A flat plate made of cast iron is exposed to a solar flux of 600 W/m² at an ambient temperature of 25°C. Assume that the entire solar flux is absorbed by the plate. Cast iron has a low temperature absorptivity of 0.21.Use Stefan-Boltzmann constant = 5.669 ×10^–8 W/m²–K⁴. Neglect all other modes of heat transfer except radiation. Under the aforementioned conditions, the radiation equilibrium temperature of the plate is __________ °C (round off to the nearest integer)

The correct answer is 218.

Step-by-Step Explanation:

To find the radiation equilibrium temperature of the cast iron plate, we set up an energy balance on the plate under steady-state conditions.

According to the problem description, we neglect all modes of heat transfer except radiation. Therefore, the solar radiation absorbed by the plate must be balanced by the net thermal radiation exchanged between the plate and the ambient environment.

1. Identify the given parameters:
- Solar flux incident on the plate, $G_s=600{\text{W/m}}^2$
- Absorptivity of the plate (which equals the emissivity $\epsilon$ by Kirchhoff's law for low-temperature radiation exchange), $\alpha =\epsilon =0.21$
- Ambient temperature, $T_a={25}^{\circ }\text{C}=25+273.15\approx 298\text{K}$
- Stefan-Boltzmann constant, $\sigma =5.669\times {10}^{-8}{\text{W/m}}^2·{\text{K}}^4$

2. Formulate the energy balance equation:
Since the entire solar flux is absorbed by the plate, the rate of solar energy absorbed per unit area is:
$q_{\text{absorbed}}=G_s=600{\text{W/m}}^2$

At equilibrium, this is equated to the net thermal radiation heat loss from the plate to the ambient surroundings:
$q_{\text{rad, net}}=\epsilon \sigma \left(T_s^4-T_a^4\right)$

Equating the heat gain and heat loss yields:
$G_s=\epsilon \sigma \left(T_s^4-T_a^4\right)$

3. Substitute the values and solve for the surface temperature $T_s$:
$600=0.21\times 5.669\times {10}^{-8}\left(T_s^4-{298}^4\right)$

First, compute the factor in front of the parenthesis:
$0.21\times 5.669\times {10}^{-8}=1.19049\times {10}^{-8}$

Now, divide both sides by this factor:
$T_s^4-{298}^4=\frac{600}{1.19049\times {10}^{-8}}\approx 5.0399\times {10}^{10}$

Calculate ${298}^4$:
${298}^4\approx 7.8875\times {10}^9$

Add this to both sides to solve for $T_s^4$:
$T_s^4\approx 5.0399\times {10}^{10}+0.78875\times {10}^{10}=5.82865\times {10}^{10}$

Taking the fourth root gives the temperature in Kelvin:
$T_s={\left(5.82865\times {10}^{10}\right)}^{0.25}\approx 491.1\text{K}$

4. Convert the temperature to Celsius:
$T_s\left({}^{\circ }\text{C}\right)=491.1-273.15\approx 218{}^{\circ }\text{C}$

Rounding to the nearest integer, we find the radiation equilibrium temperature of the plate is 218°C.