Answer: the cancer risk for an adult male drinking the water for 10 years is 2.88 × 10⁻⁶
Explanation:
firstly, we find the time t required to travel for the contaminant to the well;
Given that, contamination flowing rate = 0.5 ft/day
Distance of well from the site = 1 mile = 5280 ft
so t = 5280 / 0.5 = 10560 days
k is given as 1.94 x 10⁻⁴ 1/day
next we find the Pollutant concentration Ct in the well
Ct = C₀ × e^-( 1.94 x 10⁻⁴ × 10560)
Ct = 0.3 x e^-(kt)
Ct= 0.0386 mg/L
next we determine the chronic daily intake, CDI
CDI = (C x CR x EF x ED) / (BW x AT)
where C is average concentration of the contaminant(0.0368mg/L), CR is contact rate (2L/day), EF is exposure frequency (350days/Year), ED is exposure duration (10 years), BW is average body weight (70kg).
now we substitute
CDI = (0.0368 x 2 x 350 x 10) / ((70x 365) x 70)
= 257.7 / 1788500
= 0.000144 mg/Kg.day
CDI = 1.44 x 10⁻⁴ mg/kg.day
Finally we calculate the cancer risk, R
Slope factor SF is given as 0.02 Kg.day/mg
Risk, R = I x SF
= 1.44 x 10⁻⁴ mg/kg.day x 0.02Kg.day/mg
R = 2.88 × 10⁻⁶
therefore the cancer risk for an adult male drinking the water for 10 years is 2.88 × 10⁻⁶
