/*! 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 87 . A thermos for liquid helium. A... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 87

. A thermos for liquid helium. A physicist uses a cylindrical metal can 0.250 \(\mathrm{m}\) high and 0.090 \(\mathrm{m}\) in diameter to store liquid helium at \(4.22 \mathrm{K} ;\) at that temperature the heat of vaporization of helium is \(2.09 \times 10^{4} \mathrm{J} / \mathrm{kg} .\) Completely surrounding the metal can are walls maintained at the temperature of liquid nitrogen, 77.3 \(\mathrm{K}\) , with vacuum between the can and the surrounding walls. How much helium is lost per hour? The emissivity of the metal can is 0.200 . The only heat transfer between the metal can and the surrounding walls is by radiation.

Short Answer

Expert verified
Approximately 0.0579 kg of helium is lost per hour.

Step by step solution

01

Calculate the Surface Area of the Metal Can

The surface area, \(A\), of a cylinder is given by the formula \(A = 2\pi r h + 2\pi r^2\), where \(r\) is the radius and \(h\) is the height of the cylinder. Here, the diameter \(d = 0.090 \mathrm{m}\), so the radius \(r = \frac{d}{2} = 0.045 \mathrm{m}\). The height \(h = 0.250 \mathrm{m}\). Therefore:\[ A = 2\pi (0.045) (0.250) + 2\pi (0.045)^2 = 0.07065 + 0.01272 = 0.08337 \text{ m}^2 \]
02

Calculate the Power of Radiation Lost from the Can

The power lost due to radiation, \(P\), is given by the Stefan-Boltzmann Law: \(P = \varepsilon \sigma A (T_1^4 - T_2^4)\). Here, \(\varepsilon = 0.200\) is the emissivity, \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m}^2\mathrm{K}^4\) is the Stefan-Boltzmann constant, \(A = 0.08337 \text{ m}^2\), and \(T_1 = 77.3 \mathrm{K}\) and \(T_2 = 4.22 \mathrm{K}\) are the temperatures. \[ P = 0.200 \times 5.67 \times 10^{-8} \times 0.08337 \times ((77.3)^4 - (4.22)^4) \]Calculating this, we get:\[ (77.3)^4 = 35637673.41 \] and \[ (4.22)^4 = 316.03 \] thus, \[ P = 0.200 \times 5.67 \times 10^{-8} \times 0.08337 \times (35637673.41 - 316.03) \approx 0.200 \times 5.67 \times 10^{-8} \times 0.08337 \times 35637357.38 \approx 0.336 \text{ W} \]
03

Calculate the Energy Lost in One Hour

Energy lost, \(E\), in one hour can be calculated by multiplying the power by the time in seconds (since \(P = 0.336 \text{ W} = 0.336 \text{ J/s}\)), so:\[ E = P \times ext{time} \]The time for one hour is \(3600\) seconds:\[ E = 0.336 \times 3600 = 1209.6 \text{ J} \]
04

Calculate the Mass of Helium Lost

To find the mass of helium lost, we use the heat of vaporization formula:\[ Q = mL \]where \(Q\) is the heat energy, \(m\) is the mass, and \(L = 2.09 \times 10^4 \mathrm{J/kg}\) is the heat of vaporization of helium. Solving for \(m\):\[ m = \frac{Q}{L} = \frac{1209.6 \text{ J}}{2.09 \times 10^4 \text{ J/kg}} \approx 0.0579 \text{ kg} \]

