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Consider a 150-W incandescent lamp. The filament of the lamp is \(5-\mathrm{cm}\) long and has a diameter of \(0.5 \mathrm{~mm}\). The diameter of the glass bulb of the lamp is \(8 \mathrm{~cm}\). Determine the heat flux, in W/ \(\mathrm{m}^{2},(a)\) on the surface of the filament and \((b)\) on the surface of the glass bulb, and (c) calculate how much it will cost per year to keep that lamp on for eight hours a day every day if the unit cost of electricity is \(\$ 0.08 / \mathrm{kWh}\).

Short Answer

Expert verified
Answer: (a) The heat flux on the surface of the filament is approximately 1,909,842 W/m². (b) The heat flux on the surface of the glass bulb is approximately 7,463 W/m². (c) The cost per year to keep the lamp on for eight hours a day every day is $34.84.

Step by step solution

01

Calculate the surface area of the filament

The filament is a cylinder with length \(l=5\mathrm{~cm}=0.05\mathrm{~m}\) and diameter \(d=0.5\mathrm{~mm}=0.0005\mathrm{~m}\). The radius of the cylinder is \(r=\frac{d}{2}=0.00025\mathrm{~m}\). The surface area of a cylinder (not including the bases) is given by the formula \(A_\mathrm{filament}=2\pi rh\), where \(h\) is the height (length) of the cylinder. So, we have: \(A_\mathrm{filament}=2\pi(0.00025\mathrm{~m})(0.05\mathrm{~m}) \approx 7.854 \times 10^{-5}\mathrm{~m}^2\)
02

Calculate the heat flux on the surface of the filament

The heat flux is the power divided by the surface area. Since we are given the power of the lamp as \(150\mathrm{~W}\), the heat flux on the surface of the filament is: \(q_\mathrm{filament}=\frac{150\mathrm{~W}}{7.854 \times 10^{-5}\mathrm{~m}^2} \approx 1,909,842\mathrm{~W/m}^2\)
03

Calculate the surface area of the glass bulb

The glass bulb is a sphere with diameter \(D=8\mathrm{~cm}=0.08\mathrm{~m}\). The radius of the sphere is \(R=\frac{D}{2}=0.04\mathrm{~m}\). The surface area of a sphere is given by the formula \(A_\mathrm{sphere}=4\pi R^2\). So, we have: \(A_\mathrm{glass_bulb}=4\pi(0.04\mathrm{~m})^2 \approx 2.0106 \times 10^{-2}\mathrm{~m}^2\)
04

Calculate the heat flux on the surface of the glass bulb

Using the same formula for heat flux, the heat flux on the surface of the glass bulb is: \(q_\mathrm{glass_bulb}=\frac{150\mathrm{~W}}{2.0106 \times 10^{-2}\mathrm{~m}^2} \approx 7,463\mathrm{~W/m}^2\)
05

Calculate the energy consumption per day

The lamp is on for 8 hours a day, so the energy consumption per day is: \(E_\mathrm{daily}=150\mathrm{~W} \times 8\mathrm{~h} = 1200\mathrm{~Wh} = 1.2\mathrm{~kWh}\)
06

Calculate the cost per year

The unit cost of electricity is \(0.08\mathrm{~\$/kWh}\). First, calculate the annual energy consumption by multiplying the daily consumption by the number of days in a year: \(E_\mathrm{annual}=1.2\mathrm{~kWh} \times 365 = 438\mathrm{~kWh}\) Then, calculate the cost per year by multiplying the annual energy consumption by the unit cost: \(C_\mathrm{annual}=438\mathrm{~kWh} \times 0.08\mathrm{~\$/kWh} = \$34.84\) So, the answers are: (a) the heat flux on the surface of the filament is approximately \(1,909,842\mathrm{~W/m}^2\), (b) the heat flux on the surface of the glass bulb is approximately \(7,463\mathrm{~W/m}^2\), and (c) the cost per year to keep the lamp on for eight hours a day every day is $34.84.

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Key Concepts

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

Incandescent Lamp
Incandescent lamps have been a common choice for lighting for many years. They work by heating a filament inside the bulb until it glows, producing light. The filament, usually made of tungsten, gets hot enough to emit visible light when electric current passes through it.

These lamps are often characterized by their wattage, which indicates the power consumption of the lamp. In our exercise, the lamp has a power rating of 150 watts, meaning it uses 150 watts of power when turned on. However, incandescent lamps are not very efficient at converting electricity into visible light, as much of the energy is lost in the form of heat.
  • They typically have a relatively short lifespan compared to other lighting technologies, such as LEDs.
  • Much of the energy used by incandescent lamps is converted to heat, making them less energy-efficient.
Despite their inefficiency, they were widely used due to their simple design and the pleasant warm light they produce.
Surface Area Calculation
Understanding how to calculate surface area is essential in physics, especially when dealing with heat exchange problems like the one in the exercise. Calculating the surface area of the incandescent lamp's components—the filament and the glass bulb—is a crucial step in determining the heat flux.

