---
title: "Power measurement with static calorimeter"
subject: "RF and Microwave Engineering"
chapter: 8
key: "semester_7/rf_and_microwave_engineering:static-calorimeter-power"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/static-calorimeter-power
site: BCT Engineering
---

# Power measurement with static calorimeter

RF and Microwave Engineering · Chapter 8: RF and Microwave Measurements

- Asked in: 2077 Chaitra, 6 marks; 2071 Magh, 5 marks (asked 2 times)
- Video: [Power Measurement of Microwave Signal | Microwave Engineering | R K Classes | Lec 63 | Join Telegram](https://www.youtube.com/watch?v=i6x2HIbO0hA) — R K Classes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/static-calorimeter-power
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-8

## Question

Explain how microwave power is measured with static calorimeter.

## Short answer

A static calorimeter absorbs microwave power in a fixed thermally insulated mass and measures temperature rise. Power is P = m c DeltaT / t, with DC substitution used for calibration. Example: 0.1 kg water rising 2.4 C in 60 s gives 16.74 W.

## Answer

A **calorimeter** measures high microwave power by converting it completely into heat in a load and measuring the temperature rise. In a **static calorimeter**, a fixed (non-flowing) mass of absorbing material is heated.

#### Principle

```text
P = m·c·ΔT / t
m  = mass of absorbing load (kg)
c  = specific heat (J/kg·°C)
ΔT = temperature rise (°C) in time t (s)
```

#### Construction and working

```text
 RF in      insulated (thermally isolated) load
 ======[ waveguide ]==[ absorbing load ]
                         |  thermometer /
                         |  thermocouple
                      [ ΔT readout ]
```

1. A matched, thermally insulated load (water, lossy dielectric or absorbing block) terminates the waveguide.
2. The microwave power is applied for a known time t; the load absorbs all of it.
3. The temperature rise ΔT is measured with a thermometer or thermocouple.
4. Power is calculated from P = m·c·ΔT/t. To remove the need to know m and c exactly, the same load is calibrated by applying a known DC or 50 Hz power and noting its temperature rise (**substitution method**):

```text
P_RF = P_DC × (ΔT_RF / ΔT_DC)   (same time)
```

#### Example

A 0.1 kg water load (c = 4186 J/kg·°C) rises by 2.4 °C in 60 s:

```text
P = 0.1 × 4186 × 2.4 / 60 = 16.74 W
```

#### Features

- Suitable for medium to high power (watts to kilowatts); accurate (about ±1 %) because it is an absolute method.
- Slow; heat losses through insulation and the mount cause error, so good thermal isolation is needed.
- For continuous high power, the circulating (flow) calorimeter is used instead.

## Check yourself

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

1. What is the basic formula for power in a static calorimeter?
   A. P = m c DeltaT / t
   B. P = m c t / DeltaT
   C. P = DeltaT / (m c t)
   D. P = m DeltaT t / c
   Answer: A. Absorbed energy m c DeltaT divided by time t gives power. (Key point: Calorimeter principle)

2. A 0.1 kg water load (c = 4186 J/kg.C) rises 2.4 C in 60 s. What is the power?
   A. 16.74 W
   B. 1.674 W
   C. 167.4 W
   D. 60.3 W
   Answer: A. P = 0.1 x 4186 x 2.4 / 60 = 16.74 W. (Key point: Numerical example)

3. Why is the load calibrated with a known DC power?
   A. To avoid needing exact m and c: P_RF = P_DC x (DeltaT_RF / DeltaT_DC) for the same time
   B. To convert RF to DC permanently
   C. To measure the wavelength
   D. To increase the heat losses
   Answer: A. Comparing temperature rises under RF and known DC power gives RF power without knowing m and c exactly. (Key point: Substitution calibration)

4. What is a limitation of the static calorimeter?
   A. It is slow and heat losses through insulation cause error
   B. It cannot measure high power
   C. It is accurate only in microwatts
   D. It measures instantaneous peak power
   Answer: A. The method is an accurate absolute method for medium to high power but needs good thermal isolation and is slow. (Key point: Limitations)

## Related in RF and Microwave Engineering

- [Microwave vs low frequency measurement; static calorimeter](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Ameasurement-differences-static-calorimeter) — RF and Microwave Engineering, chapter 8
- [Power measurement with circulating calorimeter](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Acirculating-calorimeter-power) — RF and Microwave Engineering, chapter 8, asked 2×
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