Question

The surface temperature of a blackbody is 400 K. What will be the emissive power?

• A. 1451.26 Wm$^{-2}$
• B. 4521.56 Wm$^{-2}$
• C. 2890 Wm$^{-2}$
• D. 5500 Wm$^{-2}$

Hint:

Use the Stefan-Boltzmann law $E = \sigma T^4$ to calculate blackbody radiation. Always ensure the temperature is in Kelvin.

Answer 1

The Correct Option is A

The emissive power $E$ of a blackbody is calculated using the Stefan-Boltzmann law:
\[ E = \sigma T^4 \]
Where:
- $E$ = emissive power in W/m$^2$
- $\sigma$ = Stefan-Boltzmann constant = $5.67 \times 10^{-8}$ W/m$^2$K$^4$
- $T$ = temperature in Kelvin = 400 K
Substitute the values into the formula:
\[ E = 5.67 \times 10^{-8} \times (400)^4 \]
First, calculate $400^4$:
\[ 400^4 = (4 \times 10^2)^4 = 4^4 \times 10^8 = 256 \times 10^8 = 2.56 \times 10^{10} \]
Now, calculate emissive power:
\[ E = 5.67 \times 10^{-8} \times 2.56 \times 10^{10} = (5.67 \times 2.56) \times 10^{2} = 14.5152 \times 10^2 = 1451.52\ \text{W/m}^2 \]
Rounding appropriately, the emissive power is:
\[ \boxed{1451.26\ \text{W/m}^2} \]