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Suppose that you intercept \(5.0 \times 10^{-3}\) of the energy radiated by a hot sphere that has a radius of \(0.020 \mathrm{~m},\) an emissivity of 0.80 , and a surface temperature of \(500 \mathrm{~K}\). How much energy do you intercept in \(2.0 \mathrm{~min} ?\)

Short Answer

Expert verified
The energy intercepted in 2.0 minutes is approximately 0.857 J.

Step by step solution

01

Identify Given Values

First, let's identify the given values from the problem. We have:- Radius, \( r = 0.020 \, \mathrm{m} \).- Emissivity, \( \varepsilon = 0.80 \).- Surface temperature, \( T = 500 \, \mathrm{K} \).- Fraction of energy intercepted, \( f = 5.0 \times 10^{-3} \).- Time duration, \( t = 2.0 \, \mathrm{min} = 120 \, \mathrm{s} \).
02

Determine the Surface Area

The surface area \( A \) of the sphere is calculated using the formula \( A = 4 \pi r^2 \). Here, \( r = 0.020 \, \mathrm{m} \), so:\[A = 4 \pi (0.020)^2 = 4 \pi \times 0.0004 = 0.0050265 \, \mathrm{m}^2\]
03

Use Stefan-Boltzmann Law

The power radiated by the sphere can be found using the Stefan-Boltzmann law:\[ P = \varepsilon \sigma A T^4 \]where \( \sigma = 5.67 \times 10^{-8} \, \mathrm{W/m^2K^4} \). Plugging in the values, we get:\[P = 0.80 \times 5.67 \times 10^{-8} \times 0.0050265 \times (500)^4\]\[ \Rightarrow P = 0.80 \times 5.67 \times 10^{-8} \times 0.0050265 \times 62500000 \]\[ \Rightarrow P = 0.80 \times 5.67 \times 10^{-8} \times 314.15625\]\[ \Rightarrow P = 1.4281 \, \mathrm{W}\]
04

Calculate Energy Intercepted

In time \( t = 120 \, \mathrm{s} \), the sphere radiates energy \( E \) given by:\[ E = P \times t = 1.4281 \times 120 = 171.372 \, \mathrm{J} \]The intercepted energy \( E_i \) is a fraction \( f \) of this energy:\[E_i = f \times E = 5.0 \times 10^{-3} \times 171.372 = 0.85686 \, \mathrm{J}\]
05

Conclusion

Therefore, the amount of energy intercepted in 2.0 minutes is approximately \( 0.857 \, \mathrm{J} \).

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Key Concepts

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

Understanding Thermal Radiation
Thermal radiation is a type of electromagnetic radiation emitted by all objects with a temperature above absolute zero. This radiation is generated by the thermal motion of charged particles within matter. All matter, therefore, emits thermal radiation depending on its temperature.

Importantly, the amount and spectrum of radiation emitted depends on the temperature of the object and its surface characteristics. The higher the temperature, the more energy is radiated. Thermal radiation is a continuous source of energy transfer and plays a crucial role in heating processes in everyday life.

In the context of a hot sphere, as given in the problem, this radiation transfers energy from the surface of the sphere into the surroundings.
What is Emissivity?
Emissivity is a measure of how effectively a surface emits thermal radiation. It ranges between 0 and 1, where:
  • An emissivity of 1 represents a perfect emitter, known as a black body, which emits the maximum amount of radiation possible at a given temperature.
  • An emissivity of 0 indicates a perfect reflector, which does not emit any thermal radiation.
Most real-world objects, like the hot sphere in our problem, will have emissivities between these two extremes.

In our exercise, the sphere has an emissivity of 0.8, which means it is an efficient radiator of thermal energy but not perfect. This value directly influences how much energy the sphere radiates, as shown in the use of the Stefan-Boltzmann Law.
Calculating the Surface Area
To determine the thermal radiation emitted by an object, we need to calculate its surface area. For a sphere, the formula is:
  • \[A = 4 \pi r^2\]
Where \(A\) is the surface area, and \(r\) is the radius of the sphere.

