/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Liquid helium is stored at its b... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
About 31.26 kg of liquid helium boils away in one hour.

Step by step solution

01

Understand the Problem

We are tasked with finding the mass of liquid helium that boils away from a container in one hour due to radiation heat transfer. We know the container acts as a blackbody radiator and there's a temperature difference between the container and the surrounding shield.
02

Apply the Stefan-Boltzmann Law

For a blackbody, the power radiated (or absorbed) per unit area is given by the Stefan-Boltzmann Law: \[ P = \sigma \cdot A \cdot (T_2^4 - T_1^4) \]where \( \sigma = 5.67 \times 10^{-8} \mathrm{W/m^2 \cdot K^4} \) is the Stefan-Boltzmann constant, \( A \) is the surface area, \( T_2 = 77 \mathrm{K} \) is the temperature of the shield, and \( T_1 = 4.2 \mathrm{K} \) is the temperature of the helium.
03

Calculate the Surface Area of the Sphere

The surface area \( A \) of a sphere with radius \( r \) is given by the formula: \[ A = 4 \pi r^2 \]Substitute \( r = 0.30 \mathrm{m} \) to find \( A \):\[ A = 4 \pi (0.30)^2 \approx 1.13 \mathrm{m^2} \]
04

Calculate the Power Radiated

Substitute the known values into the Stefan-Boltzmann Law:\[ P = 5.67 \times 10^{-8} \cdot 1.13 \cdot ((77)^4 - (4.2)^4) \]Calculate to find the power:\[ P = 5.67 \times 10^{-8} \cdot 1.13 \cdot (28257601 - 311.213) \approx 182.35 \mathrm{W} \]
05

Find the Energy Lost in One Hour

Since power \( P \) is the rate of energy transfer in watts (joules per second), we can find the total energy lost over one hour by multiplying \( P \) by the number of seconds in an hour (3600 seconds):\[ E = 182.35 \times 3600 \approx 656460 \mathrm{J} \]
06

Calculate the Mass of Helium Boiled Off

Use the latent heat of vaporization \( L = 2.1 \times 10^4 \mathrm{J/kg} \) to find the mass \( m \) of helium that boils away:\[ m = \frac{E}{L} = \frac{656460}{2.1 \times 10^4} \approx 31.26 \mathrm{kg} \]

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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 critical concept in thermal radiation when dealing with blackbodies. It quantifies the power emitted by a blackbody in terms of its temperature. In essence, this law states that the power radiated per unit area from the surface of a blackbody is proportional to the fourth power of its absolute temperature. Mathematically, it can be expressed as:
\[ P = \sigma \cdot A \cdot (T^4) \]
where:
  • \( P \) is the power emitted per unit area, measured in watts per square meter \( \mathrm{W/m^2} \).
  • \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \mathrm{W/m^2 \cdot K^4} \).
  • \( A \) is the surface area of the blackbody, in square meters.
  • \( T \) is the absolute temperature, measured in kelvins \( \mathrm{K} \).
This law is pivotal for understanding how energy is exchanged between objects purely through thermal radiation, especially when no physical substance or medium is transferring the energy, such as in a vacuum.
In the context of the exercise, the Stefan-Boltzmann Law helps calculate how much power is transferred between the helium container and its surrounding shield, both considered black bodies with different temperatures.
Blackbody Radiation
Blackbody radiation refers to the theoretical idealization of objects that perfectly absorb all radiation incident upon them and re-emit energy according to their temperature. Hence, the emitted radiation spectrum is determined solely by the object's temperature, making a blackbody an ideal emitter and absorber. In physics, the concept of a blackbody helps us understand the nature of electromagnetic radiation emitted by all objects with temperature. The intensity and type of radiation follow Planck's radiation law, outlining the distribution of emitted radiation over various wavelengths.
Some important points about blackbody radiation include:
  • It is only dependent on temperature, not on the material or surface characteristics.
  • The emitted radiation spans a continuous range of wavelengths.
  • Real-world objects approximate blackbody conditions; however, perfect blackbody behavior is conceptual.
In the given problem, the helium container is treated as a blackbody. This assumption allows us to apply the Stefan-Boltzmann Law directly to calculate energy transfer due to radiation, aiding in discovering how much helium boils off over time.
Latent Heat of Vaporization
Latent heat of vaporization is a crucial concept in understanding phase changes, particularly when substances transition from liquid to vapor. It is defined as the amount of heat energy required to convert a unit mass of a liquid into vapor without a change in temperature.
In equations, it is often denoted by \( L \), and it's measured in joules per kilogram (\( \mathrm{J/kg} \)). This parameter varies among substances, influencing how readily they evaporate once heat is applied.
  • It explains why water requires significant energy to boil and become steam.
  • In cryogenics, such as handling liquid helium at low temperatures, knowing the latent heat helps calculate energy requirements for cooling and vaporization.
  • It is vital in determining energy losses due to phase changes in environmental and industrial processes.
In this exercise, the latent heat of vaporization for helium is used to calculate the mass of helium that turns into vapor, given the energy transferred over one hour. The energy calculated from Stefan-Boltzmann's Law is divided by latent heat to find the mass of helium boiled away.

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Most popular questions from this chapter

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?

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 \(0.65 ?\)

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?

In a house the temperature at the surface of a window is \(25^{\circ} \mathrm{C} .\) The temperature outside at the window surface is \(5.0^{\circ} \mathrm{C} .\) Heat is lost through the window via conduction, and the heat lost per second has a certain value. The temperature outside begins to fall, while the conditions inside the house remain the same. As a result, the heat lost per second increases. What is the temperature at the outside window surface when the heat lost per second doubles?

Sirius \(\mathrm{B}\) is a white star that has a surface temperature (in kelvins) that is four times that of our sun. Sirius \(\mathrm{B}\) radiates only 0.040 times the power radiated by the sun. Our sun has a radius of \(6.96 \times 10^{8} \mathrm{m} .\) Assuming that Sirius B has the same emissivity as the sun, find the radius of Sirius B.

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