Problem 115 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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:\[ A = \pi \times 0.090 \times 0.250 = 0.0707 \text{ m}^2 \]

Step 4: 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 \((4.22)^4 \approx 317.4\):\[ Q = 0.200 \times 5.67 \times 10^{-8} \times 0.0707 \times (3.60 \times 10^6 - 317.4) \approx 0.200 \times 5.67 \times 10^{-8} \times 0.0707 \times 3.59968 \times 10^6 \]\[ Q \approx 28.67 \text{ J/s} \]

Step 5: Convert to Energy Loss per Hour

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

Step 6: Calculate Mass of Helium Evaporated

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

Key Concepts

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.