Problem 117 - 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

The helium lost per hour is approximately 0.00064 kg.

Step 1: 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).

Step 2: 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.

Step 3: 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.

Step 4: 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.

Step 5: 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.

Key Concepts

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.