.
Sources: OpenStax College Physics 2e, 15.5 Heat Pumps and Refrigerators (Heat-pump cycle and heating-mode coil roles; COPhp = Qh/W) · OpenStax College Physics 2e, 14.4 Heat Transfer Methods (Section 14.4).
Question 36 of 55 · Heat pumps
Why does an air-source heat pump need a defrost cycle in heating mode?
- A. Ice forms on the indoor coil
- B. Defrost is only needed in cooling
- C. Frost on the outdoor coil blocks airflow and slows heat transfer, cutting heating capacity
- D. Frost raises head pressure too high
Show answer and explanation
Answer: C. Frost on the outdoor coil blocks airflow and slows heat transfer, cutting heating capacity
When the outdoor coil runs below freezing in humid air, frost can build up. It restricts airflow, adds pressure drop across the coil and puts an insulating layer between the air and the refrigerant, so the coil picks up less heat. A and B put the frost in the wrong coil or mode. D doesn't describe the frost problem.
Source: Nawaz and Fricke, A Critical Literature Review of Defrost Technologies for Heat Pumps and Refrigeration Systems (Repository abstract).
Question 37 of 55 · Heat pumps
During reverse-cycle defrost, the outdoor coil temporarily acts as the:
- A. Condenser
- B. Evaporator
- C. Accumulator
- D. Metering device
Show answer and explanation
Answer: A. Condenser
The reversing valve switches the system into cooling mode, sending hot discharge gas to the outdoor coil. The outdoor coil becomes the condenser and melts the frost, while the indoor coil acts as the evaporator. B is the outdoor coil's normal heating-mode job. C and D aren't coil roles.
Sources: Keene et al., Evaluating the impacts of defrost mode in cold climate residential heat pumps (Heat pump operation section) · OpenStax College Physics 2e, 15.5 Heat Pumps and Refrigerators (Heat-pump cycle and heating-mode coil roles; COPhp = Qh/W).
Question 38 of 55 · Heat pumps
On a heat pump thermostat, which terminal typically controls auxiliary (second-stage) heat?
- A. Y
- B. G
- C. AUX/W2
- D. O
Show answer and explanation
Answer: C. AUX/W2
In the cited Resideo terminal chart, AUX/W2 denotes auxiliary heat or heat stage 2. Y is the compressor, G the fan and O the changeover valve. Confirm against the equipment's wiring diagram.
Source: Resideo T6 Pro installation instructions (33-00181EFS) (Printed pp. 3 and 5: terminal designations and heat-pump wiring; O/B footnote 7).
Question 39 of 55 · Heat pumps
A heat pump running at a COP of 3.0 delivers 30,000 BTU/h. About how much electrical power is it using?
- A. 10 kW
- B. 2.9 kW
- C. 8.8 kW
- D. 90 kW
Show answer and explanation
Answer: B. 2.9 kW
Input = output ÷ COP = 30,000 ÷ 3.0 = 10,000 BTU/h. Convert: 10,000 ÷ 3,412 ≈ 2.93 kW. A treats 10,000 BTU/h as 10,000 W instead of converting units. C is what plain resistance heat (COP 1) would need for the full 30,000 BTU/h. D multiplies by COP and also mishandles the units.
Sources: OpenStax College Physics 2e, 15.5 Heat Pumps and Refrigerators (Heat-pump cycle and heating-mode coil roles; COPhp = Qh/W) · EIA, British thermal units (Btu) (Definition; Sample Btu conversion factors).
Question 40 of 55 · Heat pumps
In a dual-fuel (hybrid) system, how are the heat pump and the gas furnace normally used?
- A. Both always run together
- B. The furnace handles mild weather and the heat pump handles cold weather
- C. The furnace only runs in cooling mode
- D. The heat pump handles milder weather, and the furnace takes over when it gets colder
Show answer and explanation
Answer: D. The heat pump handles milder weather, and the furnace takes over when it gets colder
Trane describes dual-fuel systems as letting the heat pump carry most of the heating in milder weather, with the furnace taking over at colder temperatures. B reverses the roles. A is not the operating sequence described. C is wrong — a furnace doesn't cool.
