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Chapter 17: Problem 115

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 4.93 kg of helium is lost per hour due to radiation heat transfer.

Step by step solution

01

Understand the Problem

We need to calculate the amount of liquid helium lost per hour due to heat transfer by radiation from the surrounding walls at 77.3 K to the metal can at 4.22 K. The emissivity is given, which will help in calculating the radiant energy transfer.
02

Use the Stefan-Boltzmann Law

Heat transfer by radiation is calculated using the Stefan-Boltzmann Law:\[ Q = \varepsilon \sigma A (T_1^4 - T_2^4) \]where \( \varepsilon \) is the emissivity (0.200), \( \sigma \) is the Stefan-Boltzmann constant \( 5.67 \times 10^{-8} \text{W/m}^2\text{K}^4 \), \( A \) is the surface area of the can, and \( T_1 \) and \( T_2 \) are the temperatures of the surroundings and the can respectively.
03

Calculate the Surface Area

The can is cylindrical, so we calculate the surface area excluding the top and bottom as it does not contribute to the heat lost through the side:\[ A = \pi d h \]where \( d = 0.090 \text{ m} \) is the diameter and \( h = 0.250 \text{ m} \) is the height. So, the area \( A \) is:\[ A = \pi \times 0.090 \times 0.250 = 0.0707 \text{ m}^2 \]
04

Calculate the Heat Transfer Rate

Substitute the values into the Stefan-Boltzmann Law:\[ Q = 0.200 \times 5.67 \times 10^{-8} \times 0.0707 \times ((77.3)^4 - (4.22)^4) \]Calculate \((77.3)^4 \approx 3.60 \times 10^6\) and \((4.22)^4 \approx 317.4\):\[ Q = 0.200 \times 5.67 \times 10^{-8} \times 0.0707 \times (3.60 \times 10^6 - 317.4) \approx 0.200 \times 5.67 \times 10^{-8} \times 0.0707 \times 3.59968 \times 10^6 \]\[ Q \approx 28.67 \text{ J/s} \]
05

Convert to Energy Loss per Hour

Convert the energy loss from joules per second to joules per hour:\[ Q_{hour} = 28.67 \text{ J/s} \times 3600 \text{ s/hr} = 1.03 \times 10^5 \text{ J/hr} \]
06

Calculate Mass of Helium Evaporated

Use the latent heat of vaporization to find the mass evaporated per hour:\[ m = \frac{Q_{hour}}{L_v} = \frac{1.03 \times 10^5 \text{ J/hr}}{2.09 \times 10^4 \text{ J/kg}} \approx 4.93 \text{ kg/hr} \]

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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 fundamental principle that describes how energy is radiated from a surface. It helps us understand how energy is transferred as heat in the form of radiation. This law states that the power radiated from a black body is proportional to the fourth power of its absolute temperature. The formula is expressed as:
\[ Q = \varepsilon \sigma A (T_1^4 - T_2^4) \]
where:
  • \( Q \) is the rate of heat transfer in watts (W).
  • \( \varepsilon \) is the emissivity of the surface, ranging from 0 to 1, representing how efficient a surface is at emitting thermal radiation.
  • \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \).
  • \( A \) is the surface area of the radiating body, measured in square meters.
  • \( T_1 \) and \( T_2 \) are the absolute temperatures of the two surfaces in Kelvin.
In our context, this law applies to the radiation heat transfer between the walls surrounding the helium and the metal can. It's crucial for calculating how much heat is radiating into the system due to the temperature difference.
latent heat of vaporization
Latent heat of vaporization refers to the amount of heat required to convert a unit mass of a liquid into gas without a change in temperature. It's intrinsic to phase changes that occur at a specific temperature and pressure.
For helium, the latent heat of vaporization is given as \( 2.09 \times 10^4 \, \text{J/kg} \). This value is essential to determine how much liquid helium evaporates when a given amount of heat is absorbed.
  • This heat essentially "breaks" the molecular bonds of helium, allowing it to transition from liquid to gas.
  • In the problem, the calculation involves converting the energy loss from radiation into the mass of helium lost through evaporation.
Using the calculated energy loss due to radiation, you can determine how much of the liquid helium has been vaporized by dividing the total energy by the latent heat value.
surface area calculation
When calculating heat transfer, determining the surface area is a crucial step. This is particularly true when dealing with objects of complex shapes, like the cylindrical metal can storing helium.
To calculate the surface area of a cylinder, particularly when interested in the lateral surface, we exclude the top and bottom.
  • The formula used is: \[ A = \pi d h \]
  • \( d \) is the diameter of the cylinder.
  • \( h \) is the height of the cylinder.
Given \( d = 0.090 \text{ m} \) and \( h = 0.250 \text{ m} \), the lateral surface area becomes \( \pi \times 0.090 \times 0.250 \), which calculates to \( 0.0707 \text{ m}^2 \). This calculated area is then used in the Stefan-Boltzmann Law to find how much heat transfer occurs between the surface of the can and its surroundings.

