A cylindrical metal can, blackened on the outside, \(0.10 \mathrm{~m}\) high and \(0.05 \mathrm{~m}\) in diameter, contains liquid \({ }^{4} \mathrm{He}\) at its normal boiling point of \(4.22 \mathrm{~K}\), at which its heat of vaporization is \(20.4 \mathrm{~kJ} / \mathrm{kg}\). Completely surrounding the helium can are walls maintained at the temperature of liquid nitrogen \((77.35 \mathrm{~K})\), and the intervening space is continuously evacuated to a very low pressure. How much helium is lost per hour?

The Stefan-Boltzmann Law is fundamental in thermodynamics for understanding heat transfer through radiation. It states that the power radiated per unit area of a black body is proportional to the fourth power of the temperature of the body. The formula can be expressed as:

\[ Q = \varepsilon \sigma A (T_1^4 - T_2^4) \]

Where:

• \( Q \) is the heat transfer per second,
• \( \varepsilon \) (epsilon) is the emissivity of the material, with a value of 1 for a perfect black body,
• \( \sigma \) (sigma) is the Stefan-Boltzmann constant, approximately \( 5.67 x 10^{-8} W/m^2K^4 \),
• \( A \) is the surface area,
• \( T_1 \) and \( T_2 \) are the Kelvin temperatures of the surroundings and the object, respectively.

This equation is key to calculating how much heat is radiated from an object, which in turn helps in determining the heat transfer in thermodynamic problems, such as the one involving the helium can.

Calculating the surface area is a crucial step in determining the heat transfer via the Stefan-Boltzmann Law. For a cylindrical can, the surface area includes both the cylindrical surface and the two circular ends. The formula for the total surface area, \( A \), of a cylinder is:

\[ A = 2 \pi r h + 2 \pi r^2 \]

Where:

• \( r \) is the radius of the cylinder, which is half of the diameter,
• \( h \) is the height of the cylinder,
• \( \pi \) is a constant (approximately 3.14159).

In the given problem, the radius \( r \) is \( 0.025 m \) (half of the diameter \( 0.05 m \)) and the height \( h \) is \( 0.10 m \). Therefore, the total surface area is computed as:

\[ A = 2 \pi (0.025 m)(0.10 m) + 2 \pi (0.025 m)^2 \]

This yields the surface area necessary for further calculations in determining the rate of heat transfer.

The heat of vaporization is the amount of heat required to convert a liquid into a gas at its boiling point without a change in temperature. In this exercise, we focus on the heat of vaporization of liquid helium (\textsuperscript{4} He). The value for helium is given as \( 20.4 kJ/kg \) (or \( 20400 J/kg \)). This parameter is crucial for calculating the mass of helium lost due to vaporization.

Once we determine the total heat energy (Q) absorbed by the helium can in one hour using the Stefan-Boltzmann Law, we can calculate the mass of helium lost (m) using:

\[ m = \frac{Q}{L} \]

Where:

• \( m \) is the mass of helium lost,
• \( Q \) is the total heat transferred in one hour,
• \( L \) is the heat of vaporization (\( 20400 J/kg \)).

This formula shows the relationship between the energy absorbed and the amount of mass that is vaporized.

Thermodynamic equilibrium is a state where all parts of a system are at the same temperature, and no net heat transfer occurs within the system. For the problem at hand, understanding thermodynamic equilibrium helps explain why heat flows from the surrounding walls to the helium can.

The walls surrounding the can are at a higher temperature (77.35 K) compared to the helium (4.22 K). To reach equilibrium, heat transfers from the walls to the helium. Since the helium is at its boiling point, this heat causes it to vaporize.

In such exercises, ensuring that all calculations assume the system is not in equilibrium helps underscore the context and the need to find how much helium is lost as it tries to equilibrate with its surroundings. Since the intervening space is evacuated to low pressure, heat transfer occurs primarily through radiation, emphasizing the role of the Stefan-Boltzmann Law in our solution.