Problem 87 - College 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 0.0579 kg of helium is lost per hour.

Step 1: Calculate the Surface Area of the Metal Can

The surface area, \(A\), of a cylinder is given by the formula \(A = 2\pi r h + 2\pi r^2\), where \(r\) is the radius and \(h\) is the height of the cylinder. Here, the diameter \(d = 0.090 \mathrm{m}\), so the radius \(r = \frac{d}{2} = 0.045 \mathrm{m}\). The height \(h = 0.250 \mathrm{m}\). Therefore:\[ A = 2\pi (0.045) (0.250) + 2\pi (0.045)^2 = 0.07065 + 0.01272 = 0.08337 \text{ m}^2 \]

Step 2: Calculate the Power of Radiation Lost from the Can

The power lost due to radiation, \(P\), is given by the Stefan-Boltzmann Law: \(P = \varepsilon \sigma A (T_1^4 - T_2^4)\). Here, \(\varepsilon = 0.200\) is the emissivity, \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m}^2\mathrm{K}^4\) is the Stefan-Boltzmann constant, \(A = 0.08337 \text{ m}^2\), and \(T_1 = 77.3 \mathrm{K}\) and \(T_2 = 4.22 \mathrm{K}\) are the temperatures. \[ P = 0.200 \times 5.67 \times 10^{-8} \times 0.08337 \times ((77.3)^4 - (4.22)^4) \]Calculating this, we get:\[ (77.3)^4 = 35637673.41 \] and \[ (4.22)^4 = 316.03 \] thus, \[ P = 0.200 \times 5.67 \times 10^{-8} \times 0.08337 \times (35637673.41 - 316.03) \approx 0.200 \times 5.67 \times 10^{-8} \times 0.08337 \times 35637357.38 \approx 0.336 \text{ W} \]

Step 3: Calculate the Energy Lost in One Hour

Energy lost, \(E\), in one hour can be calculated by multiplying the power by the time in seconds (since \(P = 0.336 \text{ W} = 0.336 \text{ J/s}\)), so:\[ E = P \times ext{time} \]The time for one hour is \(3600\) seconds:\[ E = 0.336 \times 3600 = 1209.6 \text{ J} \]

Step 4: Calculate the Mass of Helium Lost

To find the mass of helium lost, we use the heat of vaporization formula:\[ Q = mL \]where \(Q\) is the heat energy, \(m\) is the mass, and \(L = 2.09 \times 10^4 \mathrm{J/kg}\) is the heat of vaporization of helium. Solving for \(m\):\[ m = \frac{Q}{L} = \frac{1209.6 \text{ J}}{2.09 \times 10^4 \text{ J/kg}} \approx 0.0579 \text{ kg} \]

Key Concepts

Stefan-Boltzmann Law

In the fascinating world of radiation heat transfer, the Stefan-Boltzmann Law is a key concept. It describes how much thermal energy is radiated from a blackbody in terms of its temperature. The Stefan-Boltzmann Law is formulated as \( P = \varepsilon \sigma A (T_1^4 - T_2^4) \). Here,

• \( P \) is the power radiated due to heat transfer
• \( \varepsilon \) represents the emissivity of the material
• \( \sigma \) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8} \mathrm{W/m}^2\mathrm{K}^4\)
• \( A \) is the surface area of the radiating body
• \( T_1 \) and \( T_2 \) are the absolute temperatures of the body and its surroundings in Kelvin

The law tells us that the power radiated is proportional to the difference of the fourth powers of the object's surface temperature and the surroundings temperature. This fourth power makes the formula especially sensitive to temperature changes, meaning small variations can result in large changes in radiated energy.

emissivity

Emissivity is a fundamental concept in radiation heat transfer. It measures how effectively a surface emits thermal radiation compared to a perfect emitter, known as a blackbody. Emissivity values range from 0 to 1, where:

• 0 represents a perfect reflector, meaning no radiation is emitted
• 1 indicates a perfect emitter, radiating energy most efficiently

In most materials, such as the metal can storing helium, emissivity is less than 1, which means they are not perfect emitters. For example, the emissivity of the metal can in the exercise is \(0.200\).
A lower emissivity indicates that the surface emits less radiation, which can be advantageous when minimizing heat transfer, such as in insulating purposes or preventing a liquid from boiling away quickly. Accurately measuring or estimating emissivity is crucial for precise calculations in thermodynamics applications, particularly when leveraging the Stefan-Boltzmann Law.

heat of vaporization

Understanding the heat of vaporization is essential when dealing with phase changes in substances, especially within thermodynamics. The heat of vaporization is the amount of energy required to change 1 kilogram of a substance from a liquid to a gaseous state without changing its temperature. For helium, this value is \( 2.09 \times 10^4 \mathrm{J/kg} \).
This significant value implies that a considerable amount of energy is needed for even a small quantity of helium to evaporate. When calculating how much liquid helium evaporates due to heat transfer, we need to know both the energy lost from the helium and the heat of vaporization.
Using the formula \( Q = mL \), where:

• \( Q \) is the energy required for vaporization
• \( m \) is the mass of the liquid that has evaporated
• \( L \) is the heat of vaporization

This calculation is crucial in applications such as cryogenics, where minimizing the loss of liquid helium is of the utmost importance. It guides physicists in designing storage containers to control and reduce heat transfer efficiently.