HERS Rater Certification | RESNET Examination|120 Questions with
Correct Answers| (2026|2027)
SECTION 1: BUILDING SCIENCE PRINCIPLES (Questions 1-15)
1. A home in Climate Zone 5 has an unconditioned attic with a ceiling plane consisting of R-38
fiberglass batts installed over drywall. During a winter inspection, the rater observes frost on
the underside of the roof sheathing near the eaves. What is the most likely cause of this
condition?
A. Inadequate attic ventilation causing moisture buildup
B. Warm, moist air from the conditioned space leaking into the attic and condensing on cold
surfaces
C. Vapor barrier installed on the wrong side of the insulation
D. Excessive solar gain melting snow on the roof
Correct Answer: B
Explanation: Frost on roof sheathing during winter indicates warm, moist air is reaching a cold
surface. The thermal boundary is at the ceiling plane, and air leakage bypasses (thermal
bypasses) allow conditioned air carrying moisture to enter the cold attic. When this moist air
contacts the cold roof sheathing (below dew point), condensation occurs and freezes. Option A
is incorrect because ventilation alone doesn't create frost—moisture source is required. Option
C is incorrect because vapor barriers control diffusion, not bulk air leakage which is the primary
moisture transport mechanism. Option D is incorrect as solar gain would not cause frost
formation. This demonstrates the importance of air sealing the thermal boundary per RESNET
standards.
2. A rater measures the following conditions in a home: Indoor temperature 70°F, indoor
relative humidity 40%, outdoor temperature 20°F. Using the psychrometric chart, what is the
approximate dew point temperature of the indoor air?
A. 45°F
B. 55°F
C. 65°F
D. 75°F
Correct Answer: A
,Explanation: At 70°F and 40% relative humidity, the dew point is approximately 45°F. This is
calculated using the psychrometric relationship or chart: the temperature at which air becomes
saturated when cooled at constant moisture content. This matters because any surface below
45°F in this home will condense moisture. Option B (55°F) would correspond to roughly 60% RH,
Option C (65°F) to near-saturation, and Option D (75°F) is impossible as dew point cannot
exceed dry bulb temperature. Understanding dew point is critical for identifying condensation
risks on windows, thermal bridges, and within building assemblies.
3. During a blower door test, a rater measures -50 Pascals with respect to outside. What does
this negative pressure indicate?
A. The house is under positive pressure relative to outside
B. The house is under negative pressure relative to outside
C. The reference pressure gauge is malfunctioning
D. The blower door fan is running in reverse
Correct Answer: B
Explanation: In building science, pressure is always referenced to outside. A negative value (-50
Pa) indicates the house is at lower pressure than outdoors, meaning air is being exhausted from
the building (suction). This is the standard configuration for depressurization testing per RESNET
standards. Option A describes positive pressure. Option C is incorrect as negative readings are
normal and expected. Option D is incorrect because fan direction determines pressurization vs.
depressurization, not the sign of the pressure reading itself. The convention is: negative =
depressurization, positive = pressurization.
4. The stack effect in a three-story home is most pronounced during which condition?
A. Hot, windy summer days
B. Cold winter conditions with indoor heating
C. Mild spring days with minimal temperature difference
D. During mechanical ventilation operation only
Correct Answer: B
Explanation: Stack effect is driven by temperature differences (density differences) between
indoor and outdoor air. In winter, warm indoor air is less dense and rises, creating positive
pressure at the top of the building and negative pressure at the bottom, while cold outdoor air
is drawn in at lower levels. This temperature difference (ΔT) creates the stack pressure. Option A
,involves wind effect, not stack effect. Option C minimizes stack effect due to small ΔT. Option D
is incorrect because stack effect is a natural phenomenon independent of mechanical systems,
though mechanical systems can interact with it. The pressure difference follows: ΔP = ρ × g × h ×
(ΔT/T), making winter conditions most severe.
5. A home has a wall assembly consisting of: interior gypsum board, 2x6 studs with R-19
fiberglass batts, OSB sheathing, and vinyl siding. What is the approximate whole-wall R-value
considering thermal bridging?
A. R-19
B. R-15
C. R-13
D. R-22
Correct Answer: C
Explanation: The whole-wall R-value is significantly less than the cavity insulation R-19 due to
thermal bridging through the wood studs (R-6.88 per inch, approximately R-6.88 for 2x6 actual
depth). Wood studs comprise approximately 15-20% of the wall area. Using parallel path
calculation or hot-box testing, a 2x6 wall with R-19 batts typically achieves R-13 to R-14 whole-
wall. Option A ignores thermal bridging. Option B might represent a 2x4 wall calculation. Option
D is impossible as it exceeds the cavity insulation value. This demonstrates why continuous
insulation (ci) is specified in energy codes to mitigate thermal bridging.
