Problem 82 - College Physics

Question

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 0.0489 kg of helium is lost per hour.

Step 1: Determine the Surface Area of the Cylinder

The surface area of a cylinder without the top and bottom (as the sides are directly exposed to the heat) is given by the formula: \[ A = 2\pi r h \] Where \( r = 0.045 \text{ m} \) \( (\text{radius is half the diameter}) \) and \( h = 0.250 \text{ m} \). Substituting these values, the surface area is: \[ A = 2 \pi (0.045 \text{ m}) (0.250 \text{ m}) \approx 0.0707 \text{ m}^2 \]

Step 2: Use Stefan-Boltzmann Law for Heat Transfer via Radiation

The heat transfer via radiation can be calculated using: \[ Q = \epsilon \sigma A (T_1^4 - T_2^4) \] where \( \epsilon = 0.200 \) is the emissivity, \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \) (the Stefan-Boltzmann constant), and \( T_1 = 77.3 \text{ K} \), \( T_2 = 4.22 \text{ K} \) are the temperatures of the surroundings and the helium, respectively. Substitute the given values to find the heat loss \( Q \):\[ Q = 0.200 \times 5.67 \times 10^{-8} \times 0.0707 \times ((77.3)^4 - (4.22)^4) \]

Step 3: Calculate the Heat Transfer

Calculating the above expression: \[ Q = 0.200 \times 5.67 \times 10^{-8} \times 0.0707 \times (356078 - 318) \approx 0.200 \times 5.67 \times 10^{-8} \times 0.0707 \times 355760 \approx 0.284 \text{ W} \] Thus, the rate of heat loss is approximately \( 0.284 \text{ W} \).

Step 4: Determine the Mass of Helium Lost

The heat loss per second is 0.284 J/s. To find out how much helium is lost per hour, calculate the total heat transfer over an hour: \[ ext{Total heat loss in one hour} = 0.284 \times 3600 = 1022.4 \text{ J} \]To find the mass of helium lost, use the relation: \[ ext{mass} = \frac{\text{Total heat loss}}{\text{Latent Heat}} = \frac{1022.4}{2.09 \times 10^4} \approx 0.0489 \text{ kg} \]

Step 5: Conclusion

The mass of helium lost per hour due to the radiation heat transfer is approximately \( 0.0489 \text{ kg} \).

Key Concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is key to understanding heat transfer through thermal radiation, especially in situations involving significant temperature differences, like our problem with liquid helium. This law states that the power radiated by a black body per unit area is proportional to the fourth power of its temperature. The mathematical expression is \( P = \sigma A T^4 \), where \( \sigma \) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4) \), \( A \) is the surface area, and \( T \) is the temperature in Kelvin.
In the problem, the metal can containing helium is surrounded by walls at a higher temperature. The Stefan-Boltzmann Law helps us calculate how much heat energy transfers from the warmer walls to the cooler can through radiation. By using the difference \( (T_1^4 - T_2^4) \), where \( T_1 \) and \( T_2 \) are the temperatures of the walls and the can respectively, we find the net heat transfer.
This calculation is fundamental in physics for evaluating energy exchanges between objects, especially when conduction and convection are not involved, like in a vacuum.

Emissivity

Emissivity describes how well an object radiates energy compared to a perfect black body. It is a dimensionless value between 0 and 1. In our helium problem, the metal can has an emissivity of 0.200, indicating that it emits only 20% of the energy a perfect emitter would at the same temperature.
Photo by iStock Mathematically, the emissivity factor is bi-fold in our scenario: it scales down the theoretical maximum energy radiated according to the Stefan-Boltzmann Law, making it more realistic for non-ideal materials. This factor is crucial for precise thermal calculations.
Understanding emissivity helps physicists measure how different materials interact with thermal radiation, providing insights into energy loss mechanisms. In practical terms, it means carefully matching materials to applications based on how they emit or absorb radiant energy.

Vaporization Heat of Helium

The heat of vaporization is a critical concept when dealing with phase changes, like the conversion of liquid helium to gas. This property signifies the amount of energy required to transform 1 kg of a substance from liquid to vapor without temperature change. For helium, this value is substantial at \( 2.09 \times 10^4 \, \text{J/kg} \).
Understanding vaporization energy is essential when calculating how much of a substance will change states under given thermal conditions. In the exercise, knowing the heat of vaporization helps determine how much helium evaporates due to the heat absorbed through radiation. The latent heat calculation is used to find the mass of helium lost based on the total heat gained, highlighting the direct relationship between absorbed energy and mass vaporized.

Physics Problem Solving

Solving physics problems, like the one involving helium, requires a systematic approach to ensure success. Here’s how to approach complex problems:

• Identify the knowns and unknowns: Gather all provided data, like dimensions, emission rates, and temperatures.


• Understand the concepts: Familiarize yourself with core ideas such as the Stefan-Boltzmann Law and emissivity.


• Set up the equations: Apply the right formulas. In our case, starting with the surface area calculation was vital.


• Solve step-by-step: Tackle the problem in chunks, like computing heat transfer before finding mass loss.


• Review and check: Ensure every step adheres to the problem's framework, verifying units and logic.

This structured methodology, paired with a solid grasp of physics concepts, allows for effective problem-solving, leading to precise and meaningful results.