Source: Trane, Hybrid Heat Systems: How Do They Work? (What is a Hybrid Heat System?).
Gas heat and carbon monoxide safety (Questions 41–48)
Question 41 of 55 · Gas heat and carbon monoxide safety
In a gas furnace, which part transfers heat to the house air while keeping combustion gases separate from it?
- A. The blower
- B. The heat exchanger
- C. The gas valve
- D. The ignitor
Show answer and explanation
Answer: B. The heat exchanger
The heat exchanger passes heat from combustion to the circulating air while the flue gases stay inside it and get vented outside. That separation is why a damaged heat exchanger is a safety concern. A moves the air, C controls the flow of fuel, and D lights the burners.
Source: Carrier, How does a gas furnace work? (Heat exchanger; key parts).
Question 42 of 55 · Gas heat and carbon monoxide safety
What does a furnace rating of 95% AFUE mean?
- A. The furnace runs 95% of the time
- B. The flue gas is 95% carbon dioxide
- C. The heat exchanger is 95% sealed
- D. About 95% of the fuel's energy becomes useful furnace heat over a year of typical use
Show answer and explanation
Answer: D. About 95% of the fuel's energy becomes useful furnace heat over a year of typical use
AFUE (annual fuel utilization efficiency) measures how well a furnace or boiler turns the energy in its fuel into heat over a year. The other options misread the term.
Source: Carrier, What Is AFUE? (What Is AFUE?; How Is AFUE Calculated?).
Question 43 of 55 · Gas heat and carbon monoxide safety
A furnace fires continuously at an input of 80,000 BTU/h for 5 hours. How many therms of natural gas does it use?
- A. 0.8 therms
- B. 40 therms
- C. 4 therms
- D. 400 therms
Show answer and explanation
Answer: C. 4 therms
80,000 BTU/h × 5 h = 400,000 Btu. One therm is 100,000 Btu, so 400,000 ÷ 100,000 = 4 therms. A is one hour's use. B and D slip decimals.
Source: EIA, British thermal units (Btu) (Definition; Sample Btu conversion factors).
Question 44 of 55 · Gas heat and carbon monoxide safety
Using a heating value of 1,036 Btu per cubic foot, about how many cubic feet of natural gas per hour does a 100,000 BTU/h input furnace burn?
- A. 36 ft³/h
- B. 96.5 ft³/h
- C. 103.6 ft³/h
- D. 1,036 ft³/h
Show answer and explanation
Answer: B. 96.5 ft³/h
Cubic feet per hour = input ÷ heating value = 100,000 ÷ 1,036 ≈ 96.5. C multiplies 100 by 1.036 instead of dividing it. D is the heating value itself. A has no basis. On the job, use your gas utility's actual heating value; 1,036 Btu/ft³ is the sample conversion factor used in this question.
Source: EIA, British thermal units (Btu) (Definition; Sample Btu conversion factors).
Question 45 of 55 · Gas heat and carbon monoxide safety
Why can't people rely on their senses to notice carbon monoxide (CO)?
- A. It smells like rotten eggs only at high levels
- B. It's always visible as smoke
- C. It only comes from electric equipment
- D. It's colorless and odorless
Show answer and explanation
Answer: D. It's colorless and odorless
CPSC describes CO as a colorless, odorless, poisonous gas, which is why alarms and instruments are needed. A describes the odorant added to natural gas, not CO. B is wrong; CO is invisible. C is wrong; CO comes from fuel-burning equipment such as furnaces, generators and heaters.
Source: CPSC Carbon Monoxide Information Center (Protect Your Family from Carbon Monoxide Poisoning).
Question 46 of 55 · Gas heat and carbon monoxide safety
Which group includes common symptoms of carbon monoxide poisoning?
- A. Headache, dizziness, weakness, nausea and confusion
- B. Itchy skin and sneezing
- C. A runny nose, sore throat and swollen glands
- D. Fever and rash
Show answer and explanation
Answer: A. Headache, dizziness, weakness, nausea and confusion
CPSC lists headache, dizziness, weakness, nausea, vomiting, sleepiness and confusion. CDC also lists chest pain; a person does not need every listed symptom for CO exposure to be possible. B, C and D are not the listed symptom pattern. CO is odorless, so an odor warning is not required.
