Suppose an underground storage tank has been leaking for many years, contaminating a groundwater and causing a contaminant concentration directly beneath the site of 0.30 mg/L. The contamination is flowing at the rate of 0.5 ft/day toward a public drinking water well 1 mile away. The contaminant degrades with a rate constant of 1.94 x 10^-4 1/day. Draw a picture of the system. Estimate the steady-state pollutant concentration expected at the well. If the slope factor is 0.02 (mg/kg-day)^-1.

Required:
Estimate the cancer risk for an adult male drinking the water for 10 years.

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

(a) At a simple interest rate of 12% per year, determine how long it will take $5000 to increase to twice as much. (b) Compare the time it will take to double if the rate is 20% per year simple interest.