Question

This is a re-worded version of Problem 8.25 from Mihelcic and Zimmerman:

The following data were derived from an experiment in which Poliomyelitis virus (aka, Poliovirus) was inactivated with free bromine.For purposes of this experiment, we can assume that all of the bromine was present in the form of hypobromous acid (HOBr).The experiment was performed in a batch reactor, and the HOBr concentration was maintained at a constant value of 1.0 mg/L.Based on these data, calculate the coefficient of specific lethality for inactivation of Poliovirus using HOBr.Calculate the time required to achieve 4 log₁₀ units of inactivation of this virus with 2.0 mg/L of HOBr in water.

Time (s) 0.0 2.0 4.0 6.0 8.0 10.0 12.0 N viable virus particles 1000 350 78 20

HINT: To solve this problem, it would be helpful to rearrange the Chick-Watson law (as presented in class) so that you can use a plot of the data presented above, transformed in the form of ln(N/No) vs. t, along with linear regression to estimate the coefficient of specific lethality. Note that to fit these data to the model, you should force your “best-fit” line to go through the origin. Many software packages, including Excel, have this capability built-in to their regression tools.

Answer 1

At bromine of 1mg/L or 1*10^-3kg/l

Ln(1/1000) = ? *10^-3*12

(-6.9/12)*10^3 = -575 => inactivation constant

Time required for 4 as inactivation constant at bromine of 2mg/L or 2*10^-3kg/l is

Ln(1/1000) = 4*2*T

T=( -6.9/8)*10^3

=> 862.5secs

time

0

2

4

6

8

10

12

ln(N/N₀)

0

-1.049822124

-2.551046452

-3.912023005

-5.11599581

-6.214608098

-6.907755279

inactivation of poliomyelitis virus 0 2 46 8 10 12 -1 -2 No. of viable virus particles inactivation of poliomyelitis virus -4 -5 -6 -7 -8 time

The remaining part of the answer that involves forcing the line to go through the origin can be solved by statistics experts.