Sources: CPSC Carbon Monoxide Information Center (Protect Your Family from Carbon Monoxide Poisoning) · CDC, Carbon Monoxide Poisoning Basics (Symptoms).
Question 47 of 55 · Gas heat and carbon monoxide safety
You arrive on a furnace call. The CO alarm is sounding, and the homeowner has a headache. What comes first?
- A. Inspect the heat exchanger
- B. Silence and reset the alarm
- C. Open a window and keep diagnosing
- D. Get everyone outside to fresh air and call 911
Show answer and explanation
Answer: D. Get everyone outside to fresh air and call 911
When CO poisoning is suspected, CPSC says to get outside to fresh air right away and then call 911. People come first; the equipment diagnosis (A, C) waits. B ignores the warning.
Source: CPSC Carbon Monoxide Information Center (Protect Your Family from Carbon Monoxide Poisoning).
Question 48 of 55 · Gas heat and carbon monoxide safety
Where does CPSC recommend installing CO alarms in a home?
- A. On every level of the home and outside sleeping areas
- B. Only next to the furnace
- C. Only in the garage
- D. Only in the kitchen
Show answer and explanation
Answer: A. On every level of the home and outside sleeping areas
CPSC recommends battery-operated CO alarms, or alarms with battery backup, on every level and outside sleeping areas, ideally interconnected so they all sound together. B, C and D each leave sleeping areas unprotected.
Source: CPSC Carbon Monoxide Information Center (Protect Your Family from Carbon Monoxide Poisoning).
Safety and refrigerant handling (Questions 49–55)
These are study questions, not a complete servicing or lockout procedure.
Question 49 of 55 · Safety and refrigerant handling
Two authorized technicians are working on the same rooftop unit under a group lockout. Under OSHA's lockout/tagout standard, how should they protect themselves?
- A. The lead technician applies one lock for both
- B. Each technician puts a personal lock on the group lockout device or lockbox
- C. They set the thermostat to OFF and tag it
- D. One technician stands at the disconnect instead of locking it
Show answer and explanation
Answer: B. Each technician puts a personal lock on the group lockout device or lockbox
OSHA 1910.147(f)(3)(ii)(D) requires each authorized employee to put a personal lock or tag on the group device, lockbox or comparable mechanism when starting work and remove it when that person stops working on the equipment. A leaves one person depending on someone else's key. C uses a control device, which OSHA says isn't an energy-isolating device. D is a watch, not a lockout.
Source: OSHA 29 CFR 1910.147 (29 CFR 1910.147(b), energy-isolating device; (f)(3)(ii)(D), group lockout).
Question 50 of 55 · Safety and refrigerant handling
While servicing an R-410A air conditioner, a technician knowingly releases refrigerant into the air instead of recovering it. Under EPA's Section 608 rules, this is:
- A. Allowed if the system holds under 5 pounds
- B. Allowed if the technician is certified
- C. Prohibited venting
- D. Allowed for HFC refrigerants
Show answer and explanation
Answer: C. Prohibited venting
40 CFR 82.154(a)(1) bars anyone servicing, repairing or disposing of equipment from knowingly venting covered refrigerant. It covers substitute refrigerants such as HFCs unless EPA has exempted them, so D fails. A small charge (A) and certification (B) don't create an exception.
Sources: 40 CFR 82.154 (40 CFR 82.154(a)(1)–(2)) · EPA, Stationary Refrigeration: Prohibition on Venting Refrigerants (Types of Releases).
Question 51 of 55 · Safety and refrigerant handling
When servicing R-410A equipment in compliance with the applicable Section 608 requirements, which release can qualify as "de minimis" rather than a venting violation?
