Question

A surface with emissivity 0.8 and area $1\text{ m}^2$ is maintained at $500\text{ K}$ and is surrounded by air at $300\text{ K}$. Given Stefan–Boltzmann constant $\sigma = 5.67 \times 10^{-8}\text{ W/m}^2\text{K}^4$. What is the rate of radiative heat transfer?

• A. $3.5\text{ kW}$
• B. $2.47\text{ kW}$
• C. $2.0\text{ kW}$
• D. $1.5\text{ kW}$

Hint:

Always ensure temperatures are in Kelvin ($T$ in $\text{K}$) before performing any calculations involving the Stefan-Boltzmann law.
Using Celsius temperatures is a common source of error.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the net rate of radiative heat transfer from a gray body surface to its cooler surroundings.

Step 2: Key Formula or Approach: The net heat transfer rate by radiation ($Q$) between a surface and its surroundings is given by the Stefan-Boltzmann law:
\[ Q = \epsilon \sigma A (T_s^4 - T_{\text{surr}}^4) \]
where:
$\epsilon$ is the emissivity of the surface.
$\sigma$ is the Stefan-Boltzmann constant ($5.67 \times 10^{-8}\text{ W/m}^2\text{K}^4$).
$A$ is the surface area.
$T_s$ is the absolute surface temperature (in Kelvin).
$T_{\text{surr}}$ is the absolute temperature of the surroundings (in Kelvin).

Step 3: Detailed Explanation:
Given values:
$\epsilon = 0.8$
$A = 1\text{ m}^2$
$T_s = 500\text{ K}$
$T_{\text{surr}} = 300\text{ K}$
$\sigma = 5.67 \times 10^{-8}\text{ W/m}^2\text{K}^4$
Substitute these values into the formula:
\[ Q = 0.8 \times (5.67 \times 10^{-8}) \times 1 \times (500^4 - 300^4) \]
Calculate the temperature terms:
\[ 500^4 = 6.25 \times 10^{10} \]
\[ 300^4 = 8.1 \times 10^9 = 0.81 \times 10^{10} \]
\[ T_s^4 - T_{\text{surr}}^4 = (6.25 - 0.81) \times 10^{10} = 5.44 \times 10^{10} \]
Now, complete the multiplication:
\[ Q = 0.8 \times 5.67 \times 10^{-8} \times 5.44 \times 10^{10} \]
\[ Q = 4.536 \times 544 \]
\[ Q \approx 2467.58\text{ W} = 2.47\text{ kW} \]

Step 4: Final Answer:
The rate of radiative heat transfer is $2.47\text{ kW}$.