6. A rater calculates that a home has 3,500 CFM50 of envelope leakage. The home has a volume
of 21,000 cubic feet. What is the ACH50 (Air Changes per Hour at 50 Pascals)?
A. 10 ACH50
B. 6 ACH50
C. 3 ACH50
D. 1.5 ACH50
Correct Answer: A
Explanation: ACH50 = (CFM50 × 60 minutes) ÷ Volume. Calculation: (3,500 × 60) ÷ 21,000 =
210,000 ÷ 21,000 = 10 ACH50. This is a relatively leaky home. Option B would be 2,100 CFM50.
Option C would be 1,050 CFM50. Option D would be 525 CFM50. ACH50 is the standard metric
for comparing envelope tightness across homes of different sizes and is required for HERS Index
, calculations. The 2015 IECC requires 3 or 5 ACH50 depending on climate zone for new
construction.
7. Which of the following represents a convective heat transfer mechanism?
A. Heat flowing through a solid wall stud
B. Warm air rising from a heating register
C. Sunlight warming a carpeted floor
D. Heat transfer across an insulated attic floor
Correct Answer: B
Explanation: Convection is heat transfer through fluid motion (liquids or gases). Warm air rising
from a register is natural convection driven by density differences. Option A describes
conduction (solid-to-solid). Option C describes radiation (electromagnetic waves). Option D
primarily describes conduction through the insulation layer, though air leakage would involve
convection. Understanding these three heat transfer modes—conduction, convection, and
radiation—is fundamental to building science and diagnosing thermal comfort issues.
8. A home has a blower door reading of 2,400 CFM50. The rater needs to estimate the natural
air leakage rate. Using the Lawrence Berkeley Laboratory (LBL) correlation factor for a typical
two-story home, what is the approximate CFM natural?
A. 240 CFM
B. 480 CFM
C. 120 CFM
D. 720 CFM
Correct Answer: A
Explanation: The LBL correlation typically uses a factor of 0.10 to 0.20 depending on climate
and building height. For a typical two-story home, 0.10 is commonly used: 2,400 CFM50 × 0.10 =
240 CFM natural. This estimates average annual infiltration under natural weather conditions.
Option B uses 0.20 (appropriate for very tall buildings or extreme climates). Option C uses 0.05
(too low). Option D uses 0.30 (inappropriate for residential). This conversion is essential for
calculating ventilation requirements and energy impacts of infiltration in HERS ratings.
Correct Answers| (2026|2027)
SECTION 1: BUILDING SCIENCE PRINCIPLES (Questions 1-15)
1. A home in Climate Zone 5 has an unconditioned attic with a ceiling plane consisting of R-38
fiberglass batts installed over drywall. During a winter inspection, the rater observes frost on
the underside of the roof sheathing near the eaves. What is the most likely cause of this
condition?
A. Inadequate attic ventilation causing moisture buildup
B. Warm, moist air from the conditioned space leaking into the attic and condensing on cold
surfaces
C. Vapor barrier installed on the wrong side of the insulation
D. Excessive solar gain melting snow on the roof
Correct Answer: B
Explanation: Frost on roof sheathing during winter indicates warm, moist air is reaching a cold
surface. The thermal boundary is at the ceiling plane, and air leakage bypasses (thermal
bypasses) allow conditioned air carrying moisture to enter the cold attic. When this moist air
contacts the cold roof sheathing (below dew point), condensation occurs and freezes. Option A
is incorrect because ventilation alone doesn't create frost—moisture source is required. Option
C is incorrect because vapor barriers control diffusion, not bulk air leakage which is the primary
moisture transport mechanism. Option D is incorrect as solar gain would not cause frost
formation. This demonstrates the importance of air sealing the thermal boundary per RESNET
standards.
2. A rater measures the following conditions in a home: Indoor temperature 70°F, indoor
relative humidity 40%, outdoor temperature 20°F. Using the psychrometric chart, what is the
approximate dew point temperature of the indoor air?
A. 45°F
B. 55°F
C. 65°F
D. 75°F
Correct Answer: A
,Explanation: At 70°F and 40% relative humidity, the dew point is approximately 45°F. This is
calculated using the psychrometric relationship or chart: the temperature at which air becomes
saturated when cooled at constant moisture content. This matters because any surface below
45°F in this home will condense moisture. Option B (55°F) would correspond to roughly 60% RH,
Option C (65°F) to near-saturation, and Option D (75°F) is impossible as dew point cannot
exceed dry bulb temperature. Understanding dew point is critical for identifying condensation
risks on windows, thermal bridges, and within building assemblies.
3. During a blower door test, a rater measures -50 Pascals with respect to outside. What does
this negative pressure indicate?
