---
title: "Radiation properties and black and grey bodies"
subject: "Heat Transfer"
chapter: 4
key: "semester_5/heat_transfer:absorptivity-reflectivity-kirchhoff-black-grey"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/absorptivity-reflectivity-kirchhoff-black-grey
site: BCT Engineering
---

# Radiation properties and black and grey bodies

Heat Transfer · Chapter 4: Radiation Heat Transfer

- Asked in: year not given, 5 marks
- Video: [Blackbody and Graybody Radiation](https://www.youtube.com/watch?v=c2UkTeiPBpA) — Postcard Professor
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/absorptivity-reflectivity-kirchhoff-black-grey
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-4

## Question

Define absorptivity, reflectivity and transmissivity and show how they are related. Differentiate between a black body, a grey body and a real body. What is emissivity? State and prove Kirchhoff's law for total radiation.

## Short answer

Absorptivity α, reflectivity ρ, transmissivity τ satisfy α+ρ+τ=1. Black body: α=ε=1; grey: ε<1 constant; real: ε varies with λ. Emissivity ε = E/E_b. Kirchhoff's law: at thermal equilibrium, ε=α, proved by placing a small body in a black enclosure at same T, energy balance gives EA = α E_b A, so ε=α.

## Answer

#### Radiation properties
When radiation of total amount $G$ (irradiation) strikes a surface, part is absorbed, part reflected and part transmitted:
- **Absorptivity** $\alpha$ = fraction absorbed.
- **Reflectivity** $\rho$ = fraction reflected.
- **Transmissivity** $\tau$ = fraction transmitted.

$$
\alpha + \rho + \tau = 1
$$

For an opaque solid $\tau = 0$, so $\alpha + \rho = 1$. For a black body $\alpha = 1$; for a white body $\rho = 1$; for a transparent body $\tau = 1$.

#### Types of bodies
| Body | Property |
|---|---|
| Black | Absorbs all radiation, $\alpha = 1$, $\varepsilon = 1$; emits maximum at each $\lambda$ and $T$ |
| Grey | Emissivity $\varepsilon<1$ independent of wavelength; emits a constant fraction of black body |
| Real | $\varepsilon$ and $\alpha$ vary with wavelength, temperature and direction |

**Emissivity** $\varepsilon = E/E_b$ is the ratio of the emissive power of a surface to that of a black body at the same temperature.

#### Kirchhoff's law (proof)
Place a small body of area $A$, emissive power $E$ and absorptivity $\alpha$ inside a black enclosure at the same temperature $T$. The enclosure sends radiation $E_b$ per unit area on the body, of which the body absorbs $\alpha E_b A$. It emits $EA$. At thermal equilibrium (no net heat flow, since temperatures are equal):

$$
EA = \alpha E_b A\;\Rightarrow\;\frac{E}{E_b} = \alpha \;\Rightarrow\; \varepsilon = \alpha
$$

Thus at the same temperature the emissivity of a body equals its absorptivity.

## Check yourself

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

1. Radiation falls on a surface. Which relation connects absorptivity, reflectivity and transmissivity?
   A. alpha + rho + tau = 1
   B. alpha x rho x tau = 1
   C. alpha = rho + tau
   D. alpha + rho = tau
   Answer: A. All the incident radiation is absorbed, reflected or transmitted, so the three fractions add up to 1. (Key point: Radiation property relation)

2. What distinguishes a grey body from a real body?
   A. A grey body has emissivity less than 1 that does not depend on wavelength; a real body's emissivity varies with wavelength, temperature and direction
   B. A grey body absorbs all radiation; a real body absorbs none
   C. A grey body has zero emissivity; a real body has emissivity 1
   D. A grey body transmits all radiation; a real body reflects all
   Answer: A. A grey body emits a constant fraction of black-body radiation at every wavelength, whereas a real body's properties vary. (Key point: Grey body)

3. In the proof of Kirchhoff's law, the small body is placed in a black enclosure at the same temperature. Why does EA = alpha E_b A?
   A. Because the body is a black body
   B. Because at thermal equilibrium there is no net heat flow, so emitted energy equals absorbed energy
   C. Because the enclosure is hotter than the body
   D. Because the body reflects all radiation
   Answer: B. With equal temperatures there is no net exchange, so the body emits EA and absorbs alpha E_b A, which must be equal; hence emissivity = alpha. (Key point: Kirchhoff's law proof)

4. Emissivity of a surface is defined as
   A. the fraction of incident radiation absorbed
   B. the ratio of its emissive power to that of a black body at the same temperature
   C. the ratio of reflected to incident radiation
   D. the emissive power at the peak wavelength
   Answer: B. epsilon = E/E_b at the same temperature; Kirchhoff's law then gives epsilon = alpha. (Key point: Emissivity definition)

## Related in Heat Transfer

- [Black and grey body emission numerical](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Ablackbody-wien-emissive-power-numerical) — Heat Transfer, chapter 4
- [Radiation between two parallel grey plates](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Aradiation-exchange-parallel-grey-plates) — Heat Transfer, chapter 4
- [Effect of radiation shield between plates](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Aradiation-shield-between-parallel-plates) — Heat Transfer, chapter 4

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