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 0.487 m2 and is at 3.01·102 K.
a) Determine the rate of heat loss due to radiation.
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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a) To determine the rate of heat loss due to radiation, we can use the Stefan-Boltzmann Law, which states that the power radiated by a blackbody is proportional to the fourth power of its absolute temperature:
P = σ * A * ε * T^4
where:
P is the power radiated (heat loss) in watts,
σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W m^-2 K^-4),
A is the effective area of the blackbody in square meters,
ε is the emissivity of the blackbody (assumed to be 1 for a pseudo-blackbody),
T is the absolute temperature of the blackbody in Kelvin.
In this case, we have the following values:
A = 0.487 m^2
T = 3.01 x 10^2 K
Plugging these values into the equation, we can calculate the rate of heat loss due to radiation:
P = (5.67 x 10^-8 W m^-2 K^-4) * (0.487 m^2) * (1) * (3.01 x 10^2 K)^4
Simplifying the equation:
P = 0.487 * 5.67 x 10^-8 W * (3.01 x 10^2 K)^4
Calculating the power radiated:
P ≈ 2.768 W
Therefore, the rate of heat loss due to radiation is approximately 2.768 Watts.
b) To find 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 mass of the liquid helium being vaporized and then find the rate of change of volume.
Given:
Latent heat of vaporization of liquid helium = 20.9 kJ/kg
Density of liquid helium = 0.125 kg/L
Let's assume the volume of liquid helium vaporized per unit time is V. We can find the mass m of the vaporized helium using the density formula:
m = density * V = 0.125 kg/L * V
The energy required to vaporize this mass m of helium can be calculated using the specific latent heat of vaporization:
Energy = m * latent heat of vaporization = (0.125 kg/L * V) * 20.9 x 10^3 J/kg
According to the first law of thermodynamics, this energy loss is equal to the heat loss due to boiling:
Energy = heat loss = P * t
where P is the power radiated (which we calculated as 2.768 W) and t is the time in seconds.
Combining the equations:
(0.125 kg/L * V) * 20.9 x 10^3 J/kg = 2.768 W * t
Since the units of the left side of the equation are kg * J and the right side is W * s, we need to convert the units of volume from liters to cubic meters:
0.125 kg/L * V = 0.125 kg/L * (V / 1000) m^3 = 0.000125 kg/m^3 * V
Dividing both sides by 0.000125 kg/m^3:
20.9 x 10^3 J/kg * V = 2.768 W * t
Dividing both sides by 20.9 x 10^3 J/kg:
V = (2.768 W * t) / (20.9 x 10^3 J/kg)
Given that the density of liquid helium is 0.125 kg/L, we can convert V from cubic meters to liters:
V (in liters) = V (in cubic meters) * 1000
Substituting this conversion into the equation:
V (in liters) = [(2.768 W * t) / (20.9 x 10^3 J/kg)] * 1000
Therefore, the rate at which the volume of the liquid helium in the container decreases as a result of boiling off is [(2.768 W * t) / (20.9 x 10^3 J/kg)] * 1000 liters per second.