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Most popular questions from this chapter

Spacecraft Reentry. A spacecraft made of aluminum circles the earth at a speed of 7700 \(\mathrm{m} / \mathrm{s} .\) (a) Find the ratio of its kinetic energy to the energy required to raise its temperature from \(0^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\) . (The melting point of aluminum is \(660^{\circ} \mathrm{C}\) . Assume a constant specific heat of 910 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K} .\) (b) Discuss the bearing of your answer on the problem of the reentry of a manned space vehicle into the earth's atmosphere.

You are making pesto for your pasta and have a cylindrical measuring cup 10.0 \(\mathrm{cm}\) high made of ordinary glass \(\left[\beta=2.7 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1}\right]\) that is filled with olive oil \([\beta=6.8 \times\) \(10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1} ]\) to a height of 2.00 \(\mathrm{mm}\) below the top of the cup. Initially, the cup and oil are at room temperature \(\left(22.0^{\circ} \mathrm{C}\right) .\) You get a phone call and forget about the olive oil, which you inadvertently leave on the hot stove. The cup and oil heat up slowly and have a common temperature. At what temperature will the olive oil start to spill out of the cup?

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a 70.0-kg man to cool his body 1.00 \(\mathrm{C}^{\circ} ?\) The heat of vaporization of water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) The specific heat of a typical human body is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) (see Exercise 17.31\() .\) (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can \(\left(355 \mathrm{cm}^{3}\right)\).

Basal Metabolic Rate. The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. \(A 75-\mathrm{kg}(165-\mathrm{lb})\) person of height 1.83 \(\mathrm{m}(6 \mathrm{ft})\) has a body surface area of approximately 2.0 \(\mathrm{m}^{2}\) . (a) What is the net amount of heat this person could radiate per second into a room at \(18^{\circ} \mathrm{C}\) (about \(65^{\circ} \mathrm{F}\) ) if his skin's surface temperature is \(30^{\circ} \mathrm{C} ?\) (At such temperatures, nearly all the heat is infrared radiation, for which the body's emissivity is \(1.0,\) regardless of the amount of pigment.) (b) Normally, 80\(\%\) of the energy produced by metabolism goes into heat, while the rest goes into things like pumping blood and repairing cells. Also normally, a person at rest can get rid of this excess heat just through radiation. Use your answer to part (a) to find this person's basal metabolic rate.

Food Intake of a Hamster. The energy output of an animal engaged in an activity is called the basal metabolic rate \((\mathrm{BMR})\) and is a measure of the conversion of food energy into other forms of energy. A simple calorimeter to measure the BMR consists of an insulated box with a thermometer to measure the temperature of the air. The air has density 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and specific heat \(1020 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K} . \mathrm{A} 50.0-\mathrm{g}\) hamster is placed in a calorimeter that contains 0.0500 \(\mathrm{m}^{3}\) of air at room temperature. (a) When the hamster is running in a wheel, the temperature of the air in the calorimeter rises 1.60 \(\mathrm{C}^{\circ}\) per hour. How much heat does the running hamster generate in an hour? Assume that all this heat goes into the air in the calorimeter. You can ignore the heat that goes into the walls of the box and into the thermometer, and assume that no heat is lost to the surroundings. (b) Assuming that the hamster converts seed into heat with an efficiency of 10\(\%\) and that hamster seed has a food energy value of \(24 \mathrm{J} / \mathrm{g},\) how many grams of seed must the hamster eat per hour to supply this energy?
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