# DBT JRF 2020 — Question 28

Exam: DBT JRF 2020 · Section: A · Topic: Bioprocess Engineering · Difficulty: Hard

## Question

The temperature of media post sterilization drops from 100°C to 60°C in 40 min by simply keeping it on the lab bench and allowing slow atmospheric cooling to take place at an ambient temperature of 20°C. In the next 40 min, the approximate temperature (°C) of the media would be around

## Options

- **A.** 30°C
- **B.** 40°C
- **C.** 20°C
- **D.** 50°C

## Answer

Option A

## Explanation

Newton's Law of Cooling states: $\dfrac{dT}{dt} = -k(T - T_{\text{ambient}})$ , where $k$ is the cooling constant. The solution is $T(t) = T_{\text{ambient}} + (T_0 - T_{\text{ambient}})e^{-kt}$ . With $T_{\text{ambient}} = 20°C$ : At $t=0$ : $T=100°C$ (excess = 80°C above ambient). At $t=40$ min: $T=60°C$ (excess = 40°C — halved). So the excess temperature halves every 40 min. At $t=80$ min: excess = 20°C → $T = 20 + 20 = 40°C$ . The media would be approximately 40°C after the next 40 minutes.

Source: https://mypyqbuddy.com/questions/the-temperature-of-media-post-sterilization-drops-from-6a05a8a02adafec470293202