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Worksheet Grade 12 Math

Pinehaven Distribution System: Modeling Chlorine Residual Decay

A grade 12 worksheet in which students build and use an exponential decay function for chlorine residual across a fictional town distribution system.

Objectives

  1. Students will evaluate an exponential decay function to model chlorine residual over time in a distribution system.
  2. Students will solve an exponential equation for an unknown time value given a target residual level.

Vocabulary

Chlorine residual
Chlorine residual is the small amount of chlorine still left in the water after it leaves the treatment plant. It keeps protecting the water from germs all the way to your faucet.
Distribution system
A distribution system is the network of pipes, pumps, valves, and storage tanks that carries treated water from a treatment plant to homes, schools, and businesses. Most of it is buried underground beneath streets.

Also: exponential decay, decay constant

Teacher overview

Pinehaven (a fictional town used for this worksheet) treats its drinking water and sends it into a distribution system where water that left the treatment plant with a fresh chlorine residual can take hours, or even days, to reach customers. The residual does not stay constant along the way: it decays over time and distance as chlorine reacts with organic material and pipe surfaces. Federal rules set a maximum residual disinfectant level of 4.0 mg/L for chlorine, and utilities typically maintain a distribution residual between 0.2 and 2.0 mg/L, a range wide enough that engineers need a model to know how long water can travel before its residual falls too low.

This worksheet models Pinehaven's residual decay with an exponential function, C(t) = C₀ × e^(−kt), where C₀ is the residual leaving the plant, k is a decay constant specific to this fictional system, and t is hours since the water left the plant. A common misconception is expecting chlorine residual to decay by the same fixed amount every hour, the way a bank balance drops by a flat withdrawal. Exponential decay instead loses the same fraction of whatever residual remains, so the actual mg/L lost each hour keeps shrinking even as the fraction lost stays the same, which is exactly the pattern items 4 and 9 ask students to reason through.

Pinehaven Distribution System: Modeling Chlorine Residual Decay — Grade 12 Math
Name: ____________________________ Date: ______________

Student Worksheet

For this fictional scenario, water leaves the Pinehaven treatment plant with a chlorine residual of C₀ = 1.8 mg/L. Engineers modeled how the residual decays as water moves through the distribution system using C(t) = 1.8 × e^(−0.015t), where t is the number of hours since the water left the plant and the decay constant is k = 0.015 per hour.

  1. Calculate the residual, C(t), after 10 hours in the system. Round to the nearest hundredth of a mg/L.

  2. Calculate the residual after 24 hours (one full day) in the system. Round to the nearest hundredth of a mg/L.

  3. Calculate the residual after 48 hours (two full days) in the system. Round to the nearest hundredth of a mg/L.

  4. Is this decay pattern linear or exponential? Justify your answer using the definition of exponential decay: a function that changes by the same multiplying factor over equal time intervals, rather than by the same fixed amount.

  5. Pinehaven's target minimum residual is 0.5 mg/L, the low end of the typical range utilities maintain. Solve the equation 0.5 = 1.8 × e^(−0.015t) for t, the number of hours until the residual first drops to this level. Round to the nearest whole hour.

  6. A booster chlorination station in a second part of town adds a flat 0.3 mg/L directly to the decayed residual at the 24-hour mark (your answer to problem 2). What is the residual immediately after the booster station? Round to the nearest hundredth of a mg/L.

  7. A second distribution zone leaves the plant with a higher starting residual of C₀ = 2.0 mg/L but decays at the same rate, k = 0.015 per hour. Calculate this zone's residual at t = 24 hours. Round to the nearest hundredth of a mg/L.

  8. Compare your answers to problems 2 and 7. By how many mg/L does the higher-starting zone's residual exceed the original zone's residual at the same 24-hour mark? Round to the nearest hundredth.

  9. The second zone in problem 7 starts with a higher residual than the original zone but decays at the same rate k. Explain whether the second zone will take more time or less time than the original zone to reach the 0.5 mg/L target, and explain why having the same decay constant k does not mean both zones reach 0.5 mg/L at the same time.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: 1.55 mg/L.
    C(10) = 1.8 × e^(−0.015 × 10) = 1.8 × e^(−0.15) ≈ 1.8 × 0.8607 = 1.5493 mg/L, which rounds to 1.55 mg/L.

  2. Answer: 1.26 mg/L.
    C(24) = 1.8 × e^(−0.015 × 24) = 1.8 × e^(−0.36) ≈ 1.8 × 0.6977 = 1.2558 mg/L, which rounds to 1.26 mg/L.

  3. Answer: 0.88 mg/L.
    C(48) = 1.8 × e^(−0.015 × 48) = 1.8 × e^(−0.72) ≈ 1.8 × 0.4868 = 0.8762 mg/L, which rounds to 0.88 mg/L.

  4. Answer: Exponential.
    The residual does not lose a fixed number of mg/L each hour; instead it loses the same fraction of whatever remains, matching the definition of exponential decay rather than the constant-difference pattern of a linear function. This can be checked from problems 1 through 3: the residual falls by about 0.54 mg/L over the first day but only about 0.38 mg/L over the second full day even though the elapsed time is the same 24 hours, which a linear model would not produce.

  5. Answer: 85 hours.
    0.5 = 1.8 × e^(−0.015t) → 0.5 ÷ 1.8 = e^(−0.015t) → ln(0.2778) = −0.015t → −1.2809 = −0.015t → t = 85.40, which rounds to 85 hours.

  6. Answer: 1.56 mg/L.
    1.2558 mg/L (from problem 2) + 0.3 mg/L = 1.5558 mg/L, which rounds to 1.56 mg/L.

  7. Answer: 1.40 mg/L.
    C(24) = 2.0 × e^(−0.015 × 24) = 2.0 × e^(−0.36) ≈ 2.0 × 0.6977 = 1.3954 mg/L, which rounds to 1.40 mg/L.

  8. Answer: 0.14 mg/L.
    1.3954 mg/L − 1.2558 mg/L = 0.1396 mg/L, which rounds to 0.14 mg/L.

  9. Answer: The second zone will take more time.
    Because both zones decay by the same fraction each hour, the zone that starts higher (2.0 mg/L versus 1.8 mg/L) has farther to fall before reaching 0.5 mg/L, so it takes more hours even though the rate constant k is identical. The decay constant k controls how fast the residual shrinks as a fraction of its current value, not the total number of hours before a fixed target is reached, and that total time depends on the starting value too.

Differentiation

Support (below grade level)

Provide a table of e^(−0.015t) values already calculated for t = 0, 10, 24, and 48 hours, so students can focus on multiplying by C₀ rather than operating the exponent key. For problem 5, walk through the algebraic steps of isolating the exponential term and taking the natural log as a class before students solve a similar equation on their own.

Extension (above grade level)

Ask students to find the "half-life" of the Pinehaven chlorine residual, the number of hours it takes the residual to drop to exactly half its starting value, by solving 0.5 × C₀ = C₀ × e^(−0.015t) for t, and to verify that this half-life is the same regardless of what starting value C₀ they choose.

ELL supports

Pre-teach "residual," "decay," and "distribution system" using a simple diagram of water moving from a treatment plant through pipes to houses, with a shrinking color gradient representing decreasing chlorine strength. Allow students to describe the shape of the decay pattern in problem 4 using a sketch of the curve alongside their written justification.