Problem 74 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

The required power input is approximately 515 W.

Step 1: Understand the Stefan-Boltzmann Law

The Stefan-Boltzmann Law is used to calculate the power radiated by an object based on its temperature. It is given by the formula: \( P = \varepsilon \cdot \sigma \cdot A \cdot (T^4 - T_0^4) \), where \( P \) is the power, \( \varepsilon \) is the emissivity, \( A \) is the surface area of the sphere, \( T \) is the temperature of the sphere, \( T_0 \) is the temperature of the surroundings, and \( \sigma \) is the Stefan-Boltzmann constant, \( 5.67 \times 10^{-8} \, \mathrm{W/m^2 \cdot K^4} \).

Step 2: Calculate the Surface Area of the Sphere

The surface area \( A \) of a sphere is given by the formula \( A = 4 \pi r^2 \), where \( r \) is the radius of the sphere. Given that the radius \( r = 1.50 \, \mathrm{cm} = 0.015 \, \mathrm{m} \), we calculate the surface area as follows:\[ A = 4 \pi (0.015)^2 \approx 2.827 \times 10^{-3} \, \mathrm{m^2} \]

Step 3: Substitute the Values into the Stefan-Boltzmann Equation

Now substitute the given values into the Stefan-Boltzmann Law:- \( \varepsilon = 0.350 \)- \( \sigma = 5.67 \times 10^{-8} \, \mathrm{W/m^2 \cdot K^4} \)- \( A = 2.827 \times 10^{-3} \, \mathrm{m^2} \)- \( T = 3000 \, \mathrm{K} \)- \( T_0 = 290.0 \, \mathrm{K} \)The formula becomes:\[ P = 0.350 \cdot 5.67 \times 10^{-8} \cdot 2.827 \times 10^{-3} \cdot ((3000)^4 - (290)^4) \]

Step 4: Calculate the Power Output

Performing the calculation:First, calculate the temperatures raised to the fourth power:- \( 3000^4 = 8.1 \times 10^{13} \, \mathrm{K^4} \)- \( 290^4 = 7.11 \times 10^{9} \, \mathrm{K^4} \)Now, substitute these into the formula:\[ P = 0.350 \cdot 5.67 \times 10^{-8} \cdot 2.827 \times 10^{-3} \cdot ((8.1 \times 10^{13}) - (7.11 \times 10^{9})) \]This gives:\[ P \approx 0.350 \cdot 5.67 \times 10^{-8} \cdot 2.827 \times 10^{-3} \cdot 8.0992929 \times 10^{13} \]\[ P \approx 515.37 \, \mathrm{W} \]

Step 5: Conclusion and Round the Answer

The calculated power input required to maintain the sphere at 3000.0 K is approximately \( 515.37 \, \mathrm{W} \). For practical purposes, we round this to \( 515 \, \mathrm{W} \).

Key Concepts

Emissivity

Emissivity is a measure of how effectively an object emits thermal radiation. It is represented by the symbol \( \varepsilon \) and ranges from 0 to 1.
A perfect blackbody, which is an ideal emitter, has an emissivity of 1, while an object that does not emit thermal radiation at all has an emissivity of 0.
In practical terms, most materials have emissivity values between these two extremes.
The Stefan-Boltzmann Law takes emissivity into account when calculating the power radiated by an object. The formula, \( P = \varepsilon \cdot \sigma \cdot A \cdot (T^4 - T_0^4) \), includes emissivity as a factor that modifies the ideal blackbody radiation to match the actual material properties.
In this exercise, the emissivity of tungsten is given as \(0.350\), which means that it emits 35% as much thermal radiation as an ideal blackbody at the same temperature. Understanding emissivity helps in predicting how much energy an object will radiate under given circumstances.

Surface Area of a Sphere

To find out how much thermal radiation a sphere emits, you need to know its surface area.
The formula for the surface area \( A \) of a sphere is \( A = 4 \pi r^2 \), where \( r \) is the radius of the sphere.
This formula is derived from geometric principles and is essential for understanding how energy emission scales with the size of the object.
In the given problem, the sphere's radius is 1.50 cm, which we convert to meters (\( r = 0.015 \, \mathrm{m} \)) for consistency with SI units.
Plugging this into the surface area formula, we get:\[ A = 4 \pi (0.015)^2 \approx 2.827 \times 10^{-3} \, \mathrm{m^2} \]
This surface area is used in the Stefan-Boltzmann Law to calculate the power emitted by the sphere, showing the essential role geometry plays in the physics of radiation.

Power Input Calculation

To keep a tungsten sphere at a stable high temperature, one must calculate the power input needed to balance the thermal radiation it emits. The calculation involves several quantitative aspects.
The Stefan-Boltzmann Law is our tool, expressed as \( P = \varepsilon \cdot \sigma \cdot A \cdot (T^4 - T_0^4) \).
Let's break down the process:

• Use the sphere's emissivity (\( \varepsilon = 0.350 \)) and surface area (\( A = 2.827 \times 10^{-3} \, \mathrm{m^2} \))
• Apply the Stefan-Boltzmann constant \( \sigma = 5.67 \times 10^{-8} \, \mathrm{W/m^2 \cdot K^4} \)
• Insert temperatures \( T = 3000 \, \mathrm{K} \) (sphere) and \( T_0 = 290 \, \mathrm{K} \) (surroundings)

These values lead to the equation for power \( P \), which can be solved to find:\[ P \approx 515.37 \, \mathrm{W} \]
Power input equal to this value is required to maintain the given conditions, balancing the thermal radiation emitted by the sphere.

Thermal Radiation

Thermal radiation is a crucial concept in understanding how objects emit energy in the form of electromagnetic waves. Any object at a temperature above absolute zero emits thermal radiation due to the movement of its charged particles.
The key law governing this process is the Stefan-Boltzmann Law, which not only depends on temperature but also factors in the material's emissivity and surface area.
When calculating the power radiated by the tungsten sphere, the difference in temperatures accounted in the formula \( (T^4 - T_0^4) \) represents the directional flow of energy - from the hot object to the cooler surroundings.
Understanding thermal radiation helps in various applications, from designing heating systems to explaining natural phenomena like how Earth emits energy into space.

Temperature Difference

The role of temperature difference in thermal radiation can't be overstated. It dictates the net flow of energy between objects, influencing the power output.
In the Stefan-Boltzmann equation, the term \( (T^4 - T_0^4) \) represents the difference between the temperatures of the emitting body and its surroundings, raised to the fourth power.
This non-linear relationship means that even small changes in temperature can drastically alter the radiation power.
In our scenario, the sphere's temperature is 3000 K, while the surroundings are at 290 K. Despite this seeming small relativity in difference when simply subtracting, the fourth power raises the impact of these temperatures significantly.
Thus, understanding how the steep sensitivity of thermal radiation to temperature changes is vital for accurate predictions and adjustments in practical applications, like maintaining consistent heat levels in the tungsten sphere.