*Please explain it step by step so that I can understand* 3. A man weighing 70 k
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Question
*Please explain it step by step so that I can understand*
3. A man weighing 70 kg is married to a woman weighing 50 kg. They have a male child weighing 15 kg that will grow up to be 70 kg. The air they breathe contains 2 2m3 of a listed carcinogenic chemical, and the water they drink contains 6 |lg/L of the same chemical. Using the default parameters from the table below, estimate the total chemical dose per kg of body weight for each family member. Assume the body retains 80% of th air pollutant and 100% of the water pollutant (20 marks) Table 14.2 Standard default exposure factors for environmental risk assessments. These EPA quidelines are used for cakulating reasonable maximum exposure at a contaminaled site Exposure Frequncy Duration (Days/Year)Explanation / Answer
Solution:
Assumption land used is residential and agricultural.
Exposure dose (D) = (C × IR × AF × EF)/ BW
Where,
C = Contaminant concentration = Given 2 µg/m3 for air, 6 µg/L for water
IR = Intake rate of contaminant medium = From Given table 2L for water, 20m3 for air
AF = bioavailability factor = 1
EF = Exposure factor
BW = body weight
Exposure factor can be calculated as follows:
EF = (F × ED)/ AT
Where,
F = Frequency of exposure = given from table 350 days/year
ED = Exposure duration = 30 years (from table)
AT = Average time = (ED × 365)days/year
Calculation:
EF = (350 × 30) / (30 × 365) = 0.9589
D = ((2µg/m3) × (20 m3/day) × 1 × 0.9589)/ 70 = 0.548 µg/day/Kg
Man water chemical dose:
EF = (350 × 30) / (30 × 365) = 0.9589
D = ((6µg/L) × (2 L/day)× 1 × 0.9589)/ 70 = 0.164 µg/day/Kg
Total chemical dose in Man body = ( 0.80 ×0.548 ) + ( 1 × 0.164 ) = 0.6 µg/day/Kg
EF = (350 × 30) / (30 × 365) = 0.9589
D = ((2µg/m3) × (20 m3/day) × 1 × 0.9589)/ 50 = 0.77 µg/day/Kg
Woman water chemical dose:
EF = (350 × 30) / (30 × 365) = 0.9589
D = ((6µg/L) × (2 L/day)× 1 × 0.9589)/ 50 = 0.23 µg/day/Kg
Total chemical dose in woman body = (0.80 ×0.77 ) + ( 1 × 0.23 ) = 0.846 µg/day/Kg
For children average intake rate is generally half of the adult intake rate so,
EF = (350 × 30) / (30 × 365) = 0.9589
D = ((2µg/m3) × (10 m3/day) × 1 × 0.9589)/ 15 = 1.28 µg/day/Kg
Child water chemical dose:
EF = (350 × 30) / (30 × 365) = 0.9589
D = ((6µg/L) × (1 L/day)× 1 × 0.9589)/ 15 = 0.38 µg/day/Kg
Total chemical dose in child body = (0.80 ×1.28 ) + ( 1 × 0.38 ) = 1.4 µg/day/Kg
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