---
title: "Temperature effect on BOD; 5-day BOD"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:temperature-effect-decomposition"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/temperature-effect-decomposition
site: BCT Engineering
---

# Temperature effect on BOD; 5-day BOD

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2075 Bhadra, 8 marks
- Video: [Finding Five Day BOD, Ultimate BOD and BOD at Different Temperature for a Wastewater Sample|Problem](https://www.youtube.com/watch?v=M8GrD3vQkg8) — Master Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/temperature-effect-decomposition
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

Show the effect of temperature on decomposition rate of organic matters in waste water. If one day BOD of a sewage sample at 20°C is 100 mg/l, what will be its five day BOD at 20°C? Consider K20 = 0.1/day.

## Short answer

Decomposition rate increases with temperature per van't Hoff-Arrhenius relation (θ=1.047); ultimate BOD unchanged. Given 1-day BOD=100 mg/L at 20°C and K20=0.1/day, ultimate BOD=486.2 mg/L, so 5-day BOD at 20°C=332.5 mg/L.

## Answer

#### Effect of temperature on the rate of decomposition
Decomposition of organic matter is done by bacteria, and their activity increases as temperature increases (up to about 35-40°C). The rate constant follows the van't Hoff-Arrhenius relation:

$$
K_T = K_{20}\,\theta^{\,(T-20)}
$$

where $\theta$ is the temperature coefficient, commonly taken as 1.047 (some books use 1.056 for 20-30°C and 1.135 for 4-20°C). With $K_{20}=0.1$ per day and $\theta=1.047$:

| T (°C) | $K_T$ (day⁻¹) | $K_T/K_{20}$ |
|---|---|---|
| 5 | 0.0502 | 0.502 |
| 10 | 0.0632 | 0.632 |
| 15 | 0.0795 | 0.795 |
| 20 | 0.1000 | 1.000 |
| 25 | 0.1258 | 1.258 |
| 30 | 0.1583 | 1.583 |
| 35 | 0.1992 | 1.992 |

- Higher temperature: the reaction is faster, the BOD is satisfied in a shorter time, and also less oxygen dissolves in water. The ultimate BOD $L$ remains the same.
- Lower temperature: bacterial activity slows, the BOD exerted in 5 days is less, and the sewage decomposes more slowly.

```text
 BOD exerted
   |        ______________ 30 C
   |     __/  _____________ 20 C
   |   _/  __/  ___________ 10 C
   |  / __/ __/
   +--------------------------> time
```

All curves reach the same ultimate BOD $L$, but at different times. For this reason the standard BOD test is done at 20°C.

#### Numerical
Given $y_1=100$ mg/L at 20°C, $K_{20}=0.1$ day⁻¹ (base 10).

$$
\begin{aligned}
L&=\frac{y_1}{1-10^{-K t}}=\frac{100}{1-10^{-0.1}}=\frac{100}{1-0.7943}=486.2\ \text{mg/L}\\
y_5&=L\left(1-10^{-0.1\times5}\right)=486.2\left(1-0.3162\right)=332.5\ \text{mg/L}
\end{aligned}
$$

**Answer: Five-day BOD at 20°C = 332.5 mg/L** (ultimate BOD = 486.2 mg/L).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does a rise in temperature (up to about 35-40°C) affect BOD decomposition?
   A. Bacterial activity increases, so the rate constant rises and BOD is satisfied sooner
   B. The ultimate BOD increases
   C. Decomposition stops
   D. The rate constant falls
   Answer: A. Higher temperature speeds bacteria, but L is unchanged. (Key point: Effect of temperature)

2. Which relation expresses the effect of temperature on k?
   A. K_T = K20 + theta x T
   B. K_T = K20 x theta^(T-20), theta = 1.047
   C. K_T = K20 / T
   D. K_T = K20 x theta^(20-T) for T above 20°C
   Answer: B. van't Hoff-Arrhenius form with temperature coefficient theta. (Key point: Temperature relation)

3. Why is the BOD test standardised at 20°C?
   A. Bacteria die above 20°C
   B. L is highest at 20°C
   C. All curves reach the same L but at different times, so a fixed temperature allows comparison
   D. DO is zero at 20°C
   Answer: C. A standard temperature makes results comparable. (Key point: Why 20°C)

4. If y1 = 100 mg/L and K20 = 0.1 per day (base 10), what is y5 at 20°C?
   A. About 486 mg/L
   B. About 100 mg/L
   C. About 500 mg/L
   D. About 332.5 mg/L
   Answer: D. L = 100 / 0.2057 = 486.2; y5 = 486.2 x 0.6838 = 332.5 mg/L. (Key point: Numerical)

## Related in Sanitary Engineering

- [1-day BOD 300 mg/l to 5-day at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-300-to-5day-30c) — Sanitary Engineering, chapter 5
- [BOD, COD; ultimate BOD of 480 mg/l](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-cod-ultimate-bod-480) — Sanitary Engineering, chapter 5
- [Ultimate BOD and rate constant](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-ultimate-rate-constant-100-210) — Sanitary Engineering, chapter 5
- [BOD of Bagmati sample at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-thapathali-4day-30c) — Sanitary Engineering, chapter 5
- [5-day BOD and kg BOD in 3.2 ML](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-dilution-2p5-in-250-7p8) — Sanitary Engineering, chapter 5, asked 2×
- [Rate constant and remaining BOD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-rate-250-325) — Sanitary Engineering, chapter 5

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