- A. A small release while connecting or disconnecting hoses during a good-faith service or recovery effort
- B. Deliberately venting a small charge to save time
- C. Venting outdoors instead of indoors
- D. Venting while wearing proper protective equipment
Show answer and explanation
Answer: A. A small release while connecting or disconnecting hoses during a good-faith service or recovery effort
EPA recognizes small releases during good-faith efforts to recapture or recycle refrigerant, including hose connections and disconnections, only when the applicable Section 608 requirements are met. Good faith alone does not waive recovery, equipment or certification requirements. B is intentional venting, no matter the size. C and D change where or how, not whether, refrigerant is vented.
Sources: EPA, Stationary Refrigeration: Prohibition on Venting Refrigerants (Types of Releases) · 40 CFR 82.154 (40 CFR 82.154(a)(1)–(2)).
Question 52 of 55 · Safety and refrigerant handling
A service record says refrigerant was removed from a system and stored in a recovery cylinder. Nothing else is recorded. Which term describes what was done?
- A. Reclaimed
- B. Recycled
- C. Vented
- D. Recovered
Show answer and explanation
Answer: D. Recovered
EPA defines recover as removing refrigerant in any condition and storing it in an external container without necessarily testing or processing it. Recycling means cleaning it for reuse in equipment of the same owner (B). Reclaiming means reprocessing it to the AHRI 700-2016 purity specifications cited in EPA’s definition and verifying that purity (A). Neither is shown by this record. C is the opposite of storing it.
Source: EPA, Definitions of Section 608 terms (Recover; Recycle; Reclaim).
Question 53 of 55 · Safety and refrigerant handling
For an appliance subject to the Table 1 recovery requirements in 40 CFR 82.156(a), which target is listed for a high-pressure appliance with a full charge under 200 pounds, using recovery equipment manufactured or imported on or after November 15, 1993?
- A. 4 inches of Hg vacuum
- B. 0 inches of Hg vacuum (0 psig)
- C. 10 inches of Hg vacuum
- D. 25 mm Hg absolute
Show answer and explanation
Answer: B. 0 inches of Hg vacuum (0 psig)
Table 1 of 82.156(a) sets 0 inches of Hg vacuum for high-pressure appliances under 200 pounds, whether the recovery equipment is older or newer. C applies to high-pressure appliances of 200 pounds or more with newer equipment. A applies to those larger appliances with older equipment. D is the table value for low-pressure appliances. These are recovery targets, not deep-vacuum dehydration targets. Section 82.156(a) requires removal of all liquid refrigerant and verification of the applicable recovery level before opening the appliance. Small appliances, motor vehicle air conditioners (MVACs) and MVAC-like appliances have separate provisions, and the listed repair/leak exceptions must also be considered.
Source: 40 CFR 82.156 (40 CFR 82.156(a), Table 1, and exceptions in (a)(1)–(2)).
Question 54 of 55 · Safety and refrigerant handling
In ASHRAE Standard 34's safety group "A2L," what do the letter and number mean?
- A. Higher toxicity, highly flammable
- B. Lower toxicity, lower flammability — burns very slowly
- C. Lower toxicity, no flame propagation
- D. Acid-resistant, low global warming potential
Show answer and explanation
Answer: B. Lower toxicity, lower flammability — burns very slowly
The letter is toxicity (A = lower, B = higher). The number is flammability (1 = no flame propagation under standard test conditions, 2 = flammable, 3 = higher flammability). The 2L subclass marks class 2 refrigerants that burn very slowly. A describes something like B3. C describes A1. D isn't part of the system.
Source: ASHRAE/UNEP, Refrigerant Designations and Safety Classifications (PDF pp. 1–2, safety groups; p. 3, R-454B blend-table row).
Question 55 of 55 · Safety and refrigerant handling
R-454B is a blend of R-32 and R-1234yf. What is its ASHRAE safety group?
- A. A1
- B. A3
- C. B1
- D. A2L
Show answer and explanation
Answer: D. A2L
ASHRAE lists R-454B (R-32/R-1234yf, 68.9/31.1 by mass) as A2L: lower toxicity, lower flammability. A1 (A) means no flame propagation under standard test conditions, not the classification of this blend. A3 (B) is the higher-flammability group. B1 (C) is in the higher-toxicity class.
