Final answer:
The rate at which energy is transferred by radiation from the lava's surface is approximately 9.80 x 10^4 W. The thickness of the lava between the surface and interior is approximately 0.33 m.
Step-by-step explanation:
In order to calculate the rate at which energy is transferred by radiation, we can use the Stefan-Boltzmann Law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature and surface area. The power radiated can be calculated using the equation:
Power = emissivity * Stefan-Boltzmann constant * surface area * (temperature^4 - surroundings temperature^4)
Given that the emissivity is 1 (since it's a black body), the surface area is 1.00 m², the temperature of the surface is 450°C (723K), and the surroundings temperature is 27.0°C (300K), we can plug these values into the equation to calculate the power:
Power = 1 * 5.67 x 10^-8 * 1.00 * (723^4 - 300^4)
The calculated power is approximately 9.80 x 10^4 W. Therefore, the correct answer is c) 9.80 x 10^4 W.
To calculate the thickness of the lava between the 450°C surface and the 1200°C interior, we can use the equation for steady-state conduction:
Rate of heat conduction = (thermal conductivity * surface area * (interior temperature - surface temperature)) / thickness
Assuming the lava's conductivity is the same as that of brick, which is 0.80 W/(m·K), we can plug in the values to calculate the thickness:
Rate of heat conduction = (0.80 * 1.00 * (1200 - 450)) / thickness
Simplifying the equation and solving for thickness, we find that the thickness of the lava is approximately 0.33 m.
Therefore, the correct answer is b) 0.33 m.