The two cubes in the drawing are made from the same material. The radiant energy per second emitted by the small cube is 3.70 J/s. What is the energy emitted per second by the larger one?
To determine the energy emitted per second by the larger cube, we can use the concept of the Stefan-Boltzmann Law.
The Stefan-Boltzmann Law states that the power radiated by an object is proportional to its surface area and the fourth power of its temperature. Mathematically, it can be expressed as:
P ∝ A * T^4
Where P is the power or energy emitted per second (in watts), A is the surface area (in square meters), and T is the absolute temperature (in Kelvin) of the object.
Given that the two cubes are made from the same material, we can assume that they have the same emissivity (ability to emit radiation) and temperature. Therefore, the only difference between the two cubes is their surface area.
Let's say the surface area of the small cube is A_s.
Since the larger cube is simply a scaled-up version of the small cube, its surface area (A_l) is proportional to the square of the ratio of their lengths (L_l and L_s):
A_l/A_s = (L_l/L_s)^2
Assuming that the length of the small cube is L_s, and the length of the larger cube is L_l, we need to find the ratio L_l/L_s.
Since we do not have specific values for the lengths of the cubes, we cannot compute the exact surface area ratio (A_l/A_s). However, we can still calculate the energy emitted per second by the larger cube relative to the small cube using the ratio of lengths.
Based on the information given, we know that the energy emitted per second by the small cube is 3.70 J/s. Let's assume the energy emitted per second by the larger cube is P_l.
Therefore, we can set up the following equation:
P_l / 3.70 J/s = (L_l / L_s)^2
To find the energy emitted per second by the larger cube (P_l), we need to know the ratio of the lengths of the two cubes (L_l / L_s). Without this information, it is not possible to calculate the exact value.
If you have the length ratio (L_l / L_s) or any additional information, we can use it to find the energy emitted per second by the larger cube.