IICRC S500 Water Mitigation Exam
Questions and Answers Latest Update
The homeowner - ANSWER-is responsible to take the necessary steps to preserve and
protect their property from further damage
Four principles of water damage restoration: - ANSWER-Provide for health and safety
of workers and occupants; Document everything; Mitigate, and Dry
Four principles of drying - ANSWER-Extract, Evaporate, dehumidify, control temp
Extraction - ANSWER-is removing excess water and is the first step in mitigation
Evaporation - ANSWER-is achieved by using energy (heat) to transform water trapped
in porous materials into a vapor, thereby releasing the moisture from the materials.
Dehumidification - ANSWER-the process of removing water vapor from the air.
Rate of Dehumidification - ANSWER-must be at least equal to the rate of evaporation.
True or False - ANSWER-Elevated vapor pressure in a room Can cause secondary
damage to hygroscopic materials
True or False - ANSWER-The control of temperature is important to enhance both
evaporation and dehumidification for effective drying
True or False - ANSWER-When temperature increases evaporation of absorbed water
increases and water vapor suspended in air also increases.
Classification of Water Intrusion - ANSWER-described as Class 1, 2, 3, or 4. Must be
determined to calculate the amount of dehumidification required in drying process,
, based on the amount of wet surface area, permeance/porosity of materials in drying
area.
Class 1 - ANSWER-is the least amount of water absorption and evaporation load,
affecting less 5% of porous materials with minimal absorption into low evaporation
materials.
Class 2 - ANSWER-has a significant amount of water absorption and evaporation load,
affection 5-40% porous materials, with minimal absorption into low evaporated materials
Class 3 - ANSWER-has the greatest amount of water absorption and evaporation load,
affecting more than 40% porous materials and minimal absorption into low evaporated
materials
Class 4 - ANSWER-has deeply held or bound water, and significant absorption into low
evaporation materials.
Determine the Class of water for a 20' x 20' x 8 ' room, with an entirely wet floor.. -
ANSWER-Affected area/Total SF = % affected
400/1440 = .27777 x 100 = 28%
28% = Class 2
Determine the Class of Water for a 20' x 20' x 8' room, with only half the floor affected
and arriving within 24 hours of loss. - ANSWER-Affected area/total sf = % affected
200/1440 = .13888 x 100 = 14%
14% = Class 2
Determine Class of water for a 20' x 20' x 8' room, with entirely wet plywood subfloor
arriving 48 hours after loss - ANSWER-affected area/total sf = % affected
Class 4 due to deeply bound water sitting for long time
Categories of Water - ANSWER-there are four categories - determined by range of
contamination of the source and quality of water
Questions and Answers Latest Update
The homeowner - ANSWER-is responsible to take the necessary steps to preserve and
protect their property from further damage
Four principles of water damage restoration: - ANSWER-Provide for health and safety
of workers and occupants; Document everything; Mitigate, and Dry
Four principles of drying - ANSWER-Extract, Evaporate, dehumidify, control temp
Extraction - ANSWER-is removing excess water and is the first step in mitigation
Evaporation - ANSWER-is achieved by using energy (heat) to transform water trapped
in porous materials into a vapor, thereby releasing the moisture from the materials.
Dehumidification - ANSWER-the process of removing water vapor from the air.
Rate of Dehumidification - ANSWER-must be at least equal to the rate of evaporation.
True or False - ANSWER-Elevated vapor pressure in a room Can cause secondary
damage to hygroscopic materials
True or False - ANSWER-The control of temperature is important to enhance both
evaporation and dehumidification for effective drying
True or False - ANSWER-When temperature increases evaporation of absorbed water
increases and water vapor suspended in air also increases.
Classification of Water Intrusion - ANSWER-described as Class 1, 2, 3, or 4. Must be
determined to calculate the amount of dehumidification required in drying process,
, based on the amount of wet surface area, permeance/porosity of materials in drying
area.
Class 1 - ANSWER-is the least amount of water absorption and evaporation load,
affecting less 5% of porous materials with minimal absorption into low evaporation
materials.
Class 2 - ANSWER-has a significant amount of water absorption and evaporation load,
affection 5-40% porous materials, with minimal absorption into low evaporated materials
Class 3 - ANSWER-has the greatest amount of water absorption and evaporation load,
affecting more than 40% porous materials and minimal absorption into low evaporated
materials
Class 4 - ANSWER-has deeply held or bound water, and significant absorption into low
evaporation materials.
Determine the Class of water for a 20' x 20' x 8 ' room, with an entirely wet floor.. -
ANSWER-Affected area/Total SF = % affected
400/1440 = .27777 x 100 = 28%
28% = Class 2
Determine the Class of Water for a 20' x 20' x 8' room, with only half the floor affected
and arriving within 24 hours of loss. - ANSWER-Affected area/total sf = % affected
200/1440 = .13888 x 100 = 14%
14% = Class 2
Determine Class of water for a 20' x 20' x 8' room, with entirely wet plywood subfloor
arriving 48 hours after loss - ANSWER-affected area/total sf = % affected
Class 4 due to deeply bound water sitting for long time
Categories of Water - ANSWER-there are four categories - determined by range of
contamination of the source and quality of water
