Question

A circular plate of 500-mm diameter is maintained at T1 = 600 K and is positioned coaxial to a conical shape. The back side of the cone is well insulated. The plate and the cone, whose surfaces are black, are located in a large, evacuated enclosure whose walls are at 300 K.

(a) What is the temperature of the conical surface, T2?
(b) What is the electrical power that would be required to maintain the circular plate at 600K?

Answer 1

The student’s questions involve the principles of thermal radiation and heat conduction in Physics, relating to the determination of the equilibrium temperature in a radiative system and the electrical power needed to maintain a temperature. Calculations require additional data on materials and geometry to provide accurate answers.

Step-by-step explanation:

The student's questions pertain to the topics of thermal radiation and heat conduction in Physics, which are within the realm of thermodynamics. Specifically, they are asking about (a) the equilibrium temperature of a conical surface and (b) the power required to maintain a certain temperature on a circular plate. These calculations involve understanding Stefan-Boltzmann law, conduction, and power calculations.

Unfortunately, without further information or input parameters for materials and geometries, providing a definitive answer for a system like the one described (circular plate and conical surface in an enclosure) isn't possible. To determine the electrical power required, one would need to know properties like thermal conductivity, surface area, and possibly the geometry of the objects in question to find out the heat transfer rates.

Answer 2

For part a)
Since the conical surface is not exposed to the radiation coming from the walls only from the circular plate and assuming steady state, the temperature of the conical surface is also equal to the temperature of the circular plate. T2 = 600 K

For part b)
To maintain the temperature of the circular plate, the power required would be calculated using:
Q = Aσ(T₁⁴ - Tw⁴)
Q = π(500x10^-3)²/4 (5.67x10^-8)(600⁴ - 300⁴)
Q = 5410.65 W