Answer:
given,
BOD₅ = 150 mg/L at 20°C
k = 0.23/day
using formula
[tex]L = L_0(1-e^{-kt})[/tex]
a) [tex]150= L_0(1-e^{-0.23\times 5})\\150=L_0\times 0.683\\L_0=219.62\ mg/L[/tex]
L₀ = 219.62 mg/L
b) [tex]k_{15}=k_{20}(1.047)^{T-20}\\k_{15}=0.23\times (1.047)^{-5}\\k_{15}=0.183[/tex]
k = 0.183
c) BOD₅ at 15⁰C
[tex]L = L_0(1-e^{-kt})[/tex]
[tex]L =219.62(1-e^{-0.183\times 5})[/tex]
L = 131.772 mg/L
