A sphere of radius 0.350m, temperature `27.0^(@)C` , and emissivity 0.850 is located in an environment of temperature `77.0^(@)C`. At what rate does the sphere (a) emit and (b) absorb thermal radiation ? (c ) What is the sphere's net change in energy in 3.50 min ?

Answer

To solve the problem, we will break it down into three parts: (a) calculating the rate at which the sphere emits thermal radiation, (b) calculating the rate at which the sphere absorbs thermal radiation, and (c) determining the net change in energy over a specified time period.

### Given Data:
- Radius of the sphere, \( r = 0.350 \, m \)
- Temperature of the sphere, \( T_s = 27.0 \, ^\circ C \)
- Emissivity of the sphere, \( \epsilon = 0.850 \)
- Temperature of the environment, \( T_e = 77.0 \, ^\circ C \)

### Step 1: Convert Temperatures to Kelvin
To use the Stefan-Boltzmann law, we need to convert the temperatures from Celsius to Kelvin.

\[ T_s = 27.0 + 273.15 = 300.15 \, K \]
\[ T_e = 77.0 + 273.15 = 350.15 \, K \]

### Step 2: Calculate the Surface Area of the Sphere
The surface area \( A \) of a sphere is given by the formula:

\[ A = 4\pi r^2 \]

Substituting the radius:

\[ A = 4\pi (0.350)^2 \approx 1.539 \, m^2 \]

### Step 3: Calculate the Rate of Emission of Thermal Radiation (Part a)
The power emitted by the sphere can be calculated using the Stefan-Boltzmann law:

\[ P_{\text{emit}} = \sigma \epsilon A T_s^4 \]

Where:
- \( \sigma = 5.6704 \times 10^{-8} \, W/m^2K^4 \) (Stefan-Boltzmann constant)

Substituting the values:

\[ P_{\text{emit}} = (5.6704 \times 10^{-8}) \times 0.850 \times (1.539) \times (300.15)^4 \]

Calculating \( (300.15)^4 \):

\[ (300.15)^4 \approx 8.073 \times 10^9 \, K^4 \]

Now substituting back:

\[ P_{\text{emit}} \approx (5.6704 \times 10^{-8}) \times 0.850 \times (1.539) \times (8.073 \times 10^9) \approx 602 \, W \]

### Step 4: Calculate the Rate of Absorption of Thermal Radiation (Part b)
The power absorbed by the sphere from the environment is given by:

\[ P_{\text{absorb}} = \sigma \epsilon A T_e^4 \]

Substituting the values:

\[ P_{\text{absorb}} = (5.6704 \times 10^{-8}) \times 0.850 \times (1.539) \times (350.15)^4 \]

Calculating \( (350.15)^4 \):

\[ (350.15)^4 \approx 1.5 \times 10^{10} \, K^4 \]

Now substituting back:

\[ P_{\text{absorb}} \approx (5.6704 \times 10^{-8}) \times 0.850 \times (1.539) \times (1.5 \times 10^{10}) \approx 1115 \, W \]

### Step 5: Calculate the Net Change in Energy (Part c)
The net power change is given by:

\[ P_{\text{net}} = P_{\text{absorb}} - P_{\text{emit}} = 1115 \, W - 602 \, W = 513 \, W \]

To find the net change in energy over 3.50 minutes, we convert minutes to seconds:

\[ \text{Time} = 3.50 \times 60 = 210 \, s \]

Now, calculate the net energy change:

\[ \Delta E = P_{\text{net}} \times \text{Time} = 513 \, W \times 210 \, s = 107730 \, J \approx 1.08 \times 10^5 \, J \]

### Summary of Answers:
(a) Rate of emission: \( 602 \, W \)
(b) Rate of absorption: \( 1115 \, W \)
(c) Net change in energy in 3.50 minutes: \( 1.08 \times 10^5 \, J \)