The surface area of the filament is treated as a cylinder. Its formula is given by the lateral surface area of the cylinder, \[A_{\text{filament}} = 2\pi rh\]where \(r\) is the radius and \(h\) is the height (length). Plugging in the values, the calculated area is approximately \[7.854 \times 10^{-5} \, \text{m}^2.\]

In contrast, the surface area of the glass bulb, which is a sphere, requires a different formula:\[A_{\text{bulb}} = 4\pi R^2\]With the bulb's radius being \(0.04 \, \text{m}\), this gives approximately\[2.0106 \times 10^{-2} \, \text{m}^2.\]
  • The cylindrical surface area calculation is relevant for structures with length.
  • The spherical surface area is useful when dealing with spherical objects like the glass bulb.
These calculations help in determining how much of the heat from the lamp is spread over the surfaces.
Energy Cost Calculation
An important aspect of using electrical devices, like incandescent lamps, is understanding their running costs. It's crucial for planning household budgets or business expenses related to electricity usage.

To calculate the energy cost, we first determine the energy consumption of the lamp. Given that the lamp operates for 8 hours each day:\[E_{\text{daily}} = 150 \, \text{W} \times 8 \, \text{h} = 1200 \, \text{Wh} = 1.2 \, \text{kWh}\]

This means the lamp uses 1.2 kilowatt-hours (kWh) of energy per day. Over a year (365 days), the total energy used is:\[E_{\text{annual}} = 1.2 \, \text{kWh} \times 365 = 438 \, \text{kWh}\]

Finally, to find the yearly cost, multiply the total annual energy consumption by the cost per kilowatt-hour:\[C_{\text{annual}} = 438 \, \text{kWh} \times 0.08 \, \text{\\(/kWh} = \\)34.84\]
  • Calculating daily and yearly energy use helps track consumption and costs.
  • Understanding per kWh costs aids in managing electricity expenses efficiently.
This knowledge allows consumers to be more responsible with their energy usage.
Electricity Consumption
Electricity consumption is a key factor in understanding the energy efficiency of any electrical device, including incandescent lamps. This concept revolves around how much electrical energy a device uses to perform its function. In the exercise, the lamp's electricity consumption is based on its power rating and the duration it is used.

Given its power rating of 150 watts, and usage of 8 hours a day, this consumption can significantly impact energy bills, especially over extended periods. To improve efficiency and reduce electricity costs:
  • Consider using lamps with lower power ratings for similar light output.
  • Switch to more energy-efficient lighting options, such as LEDs, which use less power.
Monitoring and optimizing electricity usage ensures responsible energy consumption, aligns with eco-friendly standards, and reduces costs. Understanding these concepts helps not only in the context of this exercise but also in real-world applications where energy conservation is increasingly important.

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Most popular questions from this chapter

A 25 -cm-diameter black ball at \(130^{\circ} \mathrm{C}\) is suspended in air, and is losing heat to the surrounding air at \(25^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat transfer from the black ball is (a) \(217 \mathrm{~W}\) (b) \(247 \mathrm{~W}\) (c) \(251 \mathrm{~W}\) (d) \(465 \mathrm{~W}\) (e) \(2365 \mathrm{~W}\)

A transistor with a height of \(0.4 \mathrm{~cm}\) and a diameter of \(0.6 \mathrm{~cm}\) is mounted on a circuit board. The transistor is cooled by air flowing over it with an average heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the air temperature is \(55^{\circ} \mathrm{C}\) and the transistor case temperature is not to exceed \(70^{\circ} \mathrm{C}\), determine the amount of power this transistor can dissipate safely. Disregard any heat transfer from the transistor base.

An electronic package with a surface area of \(1 \mathrm{~m}^{2}\) placed in an orbiting space station is exposed to space. The electronics in this package dissipate all \(1 \mathrm{~kW}\) of its power to the space through its exposed surface. The exposed surface has an emissivity of \(1.0\) and an absorptivity of \(0.25\). Determine the steady state exposed surface temperature of the electronic package \((a)\) if the surface is exposed to a solar flux of \(750 \mathrm{~W} /\) \(\mathrm{m}^{2}\), and \((b)\) if the surface is not exposed to the sun.

A \(0.3\)-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating \(0.06 \mathrm{~W}\). The board is impregnated with copper fillings and has an effective thermal conductivity of \(16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to the ambient air. Determine the temperature difference between the two sides of the circuit board. Answer: \(0.042^{\circ} \mathrm{C}\)

Steady heat conduction occurs through a \(0.3\)-m-thick \(9 \mathrm{~m} \times 3 \mathrm{~m}\) composite wall at a rate of \(1.2 \mathrm{~kW}\). If the inner and outer surface temperatures of the wall are \(15^{\circ} \mathrm{C}\) and \(7^{\circ} \mathrm{C}\), the effective thermal conductivity of the wall is (a) \(0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (b) \(0.83 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (c) \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(2.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(5.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

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