/*! 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 117 A Thermos for Liquid Helium. A p... [FREE SOLUTION] | 91Ó°ÊÓ

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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
The helium lost per hour is approximately 0.00064 kg.

Step by step solution

01

Determine the Stefan-Boltzmann Law for Power Emitted

The power radiated due to the temperature difference can be found using the Stefan-Boltzmann Law:\[ P = \varepsilon \sigma A (T_4^4 - T_1^4) \]where \( \varepsilon = 0.200 \) (emissivity of the can), \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2 \cdot \text{K}^4 \) (Stefan-Boltzmann constant), \( A \) is the surface area, \( T_4 = 77.3 \text{ K} \) (temperature of nitrogen walls), and \( T_1 = 4.22 \text{ K} \) (temperature of helium).
02

Calculate Surface Area of the Cylindrical Can

Given the height \( h = 0.250 \text{ m} \) and diameter \( d = 0.090 \text{ m} \), the surface area \( A \) of the cylindrical can is the area of the sides and ends:\[ A = \pi d h + 2 \pi \left(\frac{d}{2}\right)^2 \]Substitute to find this surface area.
03

Calculate the Power Radiated

Substitute the values into the Stefan-Boltzmann Law:\[ P = 0.200 \times 5.67 \times 10^{-8} \times A \times ((77.3)^4 - (4.22)^4) \]Compute this to find the thermal power emitted by the can due to radiation.
04

Determine the Energy Lost Per Hour

The energy lost per hour by the can is given by:\[ E = P \times 3600 \]because there are 3600 seconds in an hour.
05

Calculate Mass of Helium Lost Per Hour

The mass of helium \( m \) lost per hour can be found using the heat of vaporization:\[ m = \frac{E}{2.09 \times 10^4} \]Substitute the energy lost from Step 4 into this expression to find the mass of helium lost per hour.

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Key Concepts

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

Heat Transfer
When we talk about heat transfer in thermodynamics, we are discussing the movement of thermal energy from one object or substance to another. There are three primary modes by which this can occur: conduction, convection, and radiation. In the context of this problem involving the storage of liquid helium, heat transfer through radiation is the central focus.

This particular exercise highlights the unique properties of heat transfer in a vacuum, where conduction and convection are negligible. The surrounding walls of the can are at a higher temperature compared to the helium inside, causing energy transfer to occur via radiation. This is a form of heat transfer that does not require a medium and can occur across the vacuum space, affecting the helium stored inside the can by raising its temperature, leading to vaporization.
Radiation
Radiation is one of the main methods by which heat is transferred, especially in space or across vacuum environments. Unlike conduction or convection, radiation does not need a medium to travel through. It can directly transmit energy from one object to another in the form of electromagnetic waves.

In this problem, we have a cylindrical metal can surrounded by walls at a higher temperature, transferring heat via radiation to the liquid helium inside the can. This heat transfer leads to an increase in energy within the can, contributing to the vaporization of helium. Because a vacuum surrounds the can, radiation is the only mode of heat transfer, making it crucial to understand its role in the loss of helium over time.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is essential in understanding the energy transfer via radiation in thermodynamic systems. It defines the power radiated from a black body in terms of its temperature. The law is given as:\[ P = \varepsilon \sigma A (T_4^4 - T_1^4) \]In this equation:
  • \( P \) is the power radiated.
  • \( \varepsilon \) is the emissivity of the material, representing how efficiently it radiates energy compared to a perfect black body.
  • \( \sigma \) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8}\,\text{W/m}^2\cdot\text{K}^4\).
  • \( A \) is the surface area of the object.
  • \( T_4 \) and \( T_1 \) are the temperatures of the surrounding walls and the helium, respectively.

The Stefan-Boltzmann Law helps calculate the amount of energy radiated per unit time. In this case, it determines how much thermal energy the vacuum environment's walls transfer to the can over time.
Emissivity
Emissivity is a measure of an object's ability to emit thermal radiation relative to a perfect black body. A perfect black body has an emissivity of 1, which means it radiates energy most efficiently. Most real-world materials have a lower emissivity, meaning they are not as efficient at radiating heat.

In our example, the emissivity of the metal can is 0.200. This means it emits 20% of the thermal radiation that a perfect black body would emit at the same temperature. Emissivity is crucial for calculating the radiative heat transfer from the can using the Stefan-Boltzmann Law. Knowing the emissivity allows us to understand how much radiative energy is lost from the helium over time and, consequently, how much helium vaporizes.

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Most popular questions from this chapter

In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a \(200-\mathrm{W}\) electric immersion heater in 0.320 \(\mathrm{kg}\) of water. (a) How much heat must be added to the water to raise its temperature from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C}\) ? (b) How much time is required? Assume that all of the heater's power goes into heating the water.

A 4.00 tag silver ingot is taken from a furnace, where its temperature is \(750.0^{\circ} \mathrm{C},\) and placed on a large block of ice at \(0.0^{\circ} \mathrm{C} .\) Assuming that all the heat given up by the silver is used to melt the ice, how much ice is melted?

Find the Celsius temperatures corresponding to (a) a winter night im Seattle \(\left(41.0^{\circ} \mathrm{F}\right) ;(\mathrm{b})\) a hot summer day in Palm Springs \(\left(107.0^{\circ} \mathrm{F}\right) ;(\mathrm{c})\) a cold winter day in northern Manitoba \(\left(-18.0^{\circ} \mathrm{F}\right)\) .

A technician measures the specific heat of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat transferred to the liquid for 120 \(\mathrm{s}\) at a constant rate of 65.0 \(\mathrm{W}\) . The mass of the liquid is 0.780 \(\mathrm{kg}\) , and its temperature increases from \(18.55^{\circ} \mathrm{C}\) to \(22.54^{\circ} \mathrm{C}\) (a) Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat? Explain.

A vessel whose walls are thermally insulated contains 2.40 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice, all at a temperature of \(0.0^{\circ} \mathrm{C}\) The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to \(28.0^{\circ} \mathrm{C}\) ? You can ignore the heat transferred to the container.

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