Question

An iron furnace radiates \(42.5\) calories per second through an opening of cross-section \(1\ \text{cm}^2\). If the relative emittance of the furnace is \(0.80\), the temperature of furnace is given Stefan's constant \(\sigma=1.36\times10^{-8}\ \text{cal/m}^2\text{s K}^4\):

• A. \(2500\ \text{K}\)
• B. \(273\ \text{K}\)
• C. \(373\ \text{K}\)
• D. \(500\ \text{K}\)

Hint:

For radiation problems, use \(P=e\sigma AT^4\) and convert area into SI units if \(\sigma\) is given in \(\text{m}^2\).

Answer 1

The Correct Option is A

Concept:
According to Stefan-Boltzmann law: \[ P=e\sigma AT^4 \]

Step 1: Write given values. \[ P=42.5\ \text{cal/s} \] \[ e=0.80 \] \[ A=1\ \text{cm}^2=10^{-4}\ \text{m}^2 \] \[ \sigma=1.36\times10^{-8}\ \text{cal/m}^2\text{s K}^4 \]

Step 2: Substitute in Stefan law. \[ 42.5=0.80\times1.36\times10^{-8}\times10^{-4}\times T^4 \] \[ 42.5=1.088\times10^{-12}T^4 \]

Step 3: Solve for \(T^4\). \[ T^4=\frac{42.5}{1.088\times10^{-12}} \] \[ T^4\approx3.906\times10^{13} \]

Step 4: Take fourth root. \[ T\approx2500\ \text{K} \] \[ \therefore \text{Correct Answer is (A)} \]