Question

Stefan’s law states that the total energy Et emitted by a blackbody per unit time and per unit area is given by Et 5 aT4 , where a has a value of 5.69 3 1028 Wm22 K24 . Calculate the total energy output in W/m2 for each of the blackbodies described in Problem 7-3

Short answer

When$T=4500\mathrm{K},E_t=2.33\times {10}^7\mathrm{W}/{\mathrm{m}}^2$

When$T=2500\mathrm{K},E_t=2.22\times {10}^6\mathrm{W}/{\mathrm{m}}^2$

When$T=1250\mathrm{K},E_t=1.39\times {10}^5\mathrm{W}/{\mathrm{m}}^2$

Step-by-step solution

Given information

We are given values:

$$\begin{gathered} T=4500\mathrm{K} \\ \alpha =5.69\times {10}^{-8}{\mathrm{Wm}}^{-2}{\mathrm{K}}^{-4} \end{gathered}$$

By using the given formula

$$\begin{gathered} E_t=\alpha T^4 \\ {E}_t=5.69\times {10}^{-8}\times {4500}^4 \\ {E}_t=2.33\times {10}^7\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

We are given the temperature and total energy formula. We have to find the total energy.

Given values are:

$$\begin{gathered} T=2500\mathrm{K} \\ \alpha =5.69\times {10}^{-8}{\mathrm{Wm}}^{-2}{\mathrm{K}}^{-4} \end{gathered}$$

By using the given formula

$$\begin{gathered} E_t=𝛼T^4 \\ {E}_t=5.69\times {10}^{-8}\times 2500 \\ {E}_t=2.22\times {10}^6\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

Step 3. We are given with the temperature and total energy formula. We have to find the total energy. 

The given values are,

$$\begin{gathered} T=1250\mathrm{K} \\ \alpha =5.69\times {10}^{-8}{\mathrm{Wm}}^{-2}{\mathrm{K}}^{-4} \end{gathered}$$

By putting values in the formula

$$\begin{gathered} E_t=𝛼T^4 \\ {E}_t=5.69\times {10}^{-8}\mathrm{W}/{\mathrm{m}}^2.{\mathrm{K}}^4\times 1250K \\ {E}_t=1.39\times {10}^5\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$