Please urgent help me
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.
a) To determine the rate of heat loss due to radiation, we can use the Stefan-Boltzmann law, which states that the power radiated per unit area by a blackbody is proportional to the fourth power of its temperature. The formula is given by:
P = σ * A * (T^4 - T₀^4),
where P is the power radiated, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the effective area of the shell, T is the temperature of the shell, and T₀ is the temperature of the surroundings.
Substituting the given values, we have:
P = (5.67 x 10^-8 W/m^2K^4) * (0.487 m^2) * ((3.01 x 10^2 K)^4 - (4.20 K)^4).
P ≈ 271.56 W.
Therefore, the rate of heat loss due to radiation is approximately 271.56 W.
b) To determine the rate at which the volume of the liquid helium decreases as a result of boiling off, we need to calculate the heat energy required to vaporize the liquid helium and then divide it by the latent heat of vaporization.
The heat energy required to vaporize the liquid helium is given by the formula:
Q = m * L,
where Q is the heat energy, m is the mass of the liquid helium, and L is the latent heat of vaporization.
The mass of the liquid helium can be calculated using the density and volume of the liquid:
m = ρ * V,
where ρ is the density of liquid helium and V is the volume.
Substituting the given values, we have:
m = (0.125 kg/L) * (1 L) = 0.125 kg.
Now, substituting the calculated values into the Q formula, we have:
Q = (0.125 kg) * (20.9 x 10^3 J/kg).
Q ≈ 2,612.5 J.
Therefore, the heat energy required to vaporize the liquid helium is approximately 2,612.5 J.
Since the rate at which the volume decreases is the ratio of the heat energy to the latent heat of vaporization, we can calculate it as follows:
Rate = Q / L,
where Rate is the rate at which volume decreases.
Substituting the calculated values, we have:
Rate = (2,612.5 J) / (20.9 x 10^3 J/kg).
Rate ≈ 0.125 L/s.
Therefore, the rate at which the volume of the liquid helium decreases as a result of boiling off is approximately 0.125 L/s.
To determine the rate of heat loss due to radiation, you can use the Stefan-Boltzmann law, which states that the rate of heat loss (P) from a blackbody is proportional to the effective area (A) and the fourth power of its temperature (T). The formula for heat loss due to radiation is given by:
P = ε * σ * A * (T^4 - T0^4)
Where:
P = rate of heat loss (in watts)
ε = emissivity of the shell (assuming it's 1 for a blackbody)
σ = Stefan-Boltzmann constant (5.67 × 10^(-8) W/(m^2·K^4))
A = effective area of the shell (0.487 m^2)
T = temperature of the shell (3.01 × 10^2 K)
T0 = temperature of the surroundings (assumed to be 0 K)
Let's plug in the values and calculate the rate of heat loss (P):
P = 1 * 5.67 × 10^(-8) * 0.487 * (3.01 × 10^2)^4
Now, let's solve this equation to find the value of P.
P ≈ 3.58 × 10^4 watts
Therefore, the rate of heat loss due to radiation is approximately 3.58 × 10^4 watts.
Now, let's move on to part b of your question.
The rate at which the volume of the liquid helium decreases as a result of boiling off can be calculated using the conversion factor between volume and mass, along with the given latent heat of vaporization.
The rate of decrease in volume (V) can be given by the equation:
V = m * (L / ρ)
Where:
V = rate of decrease in volume (in liters per second)
m = rate of decrease in mass (in kg per second)
L = latent heat of vaporization of liquid helium (20.9 kJ/kg = 20.9 × 10^3 J/kg)
ρ = density of liquid helium (0.125 kg/L)
First, we need to calculate the rate of decrease in mass (m). To do this, we need to know the rate of heat loss due to boiling (P) in watts, which can be found using the following equation:
P = m * L
Rearranging the equation, we get:
m = P / L
Now we can calculate the rate of decrease in mass (m):
m = (3.58 × 10^4 W) / (20.9 × 10^3 J/kg)
m ≈ 1.71 kg/s
Finally, we can calculate the rate of decrease in volume (V):
V = (1.71 kg/s) * (20.9 × 10^3 J/kg) / (0.125 kg/L)
V ≈ 2.9 × 10^4 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 2.9 × 10^4 liters per second.