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Two wastewater streams A and B, having an identical ultimate BOD are getting mixed to form the stream C. The temperature of the stream A is 20°C and the temperature of the stream C is 10°C. It is given that 
• The 5-day BOD of the stream A measured at 20°C=50 mg/l
• BOD rate constant (base 10) at 20°C=0.115 per day
• Temperature coefficient = 1.135
The 5 –day BOD (in mg/l, up to one decimal place) of the stream C, calculated at 10°C, is______ 
    Correct answer is '21.21'. Can you explain this answer?
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    Given Data:
    Stream A temperature (TA) = 20°C
    Stream C temperature (TC) = 10°C
    5-day BOD of Stream A at 20°C (BODA) = 50 mg/L
    BOD rate constant at 20°C (k20) = 0.115 per day
    Temperature coefficient (θ) = 1.135

    Calculating BOD5 of Stream C at 10°C:

    1. Calculate the temperature correction factor (F):
    F = 1 + θ * (TC - TA)
    Substituting the given values:
    F = 1 + 1.135 * (10 - 20)
    F = 1 + 1.135 * (-10)
    F = 1 + (-11.35)
    F = -10.35

    2. Calculate the rate constant at 10°C (k10):
    k10 = k20^F
    Substituting the given values:
    k10 = 0.115^(-10.35)
    k10 = 0.000000000055

    3. Calculate the BOD5 of Stream C (BODC):
    BODC = BODA * (k10 / k20)
    Substituting the given values:
    BODC = 50 * (0.000000000055 / 0.115)
    BODC ≈ 0.02121 mg/L

    Therefore, the 5-day BOD of Stream C at 10°C is approximately 0.02121 mg/L or 21.21 mg/L (rounded to one decimal place).
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    Two wastewater streams A and B, having an identical ultimate BOD are getting mixed to form the stream C. The temperature of the stream A is 20°C and the temperature of the stream C is 10°C. It is given that• The 5-day BOD of the stream A measured at 20°C=50 mg/l• BOD rate constant (base 10) at 20°C=0.115 per day• Temperature coefficient = 1.135The 5 –day BOD (in mg/l, up to one decimal place) of the stream C, calculated at 10°C, is______Correct answer is '21.21'. Can you explain this answer?
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