Radiation Heat Transfer Questions and Answers

Answer: A

The basic rate equation of radiation heat transfer is the Stefan-Boltzmann law, \(E = \sigma T^{4}\) for a black body. Newton's law of cooling is the rate equation of convection and Fourier's law is that of conduction. The Gibbs-Helmholtz equation belongs to thermodynamics and is not a heat transfer rate equation.

Answer: B

The temperature is 1000 K.

Wien's displacement law says the peak wavelength is inversely proportional to absolute temperature, \(\lambda_{\max}T = \text{constant}\):

\[ \begin{aligned} T_{2} &= T_{1}\frac{\lambda_{1}}{\lambda_{2}} \\ &= 2000 \times \frac{1.45}{2.90} = 1000\ \text{K} \end{aligned} \]

The constant here is 1.45 × 2000 = 2900 µm K, the familiar Wien constant. Doubling the peak wavelength halves the temperature; 4000 K treats the relation as direct, and 500 K and 8000 K square the ratio, as if confusing it with a power law.

Answer: C

For an opaque surface, \(\alpha+\rho=1\).

Radiation falling on any surface is absorbed, reflected or transmitted, so the three fractions add up to one:

\[ \alpha+\rho+\tau=1 \]

An opaque surface transmits nothing, so \(\tau=0\) and the relation reduces to \(\alpha+\rho=1\).

The fractions are each between 0 and 1, so their sum can never be zero, which rules out both options ending in 0. \(\alpha+\rho=\tau\) would mean the surface transmits as much as it absorbs and reflects together, which contradicts both the energy balance and the meaning of opaque.

Answer: B

Emissivity does not depend on the surface area.

Emissivity is the ratio of the emissive power of a real surface to that of a black body at the same temperature, \(\varepsilon = E/E_{b}\). Both powers are per unit area, so the area cancels: emissivity is a property of the surface material and its condition, not of its size.

It does depend on the other three. It is defined through the total emissive power E, it varies with absolute temperature for real surfaces, and by Kirchhoff's law it equals the absorptivity α at thermal equilibrium.

Answer: A

The emissivity of a white body is zero.

In radiation terms a white body is an ideal surface that reflects all the radiation falling on it, so its absorptivity is zero. By Kirchhoff's law emissivity equals absorptivity, so its emissivity is zero at every wavelength and temperature.

An emissivity of 1 describes a black body, and a constant value between 0 and 1, such as 0.6, describes a grey body. The range 0.1 to 0.4 describes no ideal body.

Answer: D

The reflectivity is 0.65.

For any surface, the incident radiation is absorbed, reflected or transmitted: \(\alpha + \rho + \tau = 1\). An opaque surface transmits nothing, so \(\tau = 0\). For a grey surface, Kirchhoff's law makes the absorptivity equal to the emissivity, whatever the source temperature:

\[ \begin{aligned} \alpha &= \varepsilon = 0.35 \\ \rho &= 1 - \alpha = 1 - 0.35 \\ &= 0.65 \end{aligned} \]

0.35 is the absorptivity, not the reflectivity, which is the trap. The source temperature of 550 K matters only to confirm that source and surface are at the same temperature; for a grey body the answer would not change anyway. 0.85 and 0.75 do not follow from the emissivity given.

Answer: C

For an absolutely white or specular body, ρ = 1 and α = τ = 0.

Incident radiation is absorbed, reflected or transmitted, so \(\alpha + \rho + \tau = 1\). An absolutely white body reflects all the radiation falling on it, diffusely, and a specular body reflects it all like a mirror. In both, the reflectivity is 1, so nothing is absorbed or transmitted.

α = 1 with nothing reflected or transmitted describes a black body. τ = 1 describes a perfectly transparent (diathermanous) body. α + ρ = 1 with τ = 0 is true of any opaque body, so it is not specific to a white body.