To determine the emissivity of the copper sphere, we need to analyze the heat transfer processes involved when the sphere is heated and when part of its surface is blackened. Emissivity is a measure of how effectively a surface emits thermal radiation compared to a perfect black body, which has an emissivity of 1. Let's break this down step by step.
Understanding Heat Transfer
The sphere is initially heated to 900 K while suspended in an evacuated chamber at 300 K. The power required to maintain this temperature is 300 W. This power is used to balance the heat lost through radiation from the sphere to its surroundings.
Calculating the Power Radiated by the Sphere
The power radiated by a surface can be calculated using the Stefan-Boltzmann Law:
Where:
- P = power radiated (W)
- ε = emissivity of the surface
- σ = Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/m²K^4)
- A = surface area of the sphere (m²)
- T = absolute temperature of the surface (K)
Initial Conditions
In the initial state, the sphere is at 900 K, and the power required to maintain this temperature is 300 W. Therefore, we can express this as:
- 300 W = ε(5.67 x 10^-8)(A)(900^4)
Blackened Half of the Sphere
When half of the surface of the copper sphere is blackened, it becomes a perfect absorber and emitter (emissivity = 1). The total power required to maintain the same temperature of the sphere increases to 600 W. The power radiated from the blackened half and the unblackened half can be expressed as:
- 600 W = (1)(5.67 x 10^-8)(A/2)(900^4) + (ε)(5.67 x 10^-8)(A/2)(900^4)
Here, the first term represents the power radiated from the blackened half, and the second term represents the power radiated from the unblackened half.
Setting Up the Equations
From the first condition, we have:
- 300 W = ε(5.67 x 10^-8)(A)(900^4)
From the second condition, we can rearrange the equation:
- 600 W = (1)(5.67 x 10^-8)(A/2)(900^4) + (ε)(5.67 x 10^-8)(A/2)(900^4)
Factoring out common terms gives us:
- 600 W = (5.67 x 10^-8)(A/2)(900^4) + (5.67 x 10^-8)(A/2)(900^4)ε
Solving the Equations
Now, we can express the second equation in terms of the first:
- 600 W = (5.67 x 10^-8)(A/2)(900^4)(1 + ε)
From the first equation, we can express A in terms of ε:
- A = (300 W) / (ε(5.67 x 10^-8)(900^4))
Substituting this into the second equation allows us to solve for ε:
- 600 W = (5.67 x 10^-8)(A/2)(900^4)(1 + ε)
After substituting and simplifying, we find:
Solving for ε gives:
Conclusion
The emissivity of the copper sphere is 1, indicating that when half of its surface is blackened, it behaves like a perfect black body in terms of thermal radiation. This result aligns with our expectations, as blackened surfaces are known to be excellent absorbers and emitters of thermal radiation.
