Use Stefan's law to calculate how much less energy is emitted per unit area of a 4500-K sunspot.
Use Stefan's law to calculate how much less energy is emitted per unit area of a 4500-K sunspot.
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![**Physics Problem Solving with Stefan's Law**
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**Problem Statement:**
Use Stefan’s law to calculate how much less energy is emitted per unit area of a 4500-K sunspot.
**Explanation:**
Stefan's Law, also known as the Stefan-Boltzmann Law, is fundamental in understanding the radiation emitted by a blackbody. The law states that the total energy radiated per unit surface area of a blackbody across all wavelengths per unit time \( (E) \) is directly proportional to the fourth power of the blackbody's absolute temperature \( (T) \). Mathematically, it is expressed as:
\[ E = \sigma T^{4} \]
where:
- \( E \) is the radiant energy emitted per unit area.
- \( \sigma \) (sigma) is the Stefan-Boltzmann constant, approximately \( 5.67 × 10^{-8} \, W \, m^{-2} \, K^{-4} \).
- \( T \) is the absolute temperature of the blackbody in Kelvin (K).
To determine how much less energy is emitted per unit area by a sunspot with a temperature of 4500 K, apply Stefan’s Law to compare it with the energy emitted by the Sun's photosphere (typically around 5778 K).
**Steps:**
1. **Calculate the Energy Emitted by the Sun’s Photosphere:**
Let \( T_{sun} \) be the temperature of the Sun’s photosphere.
\( E_{sun} = \sigma T_{sun}^{4} \)
\( E_{sun} = (5.67 \times 10^{-8} \, W \, m^{-2} \, K^{-4}) \times (5778 \, K)^{4} \)
2. **Calculate the Energy Emitted by the Sunspot:**
Let \( T_{sunspot} \) be the temperature of the sunspot.
\( E_{sunspot} = \sigma T_{sunspot}^{4} \)
\( E_{sunspot} = (5.67 \times 10^{-8} \, W \, m^{-2} \, K^{-4}) \times (4500 \, K)^{4} \)
3. **Determine the Difference:**
The difference in emitted energy per unit area:
\( \Delta E = E](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbdf4ad1c-8d6e-4c6e-ab0d-20fae6a9f061%2F5d1e481b-bf84-4d20-81a4-a0b3db4dcf5f%2Ffgsqltiq_processed.png&w=3840&q=75)
Transcribed Image Text:**Physics Problem Solving with Stefan's Law**
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**Problem Statement:**
Use Stefan’s law to calculate how much less energy is emitted per unit area of a 4500-K sunspot.
**Explanation:**
Stefan's Law, also known as the Stefan-Boltzmann Law, is fundamental in understanding the radiation emitted by a blackbody. The law states that the total energy radiated per unit surface area of a blackbody across all wavelengths per unit time \( (E) \) is directly proportional to the fourth power of the blackbody's absolute temperature \( (T) \). Mathematically, it is expressed as:
\[ E = \sigma T^{4} \]
where:
- \( E \) is the radiant energy emitted per unit area.
- \( \sigma \) (sigma) is the Stefan-Boltzmann constant, approximately \( 5.67 × 10^{-8} \, W \, m^{-2} \, K^{-4} \).
- \( T \) is the absolute temperature of the blackbody in Kelvin (K).
To determine how much less energy is emitted per unit area by a sunspot with a temperature of 4500 K, apply Stefan’s Law to compare it with the energy emitted by the Sun's photosphere (typically around 5778 K).
**Steps:**
1. **Calculate the Energy Emitted by the Sun’s Photosphere:**
Let \( T_{sun} \) be the temperature of the Sun’s photosphere.
\( E_{sun} = \sigma T_{sun}^{4} \)
\( E_{sun} = (5.67 \times 10^{-8} \, W \, m^{-2} \, K^{-4}) \times (5778 \, K)^{4} \)
2. **Calculate the Energy Emitted by the Sunspot:**
Let \( T_{sunspot} \) be the temperature of the sunspot.
\( E_{sunspot} = \sigma T_{sunspot}^{4} \)
\( E_{sunspot} = (5.67 \times 10^{-8} \, W \, m^{-2} \, K^{-4}) \times (4500 \, K)^{4} \)
3. **Determine the Difference:**
The difference in emitted energy per unit area:
\( \Delta E = E
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