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A solar collector is placed in direct sunlight where it absorbs energy at the rate of \(880 \mathrm{J} / \mathrm{s}\) for each square meter of its surface. The emissivity of the solar collector is \(e=0.75 .\) What equilibrium temperature does the collector reach? Assume that the only energy loss is due to the emission of radiation.

Short Answer

Expert verified
The equilibrium temperature is approximately 527 K.

Step by step solution

01

Understanding Energy Absorption and Emission

The solar collector absorbs energy from the sun at a rate of \(880 \text{ J/s} \) for each square meter. This energy is absorbed continuously until it reaches an equilibrium temperature. At equilibrium, the power absorbed equals the power emitted by radiation.
02

Applying Stefan-Boltzmann Law

The rate of energy emission due to radiation is given by the Stefan–Boltzmann Law: \( P_{emit} = e \sigma A T^4 \). Here, \( e \) is the emissivity, \( \sigma \) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4) \), \( A \) is the area, and \( T \) is the temperature in Kelvin.
03

Setting Up the Equation

At equilibrium, \( P_{abs} = P_{emit} \). Thus, \( 880 \text{ J/s} = 0.75 \times 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \times T^4 \).
04

Solving for Temperature

Rearrange to solve for \( T \): \[ T^4 = \frac{880}{0.75 \times 5.67 \times 10^{-8}} \] \[ T^4 = \frac{880}{4.2525 \times 10^{-8}} \] \[ T^4 \approx 2.069 \times 10^{10} \] \[ T \approx (2.069 \times 10^{10})^{0.25} \approx 527 \text{ K} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solar Energy
Solar energy is a form of renewable energy that is harnessed from the sun's radiation. It is clean and sustainable, playing a crucial role in addressing global energy needs.
Solar collectors are devices designed to absorb sunlight and convert it into usable energy, usually in the form of heat. They work by capturing the sun's rays and converting this solar radiation into thermal energy. This process is vital for various applications, such as heating water, generating electricity, and more.
  • Solar collectors absorb energy continuously, as long as they are exposed to sunlight.
  • The efficiency of energy absorption depends on factors like the surface area, angle of exposure, and the material's characteristics.
  • At the molecular level, solar radiation energizes electrons, leading to an increase in kinetic energy, which is perceived as heat.
Solar energy is not only abundant but also a key component in reducing carbon footprints and combating climate change. Understanding how it interacts with devices like solar collectors is essential for optimizing energy capture and use.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a principle in physics describing how objects emit radiation. It states that the power radiated by a black body per unit area is directly proportional to the fourth power of its absolute temperature.
This law is represented mathematically as:\[ P = \sigma\ A T^4 \]Where:
  • \(\sigma\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\).
  • \(A\) is the surface area from which the radiation is emitted.
  • \(T\) is the absolute temperature in Kelvin.
In the case of solar collectors, which are not perfect black bodies, we factor in emissivity (denoted \(e\)). This adjustment accounts for the efficiency of the actual material compared to a perfect black body.
The Stefan-Boltzmann Law helps us calculate the thermal equilibrium temperature. This is when the energy absorbed from solar radiation equals the energy emitted. Effectively, it tells us how hot a surface can get solely from solar radiation before reaching a balance, or equilibrium.
Emissivity
Emissivity is a measure of how effectively a material emits thermal radiation compared to a perfect black body. It is indicated by the symbol \(e\) and its value ranges between 0 and 1. A perfect black body has an emissivity of 1, meaning it emits thermal radiation at the maximum rate.
  • An object with an emissivity of 0 would be considered a perfect reflector, emitting no thermal radiation.
  • Real objects typically have emissivity values between 0.1 and 0.95.
  • Materials like metals often reflect more, having lower emissivity, while rough or dark surfaces have higher emissivity values, more effectively radiating energy.
In the context of solar collectors, understanding emissivity is crucial. When calculating the equilibrium temperature using the Stefan-Boltzmann Law, emissivity influences how much radiation a surface emits for a given temperature. The solar collector described, with an emissivity of 0.75, suggests it is a decent emitter, not losing heat as quickly as a perfect conductor, yet efficiently radiating energy.
Understanding emissivity is significant for designing and choosing materials in applications like solar collectors to ensure effective thermal management and energy efficiency.

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Most popular questions from this chapter

In an old house, the heating system uses radiators, which are hollow metal devices through which hot water or steam circulates. In one room the radiator has a dark color (emissivity \(=0.75\) ). It has a temperature of \(62^{\circ} \mathrm{C}\). The new owner of the house paints the radiator a lighter color (emissivity \(=\) 0.50 ). Assuming that it emits the same radiant power as it did before being painted, what is the temperature (in degrees Celsius) of the newly painted radiator?

Liquid helium is stored at its boiling-point temperature of \(4.2 \mathrm{K}\) in a spherical container \((r=0.30 \mathrm{m})\). The container is a perfect blackbody radiator. The container is surrounded by a spherical shield whose temperature is \(77 \mathrm{K}\). A vacuum exists in the space between the container and the shield. The latent heat of vaporization for helium is \(2.1 \times 10^{4} \mathrm{J} / \mathrm{kg} .\) What mass of liquid helium boils away through a venting valve in one hour?

The block in the drawing has dimensions \(L_{0} \times 2 L_{0} \times 3 L_{0},\) where \(L_{0}=0.30 \mathrm{m} .\) The block has a thermal conductivity of \(250 \mathrm{J} /\left(\mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right) .\) In drawings \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C},\) heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is \(35^{\circ} \mathrm{C}\) and that of the cooler surface is \(19^{\circ} \mathrm{C} .\) Determine the heat that flows in \(5.0 \mathrm{s}\) for each case.

A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to \(35^{\circ} \mathrm{C}\) from \(175^{\circ} \mathrm{C}\), its net radiant power decreases to \(12.0 \mathrm{W}\). What was the net radiant power of the baking dish when it was first removed from the oven? Assume that the temperature in the kitchen remains at \(22^{\circ} \mathrm{C}\) as the dish cools down.

An object is inside a room that has a constant temperature of 293 K. Via radiation, the object emits three times as much power as it absorbs from the room. What is the temperature (in kelvins) of the object? Assume that the temperature of the object remains constant.

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