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Under steady-state operation the surface temperature of a small \(20-\mathrm{W}\) incandescent light bulb is \(125^{\circ} \mathrm{C}\) when the temperature of the room air and walls is \(25^{\circ} \mathrm{C}\). Approximating the bulb as a sphere \(40 \mathrm{~mm}\) in diameter with a surface emissivity of \(0.8\), what is the rate of heat transfer from the surface of the bulb to the surroundings?

Short Answer

Expert verified
The rate of heat transfer from the bulb is approximately 0.797 W.

Step by step solution

01

Understand the Problem

We need to determine the rate of heat transfer from the surface of the light bulb to its surroundings under steady-state conditions. This involves exchanging heat primarily through radiation.
02

Collect Known Variables

We know the surface temperature of the bulb ( {T_s}) is 125掳C, the ambient temperature ( {T_ ext{inf}}) is 25掳C, the diameter of the bulb is 40 mm, emissivity ( {蔚}) is 0.8, and Stefan-Boltzmann constant ( {蟽}) is approximately 5.67 x 10^{-8} W/m虏K鈦.
03

Convert Temperatures to Kelvin

We convert the temperatures to Kelvin: {T_s} = 125掳C + 273.15 = 398.15 K {T_ ext{inf}} = 25掳C + 273.15 = 298.15 K.
04

Calculate the Surface Area of the Sphere

The diameter of the sphere is 40 mm, so the radius (r) is 20 mm, or 0.020 m. The surface area (A) of a sphere is given by: A = 4蟺r虏 = 4蟺(0.020 m)虏 = 0.00503 m虏.
05

Use the Stefan-Boltzmann Law to Find Heat Transfer Rate

The rate of heat transfer by radiation ( {Q虈}) from the bulb's surface is given by: Q虈 = 蟽蔚A(T_s^4 - T_ ext{inf}^4) = 5.67 x 10^{-8} x 0.8 x 0.00503 x (398.15鈦 - 298.15鈦).
06

Perform the Calculation

Calculate T_s^4 and T_ ext{inf}^4: T_s^4 = (398.15)鈦 = 2.52 x 10^10 K鈦 T_ ext{inf}^4 = (298.15)鈦 = 7.95 x 10^9 K鈦 Now calculate {Q虈}: Q虈 = 5.67 x 10^{-8} x 0.8 x 0.00503 x (2.52 x 10^{10} - 7.95 x 10^9) Q虈 鈮 0.797 W.

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in thermal physics. It describes the power radiated from a black body per unit area as a function of its temperature. This law is central to understanding how objects like light bulbs emit energy. The formula given by the Stefan-Boltzmann Law is:\( Q虈 = 蟽蔚A(T_s^4 - T_{\text{inf}}^4) \),where:
  • \( Q虈 \) is the rate of heat transfer or emitted power.
  • \( 蟽 \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \).
  • \( 蔚 \) is the emissivity of the material, representing its efficiency in emitting thermal radiation.
  • \( A \) is the surface area of the emitting body.
  • \( T_s \) and \( T_{\text{inf}} \) are the surface and surrounding temperatures in Kelvin, respectively.
By applying this formula, one can calculate the thermal radiation emitted by any body at a specific temperature.
This concept highlights the key role temperature plays in thermal radiation; the higher the temperature, the greater the heat transfer rate.
Thermal Radiation
Thermal radiation is the transfer of heat through electromagnetic waves. Unlike conduction or convection, it doesn't require a medium like air or water to carry the heat. This form of heat can travel through a vacuum, making it the primary form of heat transfer for many celestial bodies, including the sun. For objects like a light bulb, thermal radiation occurs from the surface of the bulb to the surroundings. The efficiency of this radiation depends on factors like temperature and emissivity. Emissivity, in particular, indicates how well a surface emits heat compared to a perfect black body. Materials with high emissivity values, like the light bulb with 0.8 emissivity, are efficient at radiating energy. Understanding thermal radiation is crucial for energy efficiency. For instance, improving the emissivity of materials can enhance heating or cooling efficacy, influencing industrial and household applications significantly.
Steady-State Conditions
Steady-state conditions imply that a system is in equilibrium: the input and output of energy are balanced. In the context of a light bulb, this means that the heat being produced by the bulb is equal to the heat being lost to the surroundings, ensuring consistent operating conditions. Under steady-state conditions, temperature remains constant over time, making it easier to analyze and predict thermal behavior. This concept is crucial in systems design as it provides a simplistic model to evaluate performance. In practical terms, achieving steady-state can involve carefully managing the balance of energy in a system. For the light bulb in our exercise, steady-state allows us to focus on the radiant heat transfer without being concerned about changes over time. It's a foundational concept in engineering and physics for explaining phenomena in a stable, unchanging environment.

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Most popular questions from this chapter

Consider two long vertical plates maintaincd at unifomm temperatures \(T_{a, 1}>T_{a, 2}\). The plates are open at their ends and are separated by the distance \(2 l\). (a) Sketch the velocity distribution in the space between the plates. (b) Write appropriate forms of the continuity, momentum, and energy cquations for laminar fow between the plates. (c) Evaluate the temperature distribution, and express your result in terms of the mean temperature, \(T_{m}=\) \(\left(T_{n 1}+T_{s .2}\right) / 2\).

A circular grill of diameter \(0.25 \mathrm{~m}\) and emissivity \(0.9\) is maintained at a constant surface temperature of \(130^{\circ} \mathrm{C}\). What electrical power is required when the room air and surroundings are at \(24^{\circ} \mathrm{C}\) ?

Certain wood stove designs rely exclusively on heat transfer by radiation and natural convection to the surtoundings. Consider a stove that forms a cubical enclosure, \(L_{4}=1 \mathrm{~m}\) on a side, in a large room. The exterior walls of the stove have an emissivity of \(\varepsilon=0.8\) and are at an operating temperature of \(T_{\text {S }}=500 \mathrm{~K}\).The stove pipe, which may be assumed to be isothermal at an operating temperature of \(T_{\varphi}=400 \mathrm{~K}\), has a diarneter of \(D_{p}=0.25 \mathrm{~m}\) and a height of \(L_{p}=2 \mathrm{~m}\), extending from steve to ceiling. The stove is in a large room whose air and walls are at \(T_{\mathrm{s}}=T_{\mathrm{a}}=300 \mathrm{~K}\). Neglecting heat transfer from the small horizontal section of the pipe and radiation exchange between the pipe and stove, estimute the rate at which heat is transferred from the stove and pipe to the surroundings.

The ABC Evening News Report in a news segment on hypothermia research sudies af the University of Minnesota claimed that heat loss from the body is 30 fimes faster in \(10^{\circ} \mathrm{C}\) watter than in air at the samme temperature. Is that a realistic statement?

Long stainless steel rods of \(50-\mathrm{mm}\) diameter are preheated to a uniform temperature of \(1000 \mathrm{~K}\) before being suspended from an overhead cenveyor for transport to a hot forming operation. The conveyor is in a large room whose walls and air are at \(300 \mathrm{~K}\). (a) Assuming the lincar motion of the rod to have a negligible effect on convecticn heat transfer from its surface, determine the average convection coefficient at the start of the transpont process. (b) If the surface emissivity of the rod is \(\varepsilon=0.40\), what is the effective radiation heat transfer coefficient at the start of the transport process?

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