Problem 75
Question
A black metal foil is warmed by radiation from a small sphere at temperature \(T\) and at a distance \(d\). It is found that the power received by the foil is \(P\). If both the temperature and distance are doubled, the power received by the foil will be (a) \(16 P\) (b) \(4 P\) (c) \(2 P\) (d) \(\underline{P}\)
Step-by-Step Solution
Verified Answer
(b) \(4P\)
1Step 1: Understanding the Stefan-Boltzmann Law
The power radiated by a black body, such as the sphere, is given by the Stefan-Boltzmann Law: \( P = \sigma A T^4 \), where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area of the sphere, and \( T \) is the temperature of the sphere in Kelvin. This is an important principle as it directly determines the radiation power output based on temperature.
2Step 2: Applying the inverse square law
The power received by the foil is also affected by the inverse square law relative to the distance \(d\) from the sphere: \( P \proportional \frac{1}{d^2} \). This means if the distance is doubled, the power received will be reduced by a factor of 4 \( \frac{1}{(2d)^2} = \frac{1}{4d^2} \).
3Step 3: Calculating the effect of doubling the temperature
When the temperature \(T\) of the sphere is doubled, the power radiated increases by \(2^4 = 16\) times, according to the Stefan-Boltzmann Law \( (2T)^4 = 16T^4 \).
4Step 4: Combining temperature and distance effects
When taking both effects into account, the new power received \( P_{new} \) when both temperature and distance are doubled can be calculated as follows: \( P_{new} = 16 \times \frac{1}{4} \times P = 4P \). Thus, the overall effect is that the power received by the foil is four times the original power.
Key Concepts
Black Body RadiationInverse Square LawThermal RadiationRadiation Power Calculation
Black Body Radiation
Black body radiation is a fundamental concept in physics. It refers to the thermal radiation emitted by an idealized object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. No real object perfectly fits this description, but many materials approximate this behavior. For our purposes, a small sphere (treated as a black body) emits radiation according to its temperature. The Stefan-Boltzmann Law, given by the equation \( P = \sigma A T^4 \), governs this process, where \( P \) is the power radiated, \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the absolute temperature in Kelvin. This law shows that the power emitted by a black body dramatically increases with temperature, specifically to the fourth power of the temperature.
Inverse Square Law
The inverse square law describes how the intensity of a physical quantity diminishes as you move away from the source. In this context, it explains how the power received by an object diminishes with distance from the radiation source. Specifically, the power received follows the relation \( P \proportional \frac{1}{d^2} \), where \( d \) is the distance from the source. This means that if the distance between the source (sphere) and observer (foil) is doubled, the power received becomes one-fourth. This is due to the spread of radiation over a larger area as distance increases, resulting in less power per unit area.
Thermal Radiation
Thermal radiation is a type of electromagnetic radiation generated by the thermal motion of charged particles in matter. All objects emit thermal radiation based on their temperature. A black body is a perfect emitter of such radiation due to its ability to absorb all incoming radiation. The radiation energy emitted is distributed across various wavelengths, with peak intensity occurring at a particular wavelength according to temperature, described by Wien's displacement law. Understanding thermal radiation is crucial for calculating how much energy is absorbed or emitted by materials in response to temperature changes in practical applications.
Radiation Power Calculation
Radiation power calculation involves determining how much power an object receives or emits through radiation, considering key factors like surface temperature, distance, and emissive properties of materials.
- Start with the Stefan-Boltzmann Law to calculate the power emitted by a black body: \( P = \sigma A T^4 \).
- Factor in distance using the inverse square law: \( P \proportional \frac{1}{d^2} \).
- Consider changes: Doubling temperature increases emitted power by \(16\) times due to the \((2T)^4\) impact, while doubling distance reduces received power by \(4\) times.
- Combine these effects to find the net power received: in the exercise, the result was \(4P\), where \(P\) is the original power.
Other exercises in this chapter
Problem 74
A black body is at a temperature of \(2880 \mathrm{~K}\). The energy of radiation emitted by this object with wavelength between \(499 \mathrm{~nm}\) and \(500
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Two bodies \(A\) and \(B\) are placed in an evacuated vessel maintained at a temperature of \(27^{\circ} \mathrm{C}\). The temperature of \(A\) is \(327^{\circ}
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The reflectance and emittance of a perfectly black body are respectively (a) 0,1 (b) 1,0 (c) \(0.5,0.5\) (d) 0,0
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The rate of emission of radiation of a black body at temperature \(27^{\circ} \mathrm{C}\) is \(E_{1}\). If its temperature is increased to \(327^{\circ} \mathr
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