In our exercise, the sphere's radius is 0.020 m. Plugging this into the formula provides a surface area necessary to calculate the total energy radiated by the sphere. Calculating the surface area helps understand how much surface contributes to radiating the energy, essential for further computations like those using the Stefan-Boltzmann Law.
Energy Interception Mechanics
Energy interception refers to capturing or absorbing a portion of the thermal energy emitted by a source. In our problem scenario, a fraction of the sphere's radiated energy is intercepted and measured.

The fraction intercepted is given as \(5.0 \times 10^{-3}\) of the total energy radiated. This fractional interception makes it essential to understand how much energy is effectively captured over a specific duration.

By knowing both the energy radiated and the fraction intercepted, we can calculate how much energy is actually gathered, which in this case involves utilizing the total energy radiated over a period and then applying the given fraction.
Sphere Radiated Energy
To determine the sphere's radiated energy, we apply the Stefan-Boltzmann Law, which quantifies the power radiated by an object as:
  • \[P = \varepsilon \sigma A T^4\]
where:
  • \(P\) is the power radiated
  • \(\varepsilon\) is the emissivity of the sphere
  • \(\sigma\) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8} \, \mathrm{W/m^2K^4}\)
  • \(A\) is the surface area of the sphere
  • \(T\) is the absolute temperature in Kelvins

By substituting the known values, including the emissivity, surface area, and temperature, we can compute the total power emitted by the sphere. Further calculations with this power allow us to evaluate how much energy the sphere radiates over a given time and thus, how much of this energy is intercepted.

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Most popular questions from this chapter

A cube of edge length \(6.0 \times 10^{-6} \mathrm{~m}\) emissivity \(0.75,\) and temperature \(-100^{\circ} \mathrm{C}\) floats in an environment at \(-150^{\circ} \mathrm{C}\). What is the cube's net thermal radiation transfer rate?

(a) In 1964 , the temperature in the Siberian village of Oymyakon reached \(-71^{\circ} \mathrm{C}\). What temperature is this on the Fahrenheit scale? (b) The highest officially recorded temperature in the continental United States was \(134^{\circ} \mathrm{F}\) in Death Valley, California. What is this temperature on the Celsius scale?

(a) Calculate the rate at which body heat is conducted through the clothing of a skier in a steady-state process, given the following data: the body surface area is \(1.8 \mathrm{~m}^{2}\), and the clothing is \(1.0 \mathrm{~cm}\) thick; the skin surface temperature is \(33^{\circ} \mathrm{C}\) and the outer surface of the clothing is at \(1.0^{\circ} \mathrm{C} ;\) the thermal conductivity of the clothing is \(0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} .\) (b) If, after a fall, the skier's clothes became soaked with water of thermal conductivity \(0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K},\) by how much is the rate of conduction multiplied?

An energetic athlete can use up all the energy from a diet of \(4000 \mathrm{Cal} /\) day. If he were to use up this energy at a steady rate, what is the ratio of the rate of energy use compared to that of a \(100 \mathrm{~W}\) bulb? (The power of \(100 \mathrm{~W}\) is the rate at which the bulb converts electrical energy to heat and the energy of visible light.)

A flow calorimeter is a device used to measure the specific heat of a liquid. Energy is added as heat at a known rate to a stream of the liquid as it passes through the calorimeter at a known rate. Measurement of the resulting temperature difference between the inflow and the outflow points of the liquid stream enables us to compute the specific heat of the liquid. Suppose a liquid of density \(0.85 \mathrm{~g} / \mathrm{cm}^{3}\) flows through a calorimeter at the rate of \(8.0 \mathrm{~cm}^{3} / \mathrm{s}\). When energy is added at the rate of \(250 \mathrm{~W}\) by means of an electric heating coil, a temperature difference of \(15 \mathrm{C}^{\circ}\) is established in steady-state conditions between the inflow and the outflow points. What is the specific heat of the liquid?

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