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Mandarinka [93]
4 years ago
12

Estimate the chronic daily intake of toluene from exposure to a city water supply that contains a toluene concentration equal to

the drinking water standard of 1 mg/L. Assume the exposed individual is an adult female who consumes water at the adult rate for 70 years, that she abhors swimming, and that she takes a long (20 minute) bath every day. Assume that the average air concentration during the bath is 1 μg/m3. Assume that the dermal uptake from water (PC) is 9.0 x 10-6m/hr and that direct dermal absorption during bathing is no more than 80% of the available toluene because she is not completely submerged. Use the EPA lifetime exposure of 75 years. Express your answer in mg/kg-day. Which exposure pathway contributes the most to the chronic daily intake?
Engineering
1 answer:
enot [183]4 years ago
4 0

Answer:

The chronic daily intake is 0.0328608 mg/kg day

Explanation:

The CDI due to ingestion is equal to:

CDI=\frac{C*IR*ED*EF}{BW*AT}

Where

C = concentration = 1 mg/L

IR = 2.3 L/day

ED = risk dying per year = 70 year

EF = number of days per year = 365 days

AT = average tenure time = 75 year

BW = weight of body = 65.4 kg

CDI=\frac{1*2.3*70*365}{65.4*365*75} =0.0328mg/kgd

The ingestion rate is:

IR = 20/60 = 0.333 h/day

The CDI due to dermal contact is equal to:

CDI_{2} =\frac{1*1.69*9x10^{-6}*0.333*70*365 }{65.4*365*75}*0.8x10^{3}  =5.41x10^{-5} mg/kgd

The CDI due to inhalation is equal to:

CDI_{3} =\frac{0.001*0.471*70*365}{65.4*365*75} =6.7x10^{-6} mg/kgd

The total is:

CDItotal = 0.0328 + 5.41x10⁻⁵ + 6.7x10⁻⁶ = 0.0328608 mg/kgday

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lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

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Here, the waters specific volume at initial condition is v_{1}, the containers volume is V, waters mass is m.

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Answer:

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Note, that the energy was taken away from the system.  

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