Problem 61 - College Physics

Question

The emissivity of tungsten is \(0.35 .\) A tungsten sphere with a radius of \(1.50 \mathrm{~cm}\) is suspended within a large evacuated enclosure whose walls are at \(290 \mathrm{~K}\). What power input is required to maintain the sphere at a temperature of \(3000 \mathrm{~K}\) if heat conduction along the supports is negligible?

Step-by-Step Solution

Answer

The power input required is approximately 112.5 W.

Step 1: Understanding Radiation Heat Transfer

To find the power input required to maintain the sphere's temperature, we need to calculate the net radiative heat transfer. According to the Stefan-Boltzmann law, the power radiated by a body is given by \( P = \varepsilon \sigma A (T^4 - T_s^4) \), where \( \varepsilon \) is the emissivity, \( \sigma \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)), \( A \) is the surface area, \( T \) is the temperature of the body, and \( T_s \) is the surrounding temperature.

Step 2: Calculate the Surface Area of the Sphere

The surface area \( A \) of a sphere is calculated using the formula \( A = 4 \pi r^2 \). For a radius \( r = 1.50 \, \text{cm} = 0.015 \, \text{m} \):\[ A = 4 \pi (0.015)^2 = 4 \pi \times 0.000225 = 0.002827 \text{ m}^2 \]

Step 3: Determine the Radiative Power Loss

Substitute the values into the Stefan-Boltzmann equation:\[ P = 0.35 \times 5.67 \times 10^{-8} \times 0.002827 \times (3000^4 - 290^4) \]Calculate \( (3000^4 - 290^4) = 8.10 \times 10^{13} \, \text{K}^4 - 7.07 \times 10^{9} \, \text{K}^4 \). Continue with the calculation:\[ P = 0.35 \times 5.67 \times 10^{-8} \times 0.002827 \times 8.0993 \times 10^{13} \]Computing further:\[ P = 0.35 \times 5.67 \times 10^{-8} \times 0.002827 \times 8.0993 \times 10^{13} \approx 112.5 \, \text{W} \]

Step 4: Final Power Input Requirement

The power input required to maintain the sphere at \(3000 \, \text{K}\) with negligible conductive heat loss is approximately \(112.5 \, \text{W}\). Thus, a minimum of \(112.5 \, \text{W}\) is needed to balance the radiative heat loss.

Key Concepts

Stefan-Boltzmann Law

Radiative heat transfer is a process where thermal energy is emitted by matter in the form of electromagnetic waves, mainly infrared radiation. A key concept to understand here is the Stefan-Boltzmann Law. This law states that the power radiated by a black body in thermal equilibrium is proportional to the fourth power of its absolute temperature. It is captured in the formula:\[ P = \varepsilon \sigma A (T^4 - T_s^4) \]where:

• \(\varepsilon\): Emissivity of the material, which measures how efficiently a body emits thermal radiation.
• \(\sigma\): Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4) \).
• \(A\): Surface area of the body.
• \(T\): Absolute temperature of the body in Kelvin.
• \(T_s\): Absolute temperature of the surroundings in Kelvin.

This equation is fundamental in understanding radiative heat transfer for objects not at absolute zero temperature.

Emissivity

Emissivity is a critical property in understanding radiative heat transfer. It is defined as the efficiency with which a surface emits thermal radiation. Represented by \(\varepsilon\), emissivity ranges from 0 to 1.

• An emissivity of 1 corresponds to a perfect black body, which emits the maximum possible radiation.
• An emissivity of 0 means the body is a perfect reflector and emits no thermal radiation.

Most real objects have an emissivity between these two extremes. In our example, tungsten has an emissivity of 0.35, meaning it emits 35% of the radiation that a black body would emit at the same temperature. This property directly influences the amount of thermal radiation your sphere emits into its surroundings and is pivotal in the calculation of radiative power loss.

Sphere Surface Area Calculation

Calculating the surface area of a sphere is essential for determining how much radiation it emits. The formula for the surface area \(A\) of a sphere is:\[ A = 4 \pi r^2 \]where \(r\) is the radius of the sphere. In our problem, the sphere's radius is given as 1.50 cm, which must be converted to meters (0.015 m) for the calculations. Applying the formula:\[ A = 4 \pi (0.015)^2 = 0.002827 \, \text{m}^2 \]This calculated surface area is then used in the Stefan-Boltzmann equation to find out how much power the sphere emits through thermal radiation. The area is a crucial aspect of radiative calculations because it directly affects the amount of radiant energy emitted by the sphere.

Net Radiative Power

Net radiative power is the difference between the power emitted by the sphere and the power received from its surroundings. It directly relates to how much energy is needed to maintain the sphere's temperature. When calculating net radiative power, consider:

• The emissivity of the material \( (\varepsilon = 0.35) \).
• The surface area of the sphere \( A = 0.002827 \, \text{m}^2 \).
• The Stefan-Boltzmann constant \(\sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \).

Substitute these values into the equation to solve for \( P \) using the temperatures: \( T = 3000 \, \text{K} \) for the sphere and \( T_s = 290 \, \text{K} \) for the surroundings:\[ P = 0.35 \times 5.67 \times 10^{-8} \times 0.002827 \times (3000^4 - 290^4) \]The resulting value, roughly \( 112.5 \, \text{W} \), is the power required to keep the sphere at the desired temperature, compensating for the energy lost through radiation. This insight is vital for any radiative heat transfer assessment.