Problem 68 - University Physics with Modern Physics

Question

The emissivity of tungsten is 0.350. A tungsten sphere with radius 1.50 cm is suspended within a large evacuated enclosure whose walls are at 290.0 K. What power input is required to maintain the sphere at 3000.0 K if heat conduction along the supports is ignored?

Step-by-Step Solution

Answer

1356.2 watts are required to maintain the sphere at 3000 K.

Step 1: Understand the Given Parameters

The problem consists of a tungsten sphere with emissivity \( \varepsilon = 0.350 \), radius \( r = 1.50 \) cm, temperature \( T = 3000.0 \) K, and surrounding wall temperature \( T_\text{walls} = 290.0 \) K. Emissivity is the efficiency of a surface in emitting thermal radiation compared to a perfect blackbody.

Step 2: Convert Units

Convert the radius from centimeters to meters for standard units in calculations. Since 1 cm = 0.01 m, the radius in meters is: \[ r = 1.50 \times 0.01 = 0.015 \text{ m} \]

Step 3: Calculate the Surface Area of the Sphere

The surface area \( A \) of a sphere is given by the formula: \[ A = 4\pi r^2 \] Substitute \( r = 0.015 \) m:\[ A = 4 \pi (0.015)^2 \] \[ A \approx 0.002827 \text{ m}^2 \]

Step 4: Apply the Stefan-Boltzmann Law

Use the Stefan-Boltzmann law to calculate the power radiated: \[ P = \varepsilon \sigma A (T^4 - T_\text{walls}^4) \] where \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \). Substitute the values: \[ P = 0.350 \times 5.67 \times 10^{-8} \times 0.002827 \times ((3000)^4 - (290)^4) \]

Step 5: Calculate the Net Power Required

Calculate the power input required to maintain the sphere's temperature against the radiation emitted to the cooler walls: First, compute the temperature terms:\( 3000^4 = 8.1 \times 10^{13} \) \( 290^4 \approx 7.1 \times 10^{9} \).Finally, compute the net power:\[ P \approx 0.350 \times 5.67 \times 10^{-8} \times 0.002827 \times (8.1 \times 10^{13} - 7.1 \times 10^{9}) \] Calculate this expression to find:\[ P \approx 1356.2 \text{ W} \]

Step 6: Interpret the Result

The power input required is approximately 1356.2 watts. This provides the necessary energy to counteract the radiation losses from the sphere at equilibrium temperature and maintain the sphere's temperature at 3000 K.

Key Concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann law explains how objects emit thermal radiation based on their temperature. It states that the total power emitted per unit surface area of a black body is proportional to the fourth power of its absolute temperature. This law is fundamental when calculating power radiated by an object, such as a tungsten sphere, in thermal environments.

For real-world objects, not all surfaces emit radiation as efficiently as a perfect black body, thus the presence of emissivity in the law's formula. Using the Stefan-Boltzmann equation, the power radiated can be expressed as:

• \( P = \varepsilon \sigma A (T^4 - T_{\text{walls}}^4) \)

Where:

• \( P \) is the power radiated.
• \( \varepsilon \) is the emissivity of the object.
• \( \sigma \) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \, \text{W/m}^2 \text{K}^{4}) \).
• \( A \) is the surface area of the object.
• \( T \) is the absolute temperature of the object and \( T_{\text{walls}} \) is the temperature of the surroundings.

In practical terms, this law helps in determining how much energy is lost or needed to keep an object at a certain temperature despite its environment.

Emissivity

Emissivity is a measure of how efficiently a surface emits thermal radiation compared to a perfect black body. A perfect black body has an emissivity of 1, meaning it emits all thermal radiation perfectly. In contrast, objects with lower emissivity radiate less energy.

• The value of emissivity \( (\varepsilon) \) ranges from 0 to 1.
• Emissivity depends on the material and surface characteristics of the object.

For example, the emissivity of tungsten is given as 0.350 in the exercise. This shows that tungsten is not as efficient in emitting thermal radiation as a black body. When calculating radiated power, this factor is essential because it adjusts the ideal power output from a theoretical black body to an actual object like the tungsten sphere.

Understanding emissivity is crucial in many applications such as thermal management in electronics, climate studies, and infrared imaging.

Temperature Conversion

Temperature conversion is vital when working with thermodynamic calculations, especially involving the Stefan-Boltzmann law. In these calculations, temperatures are typically measured in Kelvin, which is the SI unit for temperature and starts from absolute zero.

• To convert from Celsius to Kelvin, use the formula: \( T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \).
• This conversion is necessary because temperature-related calculations in physics often require absolute temperatures.

For example, if you were measuring the ambient temperature around an object, starting from Celsius, you’d need to convert it to Kelvin to perform radiation calculations accurately. It ensures that the power computed reflects the physical reality of energy exchanges within different temperature conditions.

Sphere Surface Area Calculation

When calculating the thermal properties of a sphere, determining the surface area is a major step. The surface area of a sphere is crucial for understanding how much area is available to emit or absorb thermal radiation.

• The formula for the surface area \( A \) of a sphere is \( 4 \pi r^2 \).
• Where \( r \) is the radius of the sphere.

In our example, the radius of the tungsten sphere is 1.50 cm, which must be converted to meters (\( 0.015 \) m) for compatibility with the SI units in calculations. Substituting the radius into the area formula gives us a sphere's surface that influences the total radiation emitted.

• \( A = 4 \pi (0.015)^2 \approx 0.002827 \, \text{m}^2 \).

Correct sphere surface area calculation is essential for accurate determination of how much power an object emits or needs to sustain its temperature.