Source: ASHRAE/UNEP, Refrigerant Designations and Safety Classifications (PDF pp. 1–2, safety groups; p. 3, R-454B blend-table row).
Check your score
Count the questions you got right on your first try. Write your choice before opening its answer; an answer you viewed without attempting the question counts as reviewed, not correct.
Show the full answer key
Opening the key reveals answers for the whole set. Record your first-try choices before using it.
| Section | Questions | Your first-try correct |
|---|---|---|
| Electrical and controls | 1–15 | ___ of 15 |
| Refrigeration and air conditioning | 16–25 | ___ of 10 |
| Airflow and moisture | 26–31 | ___ of 6 |
| Heat pumps | 32–40 | ___ of 9 |
| Gas heat and carbon monoxide safety | 41–48 | ___ of 8 |
| Safety and refrigerant handling | 49–55 | ___ of 7 |
| All questions | 1–55 | ___ of 55 |
Your practice percentage is 100 × first-try correct ÷ 55, rounded to one decimal place. Unanswered questions and answers viewed before an attempt do not earn a point.
Your percentage here only describes these 55 questions. It isn't an official score, a prediction of how you'll do on exam day, or proof of hands-on skill. With 6 to 15 questions per section, read section scores as a pointer to what to review next, not as a measure of what you know.
How to use your results
- Read the explanation for every miss, not just the right letter. Most wrong answers here are built from a real mistake: an unconverted unit, a skipped square, a flipped ratio, the wrong area, or a control setting treated like a disconnect.
- Name your mistake in a few words. "Used face area instead of effective area" is something you can fix. "Got it wrong" isn't.
- Rework the math from a blank page. Questions 1–10, 13, 22, 26, 28, 29, 39, 43 and 44 all turn on setting up the right formula. Change the inputs and solve it again to check that you understand the setup.
- Check your weakest section against the official subject list. HVAC Excellence publishes a Competency and Task List that its exams and several textbooks are cross-walked to. Use it to see what else sits in that topic area.
- Come back and redo only your misses before taking the whole set again.
Which HVAC Excellence exam does this match?
There isn't one "HVAC Excellence exam." HVAC Excellence offers separate exams by level and subject, so start by finding yours.
| Level | Who it's for | Questions or assessment | Passing score stated on the cited page | Prerequisite |
|---|---|---|---|---|
| Employment Ready (24 subjects, such as Electrical, Air Conditioning, Heat Pumps, Gas Heat, and System Recovery and Evacuation) | Students and entry-level technicians, usually at the end of a course or module | 50 or 100, depending on the subject | 70% | None |
| Professional Level (10 exams) | Technicians with field experience | 100 | Not stated on this page | Two years of verified field experience; the Core exam (Principles of Electrical and Refrigeration Theory) comes first |
| H.E.A.T. and H.E.A.T. Plus | High school HVACR programs | 100 (H.E.A.T. Plus adds a 25-question core exam and a hands-on assessment) | 70% | None |
| Master Specialist | Experienced technicians | Hands-on performance assessment; a related written credential is a prerequisite | Not stated on this page | Three years of field experience and the related HVAC Excellence Professional credential or an accepted equivalent listed by ESCO |
This page is mixed practice at the Employment Ready level. Here's where each section lines up with the subjects HVAC Excellence lists:
- Electrical and controls (Questions 1–15): Electrical; Electric Heat
- Refrigeration and air conditioning (Questions 16–25): Air Conditioning; Basic Refrigeration and Charging Procedures
- Airflow and moisture (Questions 26–31): Psychrometrics; Basic Refrigeration and Charging Procedures (air supply and delivery)
- Heat pumps (Questions 32–40): Heat Pumps
- Gas heat and carbon monoxide safety (Questions 41–48): Gas Heat; Carbon Monoxide Safety; Carbon Monoxide & Combustion Analysis
- Safety and refrigerant handling (Questions 49–55): System Recovery and Evacuation; Electrical (safety). Questions 54–55 add current refrigerant-safety practice; their inclusion does not claim an item-specific match to an Employment Ready blueprint.
