GATE 2024 BT Q.35: solution and answer

GATE 2024 · BTQ.35 · 1 mark · NAT

The decimal reduction time of a microbe during sterilization at 120 °C with a first order thermal death rate constant of 1 min $^{- 1}$ will be ________ min (rounded off to 1 decimal place).

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Answer and solution

ANSWER

The decimal reduction time is $D = \frac{ln 10}{k _{d}} = 2.303 min \approx 2.3 min$ .

Official answer: 2.3Source: Official answer key, GATE 2024 BT (IISc Bengaluru)

First-order thermal death: $\frac{d N}{d t} = - k_{d} N$ , so $N = N_{0} e^{- k_{d} t}$ . Here $N$ is the number of viable cells at time $t$ , $N_{0}$ is the number at $t = 0$ , and $k_{d}$ is the first-order specific death rate constant ( $k_{d} = 1 min^{- 1}$ at $120^{\circ} C$ ).
Take natural logs: $ln \frac{N _{0}}{N} = k_{d} t$ .
The decimal reduction time $D$ is the time needed to kill 90% of the viable cells, so that $N = 0.1 N_{0}$ . That is a tenfold drop: $\frac{N _{0}}{N} = 10$ at $t = D$ .
So $ln 10 = k_{d} D$ , which gives $D = \frac{ln 10}{k _{d}} = \frac{2.303}{k _{d}}$
Substitute $k_{d}$ : $D = \frac{2.303}{1 min ^{- 1}} = 2.303 min \approx 2.3 min$ (to 1 decimal place).

KEY IDEA

In first-order death kinetics, the D-value (time for a 90% kill, a one-log reduction) depends only on the death rate constant: $D = \frac{ln 10}{k _{d}} \approx \frac{2.303}{k _{d}}$ . The temperature (120 °C) only tells you which $k_{d}$ applies. It does not enter the calculation.

WATCH OUT FOR

Using $lo g_{10}$ where $ln$ belongs gives $D = 1/ k_{d} = 1.0$ min. Using $ln 2$ , which gives the half-life of 0.7 min, is the other common slip. $k_{d}$ is a natural-log rate constant, so a tenfold reduction needs $ln 10 = 2.303$ . Some students also try to use 120 °C in an Arrhenius correction, but $k_{d}$ is already given at that temperature.

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GATE 2024 BT Q.35: solution and answer

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