Fecal Coliform (FC) concentration in river water was measured as 10780 cfu/100 ml. The FC concentration after the conventional water treatment, but before chlorination, was measured as 23 cfu/100 ml. The ‘Log Kill' (inactivation) of FC due to the conventional water treatment is closest to
The 'Log Kill' represents the logarithm (base 10) of the ratio between the initial concentration and the final concentration of a contaminant, indicating the level of inactivation achieved.
The formula used to calculate Log Kill is:
$ \text{Log Kill} = \log_{10} \left( \frac{\text{Initial Concentration}}{\text{Final Concentration}} \right) $
Given:
Substitute these values into the Log Kill formula:
$ \text{Log Kill} = \log_{10} \left( \frac{10780 \text{ cfu/100 ml}}{23 \text{ cfu/100 ml}} \right) $
$ \frac{10780}{23} \approx 468.7 $
$ \log_{10}(468.7) \approx 2.6708 $
The calculated Log Kill is approximately 2.67.
The Log Kill of 2.67 indicates that the conventional water treatment process inactivated Fecal Coliform bacteria by a factor of approximately $10^{2.67}$, which is around 468 times the original count.
The dose of chlorination, beyond which any further addition of chlorine will appear as free residual chlorine, is known as:
A city that has a supply of 15000 cubic meters of water per day is treated with a chlorine dosage of 0.5 ppm. For this purpose, the requirement of 25% bleaching powder per day would be.
Which one of the following is used to kill the pathogens?
An experiment shows that a concentration of 0.1 g/m3 of free available chlorine yield a 99% kill of bacteria in 8 minutes. Calculate disinfection rate constant (1 /min)? Assume that Chick's Law and Watson's Law hold with n = 1.
Order 4 disinfectants in increasing order of their disinfection power?