GATE 2022 BT Q.63: solution and answer

GATE 2022 · BTQ.63 · 2 marks · NAT

A pilot sterilization was carried out in a vessel containing 100 m $^{3}$ medium with an initial spore concentration of $1 0^{8}$ spores/ml. The accepted level of contamination after sterilization is 1 spore in the entire vessel. The specific death rate constant for the spore is 2 min $^{- 1}$ at 121 °C. Assuming no death takes place during the heating and cooling cycles, the holding time at 121 °C (rounded off to nearest integer) is ________ min.

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

The holding time at 121 °C is about 18 min ( $ln 1 0^{16} /2 = 18.42$ min).

Official answer: 17 to 19Source: Official answer key, GATE 2022 BT (IIT Kharagpur)

Convert the vessel volume to millilitres: $V = 100 m^{3} = 100 \times 1 0^{6} mL = 1 0^{8} mL$ (since $1 m^{3} = 1 0^{6} mL$ ).
Initial number of spores in the whole vessel: $N_{0} = (1 0^{8} spores mL^{- 1}) (1 0^{8} mL) = 1 0^{16}$ spores.
Final number allowed in the whole vessel: $N = 1$ spore.
Thermal death of spores is first order: $\frac{d N}{d t} = - k_{d} N$ , which integrates at constant temperature to $ln \frac{N _{0}}{N} = k_{d} t$ . Here $k_{d} = 2 min^{- 1}$ is the specific death rate constant at $121^{\circ} C$ and $t$ is the time spent at that temperature.
Del factor needed: $\nabla = ln \frac{N _{0}}{N} = ln 1 0^{16} = 16 \times 2.3026 = 36.84$ .
Since no death is credited to heating or cooling, the whole Del factor must come from holding: $t_{h o l d} = \frac{ln ( N _{0} / N )}{k _{d}} = \frac{36.84}{2 min ^{- 1}} = 18.42 min$
Rounded to the nearest integer: $t_{h o l d} \approx 18 min$ .

KEY IDEA

Spore death is first order, so the Del factor you need is $\nabla = ln (N_{0} / N)$ , where $N_{0}$ and $N$ are the spore counts in the WHOLE vessel. If heating and cooling kill nothing, all of it has to come from holding, so $t_{h o l d} = \nabla/ k_{d}$ .

WATCH OUT FOR

Using concentrations instead of total counts: taking $ln (1 0^{8} /1)$ gives $9.2$ min. The limit of 1 spore applies to the whole $100 m^{3}$ vessel, so the volume must go into $N_{0}$ ( $1 0^{8} mL$ , not $1 0^{5}$ L mixed with a per-mL concentration). Another slip is using $lo g_{10}$ instead of $ln$ , which gives $16/2 = 8$ min.

From the GATE 2022 BT paper. Every question of this paper · Download the full paper · All 7 Upstream questions, with notes

GATE 2022 BT Q.63: solution and answer

Answer and solution

More Upstream questions