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Key Concepts

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

Stefan-Boltzmann Law
In the fascinating world of radiation heat transfer, the Stefan-Boltzmann Law is a key concept. It describes how much thermal energy is radiated from a blackbody in terms of its temperature. The Stefan-Boltzmann Law is formulated as \( P = \varepsilon \sigma A (T_1^4 - T_2^4) \). Here,
  • \( P \) is the power radiated due to heat transfer
  • \( \varepsilon \) represents the emissivity of the material
  • \( \sigma \) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8} \mathrm{W/m}^2\mathrm{K}^4\)
  • \( A \) is the surface area of the radiating body
  • \( T_1 \) and \( T_2 \) are the absolute temperatures of the body and its surroundings in Kelvin
The law tells us that the power radiated is proportional to the difference of the fourth powers of the object's surface temperature and the surroundings temperature. This fourth power makes the formula especially sensitive to temperature changes, meaning small variations can result in large changes in radiated energy.
emissivity
Emissivity is a fundamental concept in radiation heat transfer. It measures how effectively a surface emits thermal radiation compared to a perfect emitter, known as a blackbody. Emissivity values range from 0 to 1, where:
  • 0 represents a perfect reflector, meaning no radiation is emitted
  • 1 indicates a perfect emitter, radiating energy most efficiently
In most materials, such as the metal can storing helium, emissivity is less than 1, which means they are not perfect emitters. For example, the emissivity of the metal can in the exercise is \(0.200\).
A lower emissivity indicates that the surface emits less radiation, which can be advantageous when minimizing heat transfer, such as in insulating purposes or preventing a liquid from boiling away quickly. Accurately measuring or estimating emissivity is crucial for precise calculations in thermodynamics applications, particularly when leveraging the Stefan-Boltzmann Law.
heat of vaporization
Understanding the heat of vaporization is essential when dealing with phase changes in substances, especially within thermodynamics. The heat of vaporization is the amount of energy required to change 1 kilogram of a substance from a liquid to a gaseous state without changing its temperature. For helium, this value is \( 2.09 \times 10^4 \mathrm{J/kg} \).
This significant value implies that a considerable amount of energy is needed for even a small quantity of helium to evaporate. When calculating how much liquid helium evaporates due to heat transfer, we need to know both the energy lost from the helium and the heat of vaporization.
Using the formula \( Q = mL \), where:
  • \( Q \) is the energy required for vaporization
  • \( m \) is the mass of the liquid that has evaporated
  • \( L \) is the heat of vaporization
This calculation is crucial in applications such as cryogenics, where minimizing the loss of liquid helium is of the utmost importance. It guides physicists in designing storage containers to control and reduce heat transfer efficiently.

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Most popular questions from this chapter

Maintaining body temperature. While running, a 70 \(\mathrm{kg}\) student generates thermal energy at a rate of 1200 \(\mathrm{W}\) . To maintain a constant body temperature of \(37^{\circ} \mathrm{C},\) this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the heat could not flow out of the student's body, for what amount of time could a student run before irreversible body damage occurred? (Protein structures in the body are damaged irreversibly if the body temperature rises to \(44^{\circ} \mathrm{C}\) or above. The specific heat capacity of a typical human body is \(3480 \mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}),\) slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heat capacities.)

An 8.50 kg block of ice at \(0^{\circ} \mathrm{C}\) is sliding on a rough horizontal icehouse floor (also at \(0^{\circ} \mathrm{C} )\) at 15.0 \(\mathrm{m} / \mathrm{s} .\) Assume that half of any heat generated goes into the floor and the rest goes into the ice. (a) How much ice melts after the speed of the ice has been reduced to 10.0 \(\mathrm{m} / \mathrm{s} ?\) (b) What is the maximum amount of ice that will melt?

Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales: (a) the midday temperature at the surface of the moon \((400 \mathrm{K}) ;\) (b) the temperature at the tops of the clouds in the atmosphere of Saturn \((95 \mathrm{K}) ;(\mathrm{c})\) the temperature at the center of the sun \(\left(1.55 \times 10^{7} \mathrm{K}\right)\) .

Conduction through the skin. The blood plays an important role in removing heat from the body by bringing this heat directly to the surface where it can radiate away. Nevertheless, this heat must still travel through the skin before it can radiate away. We shall assume that the blood is brought to the bottom layer of skin at a temperature of \(37^{\circ} \mathrm{C}\) and that the outer surface of the skin is at \(30.0^{\circ} \mathrm{C}\) . Skin varies in thickness from 0.50 \(\mathrm{mm}\) to a few millimeters on the palms and soles, so we shall assume an average thickness of \(0.75 \mathrm{mm} . \mathrm{A} 165 \mathrm{lb}, 6 \mathrm{ft}\) person has a surface area of about 2.0 \(\mathrm{m}^{2}\) and loses heat at a net rate of 75 \(\mathrm{W}\) while resting. On the basis of our assumptions, what is the thermal conductivity of this person's skin?

\(\bullet\) One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 \(\mathrm{W}\) is required to keep the interior surface of the box 65.0 \(\mathrm{C}^{\circ}\) (about 120 \(\mathrm{F}^{\circ} )\) above the temperature of the outer surface. The total area of the box is \(2.18 \mathrm{m}^{2},\) and the wall thickness is 3.90 \(\mathrm{cm} .\) Find the thermal conductivity of the material in SI units.
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