A cryogenic storage container holds liquid helium, which boils at 4.20 K. Suppose a student painted the outer shell of the container black, turning it into a pseudo-blackbody, and that the shell has an effective area of 0.615 m2 and is at 3.01·102 K.
a) Determine the rate of heat loss due to radiation.
(in W)
b) What is the rate at which the volume of the liquid helium in the container decreases as a result of boiling off? The latent heat of vaporization of liquid helium is 20.9 kJ/kg. The density of liquid helium is 0.125 kg/L.
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To determine the rate of heat loss due to radiation (Q), we can use the Stefan-Boltzmann law:
Q = ε * σ * A * (T_shell^4 - T_surroundings^4)
Where:
Q = rate of heat loss (in watts)
ε = emissivity of the shell (assumed to be 1 for a blackbody)
σ = Stefan-Boltzmann constant (5.67 × 10^-8 W/(m^2·K^4))
A = effective area of the shell (0.615 m^2)
T_shell = temperature of the shell (3.01×10^2 K)
T_surroundings = temperature of the surroundings (4.20 K)
Substituting the given values into the equation:
Q = 1 * (5.67 × 10^-8 W/(m^2·K^4)) * (0.615 m^2) * ((3.01×10^2 K)^4 - (4.20 K)^4)
Calculating this, we find:
Q ≈ 226.53 W
Therefore, the rate of heat loss due to radiation is approximately 226.53 Watts.
To determine the rate at which the volume of the liquid helium in the container decreases as a result of boiling off, we need to calculate the amount of heat required to vaporize the liquid helium. This can be done using the latent heat of vaporization.
The rate of volume decrease (V) can be calculated using the equation:
V = Q / L
Where:
V = rate of volume decrease (in liters per second)
Q = heat loss rate from part a (226.53 W)
L = latent heat of vaporization (20.9 kJ/kg)
Now, we need to calculate the mass of the liquid helium and then convert it to volume using its density:
Mass = Volume * Density
Mass = 1 liter * (0.125 kg/L)
Mass = 0.125 kg
Substituting the given values into the equation:
V = (226.53 W) / (20.9 × 10^3 J/kg)
Calculating this, we find:
V ≈ 0.010847 L/s
Therefore, the rate at which the volume of the liquid helium in the container decreases as a result of boiling off is approximately 0.010847 liters per second.