Question

A blackbody radiant heating panel of 5 m\(^2\) surface area at 35\(^\circ\)C surface temperature is located 1 m away from a 1 m\(^2\) surface at 20\(^\circ\)C. The Stefan-Boltzmann constant is \(5.6703 \times 10^{-8}\ \text{W m}^{-2} \text{K}^{-4}\). The rate of radiant heat emission by the radiant heating panel (in W, rounded off to two decimal places) is \(\underline{\hspace{2cm}}\).

Hint:

Always convert temperatures to Kelvin before using the Stefan–Boltzmann law.

Answer 1

Correct Answer: 2550 - 2560

Using the Stefan–Boltzmann law:
\[ Q = A \sigma (T_1^4 - T_2^4) \] Given:
Area \(A = 5\ \text{m}^2\)
\(T_1 = 35^\circ\text{C} = 308\ \text{K}\)
\(T_2 = 20^\circ\text{C} = 293\ \text{K}\)
\(\sigma = 5.6703 \times 10^{-8}\ \text{W m}^{-2} \text{K}^{-4}\)
Compute fourth powers:
\[ 308^4 = 9.00 \times 10^{9}, 293^4 = 7.37 \times 10^{9} \] \[ T_1^4 - T_2^4 = 1.63 \times 10^9 \] Now heat emission:
\[ Q = 5 \times 5.6703 \times 10^{-8} \times (1.63 \times 10^9) \] \[ Q \approx 2555.40\ \text{W} \] Thus, rounded to two decimals:
\[ \boxed{2555.40\ \text{W}} \]