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How many days does it take for a perfect blackbody cube \((0.0100 \mathrm{m}\) on a side, \(30.0^{\circ} \mathrm{C}\) ) to radiate the same amount of energy that a one-hundred watt light bulb uses in one hour?

Short Answer

Expert verified
The blackbody cube takes approximately 73.5 days to radiate the same energy.

Step by step solution

01

Calculate the energy used by the light bulb

The energy used by a light bulb in one hour is calculated by multiplying power by time. For a 100 watt bulb, the energy is \( E = P \times t = 100 \, \text{W} \times 1 \, \text{hour} = 100 \, \text{W} \times 3600 \, \text{s} = 360,000 \, \text{J} \).
02

Use the Stefan-Boltzmann Law for blackbody radiation

The power radiated by a blackbody is given by the Stefan-Boltzmann Law: \( P = \sigma A T^4 \), where \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the temperature in Kelvin.
03

Convert the temperature to Kelvin

To convert Celsius to Kelvin, use the formula: \( T(K) = T(掳C) + 273.15 \). Thus, the temperature is \( 30.0 + 273.15 = 303.15 \, \text{K} \).
04

Calculate the surface area of the cube

The surface area \( A \) for a cube with side \( a \) is \( A = 6a^2 \). For a side length of \( 0.0100 \, \text{m} \), \( A = 6 \times (0.0100)^2 = 6 \times 0.0001 = 0.0006 \, \text{m}^2 \).
05

Calculate the power output of the blackbody cube

Using the surface area and the temperature, the power \( P \) radiated by the cube is \( P = (5.67 \times 10^{-8} \times 0.0006 \times (303.15)^4) \approx 0.0567 \, \text{W} \).
06

Determine the time needed to equate the energies

The energy emitted by the blackbody in time \( t \) is \( E = Pt \). Setting this equal to the energy used by the bulb, \( 360,000 = 0.0567t \). Solving for \( t \), \( t = \frac{360,000}{0.0567} \approx 6,348,500 \, \text{s} \).
07

Convert time from seconds to days

Convert the time from seconds to days by dividing by the number of seconds in a day: \( t = \frac{6,348,500}{86400} \approx 73.5 \, \text{days} \).

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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 crucial part of blackbody radiation. It tells us how much power is emitted by a perfect blackbody, which is an idealized object that absorbs all radiation. This law states that the power emitted per unit area is directly proportional to the fourth power of the blackbody's temperature. This relationship is captured by the equation: P = \sigma A T^4, where:
  • \( P \) is the radiant power emitted in watts (W).
  • \( \sigma = 5.67 \times 10^{-8} \) W/m虏K鈦 is the Stefan-Boltzmann constant.
  • \( A \) is the surface area of the blackbody in square meters (m虏).
  • \( T \) is the absolute temperature in Kelvin (K).
This formula highlights that even a small increase in temperature can lead to a significant increase in radiated power, as the temperature is raised to the fourth power. Thus, small changes in temperature have large effects on the power output of a blackbody. This principle is important in understanding how stars and other astronomical bodies emit energy.
Temperature Conversion
Temperature conversion is a simple yet important step in scientific calculations as it helps standardize measurements. In physics, we often convert temperatures from Celsius to Kelvin, since Kelvin is the SI unit for thermodynamic temperature. The conversion is straightforward: \( T(K) = T(掳C) + 273.15 \)This formula simply adds 273.15 to the Celsius temperature to convert it to Kelvin. For example, if you have a temperature of 30掳C, converting to Kelvin would be: \( T = 30 + 273.15 = 303.15 \) KThe Kelvin scale is advantageous in scientific calculations because it starts at absolute zero, the point at which all molecular motion stops. Note that increments in Celsius and Kelvin are identical, so a change of 1掳C is equivalent to a change of 1 K.
Surface Area Calculation
Surface area calculation is a fundamental part of applying the Stefan-Boltzmann Law effectively. The total power radiated by a blackbody depends on its surface area, meaning the larger the surface area, the more energy can be emitted. In this context, the surface area of a shape is the total area that the surface of the object occupies.For a cube, the formula to calculate surface area is straightforward: \( A = 6a^2 \)Where \( a \) is the length of a side of the cube. Since a cube has six faces, each with an area of \( a^2 \), the total surface area is six times the surface area of one face. If the side of the cube is 0.0100 m, then: \( A = 6 \times (0.0100)^2 = 6 \times 0.0001 = 0.0006 \) m虏Accurate calculations of surface area are essential for efficient application of formulas like Stefan-Boltzmann Law, as they impact the calculations of power and energy emissions. This calculation links geometry with physics in practical applications.

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Most popular questions from this chapter

Due to a temperature difference \(\Delta T,\) heat is conducted through an aluminum plate that is \(0.035 \mathrm{m}\) thick. The plate is then replaced by a stainless steel plate that has the same temperature difference and cross- sectional area. How thick should the steel plate be so that the same amount of heat per second is conducted through it?

A solid cylinder is radiating power. It has a length that is ten times its radius. It is cut into a number of smaller cylinders, each of which has the same length. Each small cylinder has the same temperature as the original cylinder. The total radiant power emitted by the pieces is twice that emitted by the original cylinder. How many smaller cylinders are there?

Suppose the skin temperature of a naked person is \(34^{\circ} \mathrm{C}\) when the person is standing inside a room whose temperature is \(25^{\circ} \mathrm{C} .\) The skin area of the individual is \(1.5 \mathrm{m}^{2} .\) (a) Assuming the emissivity is 0.80, find the net loss of radiant power from the body. (b) Determine the number of food Calories of energy (1 food Calorie = \(4186 \mathrm{J}\) ) that are lost in one hour due to the net loss rate obtained in part (a). Metabolic conversion of food into energy replaces this loss.

A person is standing outdoors in the shade where the temperature is 28 C. (a) What is the radiant energy absorbed per second by his head when it is covered with hair? The surface area of the hair (assumed to be flat) is \(160 \mathrm{cm}^{2}\) and its emissivity is \(0.85 .\) (b) What would be the radiant energy absorbed per second by the same person if he were bald and the emissivity of his head were \(0.65 ?\)

A wall in a house contains a single window. The window consists of a single pane of glass whose area is \(0.16 \mathrm{m}^{2}\) and whose thickness is \(2.0 \mathrm{mm}\). Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are \(18 \mathrm{m}^{2}\) and \(0.10 \mathrm{m},\) respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window?

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