A. The house is under positive pressure relative to outside
B. The house is under negative pressure relative to outside
C. The reference pressure gauge is malfunctioning
D. The blower door fan is running in reverse
Correct Answer: B
Explanation: In building science, pressure is always referenced to outside. A negative value (-50
Pa) indicates the house is at lower pressure than outdoors, meaning air is being exhausted from
the building (suction). This is the standard configuration for depressurization testing per RESNET
standards. Option A describes positive pressure. Option C is incorrect as negative readings are
normal and expected. Option D is incorrect because fan direction determines pressurization vs.
depressurization, not the sign of the pressure reading itself. The convention is: negative =
depressurization, positive = pressurization.
4. The stack effect in a three-story home is most pronounced during which condition?
A. Hot, windy summer days
B. Cold winter conditions with indoor heating
C. Mild spring days with minimal temperature difference
D. During mechanical ventilation operation only
Correct Answer: B
Explanation: Stack effect is driven by temperature differences (density differences) between
indoor and outdoor air. In winter, warm indoor air is less dense and rises, creating positive
pressure at the top of the building and negative pressure at the bottom, while cold outdoor air
is drawn in at lower levels. This temperature difference (ΔT) creates the stack pressure. Option A
,involves wind effect, not stack effect. Option C minimizes stack effect due to small ΔT. Option D
is incorrect because stack effect is a natural phenomenon independent of mechanical systems,
though mechanical systems can interact with it. The pressure difference follows: ΔP = ρ × g × h ×
(ΔT/T), making winter conditions most severe.
5. A home has a wall assembly consisting of: interior gypsum board, 2x6 studs with R-19
fiberglass batts, OSB sheathing, and vinyl siding. What is the approximate whole-wall R-value
considering thermal bridging?
A. R-19
B. R-15
C. R-13
D. R-22
Correct Answer: C
Explanation: The whole-wall R-value is significantly less than the cavity insulation R-19 due to
thermal bridging through the wood studs (R-6.88 per inch, approximately R-6.88 for 2x6 actual
depth). Wood studs comprise approximately 15-20% of the wall area. Using parallel path
calculation or hot-box testing, a 2x6 wall with R-19 batts typically achieves R-13 to R-14 whole-
wall. Option A ignores thermal bridging. Option B might represent a 2x4 wall calculation. Option
D is impossible as it exceeds the cavity insulation value. This demonstrates why continuous
insulation (ci) is specified in energy codes to mitigate thermal bridging.
6. A rater calculates that a home has 3,500 CFM50 of envelope leakage. The home has a volume
of 21,000 cubic feet. What is the ACH50 (Air Changes per Hour at 50 Pascals)?
A. 10 ACH50
B. 6 ACH50
C. 3 ACH50
D. 1.5 ACH50
Correct Answer: A
Explanation: ACH50 = (CFM50 × 60 minutes) ÷ Volume. Calculation: (3,500 × 60) ÷ 21,000 =
210,000 ÷ 21,000 = 10 ACH50. This is a relatively leaky home. Option B would be 2,100 CFM50.
Option C would be 1,050 CFM50. Option D would be 525 CFM50. ACH50 is the standard metric
for comparing envelope tightness across homes of different sizes and is required for HERS Index
, calculations. The 2015 IECC requires 3 or 5 ACH50 depending on climate zone for new
construction.
7. Which of the following represents a convective heat transfer mechanism?
A. Heat flowing through a solid wall stud
B. Warm air rising from a heating register
C. Sunlight warming a carpeted floor
D. Heat transfer across an insulated attic floor
Correct Answer: B
Explanation: Convection is heat transfer through fluid motion (liquids or gases). Warm air rising
from a register is natural convection driven by density differences. Option A describes
conduction (solid-to-solid). Option C describes radiation (electromagnetic waves). Option D
primarily describes conduction through the insulation layer, though air leakage would involve
convection. Understanding these three heat transfer modes—conduction, convection, and
radiation—is fundamental to building science and diagnosing thermal comfort issues.
8. A home has a blower door reading of 2,400 CFM50. The rater needs to estimate the natural
air leakage rate. Using the Lawrence Berkeley Laboratory (LBL) correlation factor for a typical
two-story home, what is the approximate CFM natural?
A. 240 CFM
B. 480 CFM
C. 120 CFM
D. 720 CFM
Correct Answer: A
Explanation: The LBL correlation typically uses a factor of 0.10 to 0.20 depending on climate
and building height. For a typical two-story home, 0.10 is commonly used: 2,400 CFM50 × 0.10 =
240 CFM natural. This estimates average annual infiltration under natural weather conditions.
Option B uses 0.20 (appropriate for very tall buildings or extreme climates). Option C uses 0.05
(too low). Option D uses 0.30 (inappropriate for residential). This conversion is essential for
calculating ventilation requirements and energy impacts of infiltration in HERS ratings.