The subject names and topic summaries come from Employment Ready Certifications. That's topic alignment, not full coverage. The practice set doesn't reproduce the content or weighting of any single exam, and it doesn't cover installation, troubleshooting sequences, specialty equipment or hands-on skills in depth.
A few details worth knowing before exam day:
- Question format. HVAC Excellence says its exam items typically have four choices — one correct answer and three plausible distractors — though some have up to seven (Exam Development and Test Validation Process, "Developing the Items"). That's why every question here has four.
- Time limits and fees. Your instructor or proctor can confirm the fee, time limit and rules for your exact subject and delivery route.
- Choosing a level. For a side-by-side of the levels and how to register, see our HVAC Excellence certification guide.
Preparing for a Professional Level exam?
This set is entry-level practice, not a Professional Core or specialty simulation. Professional exams are closed-book, 100 questions each, and meant for technicians with at least two years of verified field experience. Everyone starts with the Core (Professional Technician Certifications). The electrical and refrigeration sections above are a reasonable warm-up for Core topics, but they don't match its length or depth.
Looking for EPA 608 practice?
EPA Section 608 certification is a separate federal requirement for technicians doing covered work on stationary refrigeration and air-conditioning equipment that could release refrigerant. It comes in Type I, Type II, Type III and Universal; apprentices working under a certified technician’s close and continual supervision have an exemption (EPA, Section 608 Technician Certification Requirements). Passing an HVAC Excellence subject exam doesn't give you EPA 608 certification, and neither does this practice test.
ESCO Group offers free EPA 608 practice exams. These are for EPA 608, not an HVAC Excellence Employment Ready subject exam. The page lists separate manual-edition practice sets; use the one appropriate to your preparation materials and provider’s instructions.
Sources
HVAC Excellence / ESCO Group: Employment Ready Certifications · Professional Technician Certifications · H.E.A.T. Student Outcome Assessments · Master Specialist Certifications · Exam Development and Test Validation Process · Competency and Task List
Regulations and government sources: OSHA 29 CFR 1910.147 · 40 CFR 82.154 · 40 CFR 82.156 · EPA, Prohibition on Venting Refrigerants · EPA, Definitions of Section 608 Terms · EPA, Section 608 Technician Certification Requirements · CPSC Carbon Monoxide Information Center · CDC, Carbon Monoxide Poisoning Basics · EIA, British thermal units · ENERGY STAR National HVAC Commissioning Checklist, Rev. 14 · National Weather Service, Humidity
Primary research and additional technical sources: Parker Sporlan, Thermostatic Expansion Valves: Theory of Operation · Nawaz and Fricke, A Critical Literature Review of Defrost Technologies for Heat Pumps and Refrigeration Systems · Keene et al., Evaluating the impacts of defrost mode in cold climate residential heat pumps · Trane, Hybrid Heat Systems: How Do They Work? · Carrier, What Is AFUE?
Technical references: OpenStax College Physics 2e, sections 14.3, 14.4, 15.5, 20.2, 20.4, 21.1 and 23.7 · ASHRAE/UNEP refrigerant safety classification factsheet (April 2026; R-454B on p. 3) · National Refrigerants, How to Use a Two-Column Pressure-Temperature Chart · All About Circuits, Relay Construction · Fluke: How to measure resistance, How to measure capacitance, Measuring air velocity, Duct traversal airflow measurement · Carrier, How does a gas furnace work? · Resideo T6 Pro installation instructions
Last verified September 28, 2026: The cited HVAC Excellence exam levels, question counts, expressly stated passing scores, prerequisites and item format; the cited EPA, OSHA, CPSC and CDC rules and guidance; and the technical sources and calculations supporting this set. The eCFR text accessed was current through September 24, 2026. Checking sources isn't a professional technical review or a field test.
Castleport Test Prep is an independent exam prep publisher, not affiliated with, endorsed by, or approved by HVAC Excellence or ESCO Group. These are original, unofficial practice questions, not HVAC Excellence exam questions. Exam and credential names identify their subjects; trademarks belong to their respective owners. Practice doesn't award certification or guarantee an exam result.
By the Castleport Test Prep Editorial Team