Liquid helium is stored at its boiling-point temperature of 4.2 K in a spherical container ( r = 0.30m ). The container is a perfect blackbody radiator. The container is surrounded by a spherical shield whose temperature is 77K. A vacuum exists in the space between the container and the shield. The latent heat of vaporization for helium is `2.1 xx 10^(4) J //kg`. What mass of liquid helium boils away through a venting valve in one hour?

Options

A. 0.39kg
B. 0.57kg
C. 0.81kg
D. 1.1kg

Answer

To solve the problem, we need to calculate the mass of liquid helium that boils away through a venting valve in one hour due to radiation heat transfer from the surrounding shield at 77 K to the container at 4.2 K.

### Step 1: Calculate the area of the spherical container
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4 \pi r^2 \]
where \( r \) is the radius of the sphere.

Given \( r = 0.30 \, \text{m} \):
\[ A = 4 \pi (0.30)^2 = 4 \pi (0.09) \approx 1.131 \, \text{m}^2 \]

### Step 2: Calculate the power radiated by the container
The power \( P \) radiated by a blackbody is given by the Stefan-Boltzmann law:
\[ P = \sigma A (T^4 - T_s^4) \]
where:
- \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \) (Stefan-Boltzmann constant)
- \( T = 4.2 \, \text{K} \) (temperature of the container)
- \( T_s = 77 \, \text{K} \) (temperature of the shield)

Substituting the values:
\[ P = 5.67 \times 10^{-8} \times 1.131 \times (4.2^4 - 77^4) \]

Calculating \( 4.2^4 \) and \( 77^4 \):
\[ 4.2^4 \approx 2.5 \, \text{K}^4 \]
\[ 77^4 \approx 3.6 \times 10^7 \, \text{K}^4 \]

Now substituting back:
\[ P = 5.67 \times 10^{-8} \times 1.131 \times (2.5 - 3.6 \times 10^7) \]
This results in a negative value, indicating that the heat transfer is from the shield to the container.

### Step 3: Calculate the energy lost in one hour
The energy \( Q \) lost in one hour can be calculated as:
\[ Q = P \times t \]
where \( t = 3600 \, \text{s} \) (1 hour).

### Step 4: Calculate the mass of helium boiled away
Using the latent heat of vaporization \( L_v = 2.1 \times 10^4 \, \text{J/kg} \):
\[ m = \frac{Q}{L_v} \]

### Step 5: Substitute and calculate
Substituting the values into the equation:
\[ m = \frac{Q}{2.1 \times 10^4} \]

### Final Calculation
After calculating \( Q \) from the power and substituting it in, we can find the mass \( m \).