Liquid helium is stored at its boiling point temperature of 4.2K in a shperical container(r=0.30m).The container is a perfect blackbody radiator.The container is surrounede by a sherical 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 2.1X10000.What mass of liquid helium boils away thruogh a venting valve in one hour?
First you get to know the surface area of a sphere, that is A=4*pi*r^2
All the info listed down:
Emissivity,e=1 (blackbody has a good emissivity)
radius,r=0.30m
Temperature, T2=77K T1=4.2K
Latent heat of vaporization,L=2.1*10^4 J/kg
Time,t=1hour=3600sec
First, use the Steven-Boltzman Rule,
Power,P=Q/t=e*sigma*A*(T2^4-T1^4)
Thus, we get
P=1*(5.67x10^-8)*4*pi*(0.30)^2*(77^4-4.2^4)
P=2.254 J/s
Then as we found Power,and time=3600sec
Substitute into the formula, Q=Pt
Thus, we get
Q=2.254*3600 =8115.16Joule
Refer to the question the Latent heat is given, we calculate for the mass with the formula, Q=mL
m=Q/L =8115.16/(2.1*10^4) =0.386kg
hope i do get it right. :)
oo
thx dude
To calculate the mass of liquid helium that boils away through the venting valve in one hour, we need to find the rate of heat transfer from the liquid helium to the surroundings.
The rate of heat transfer (Q) is given by the equation:
Q = εσA(T^4 - T_surroundings^4)
Where:
ε = emissivity of the container (since it is a perfect blackbody radiator, ε is equal to 1)
σ = Stefan-Boltzmann constant (5.67 x 10^-8 W/(m^2K^4))
A = surface area of the container (4πr^2, as it is a spherical container)
T = temperature of the liquid helium (4.2 K)
T_surroundings = temperature of the shield (77 K)
Substituting the given values into the equation, we have:
Q = (1)(5.67 x 10^-8 W/(m^2K^4))(4π(0.30)^2)((4.2^4) - (77^4))
Now, let's calculate Q:
Q ≈ (1)(5.67 x 10^-8 W/(m^2K^4))(4π(0.09))(-20304320)
Q ≈ (1)(5.67 x 10^-8 W/(m^2K^4))(-14586.3376)
Q ≈ -0.0008278574 W
Next, we need to calculate the amount of energy required to vaporize the helium.
The latent heat of vaporization (L) for helium is given as 2.1 x 10^4 J/kg.
The mass of helium that boils away (m) can be calculated using the equation:
Q = mL
Rearranging the equation, we have:
m = Q / L
Substituting the given values, we have:
m ≈ (-0.0008278574 W) / (2.1 x 10^4 J/kg)
m ≈ -3.94265 x 10^-8 kg
Since mass cannot be negative, we take the absolute value:
m ≈ 3.94265 x 10^-8 kg
So, approximately 3.94265 x 10^-8 kg (or 39.4265 μg) of liquid helium boils away through the venting valve in one hour.
To determine the mass of liquid helium that boils away through a venting valve in one hour, we need to consider the heat transfer from the liquid helium to the surroundings.
The heat transfer can be calculated using the Stefan-Boltzmann law, which states that the power radiated by a blackbody is proportional to the fourth power of its temperature. The formula for radiant power is given by:
P = ε * σ * A * ΔT^4
Where:
P is the power radiated (in Watts)
ε is the emissivity of the blackbody (assumed to be 1 for a perfect blackbody)
σ is the Stefan-Boltzmann constant (5.67 × 10^-8 W/(m^2·K^4))
A is the surface area of the blackbody (in this case, the surface area of the container)
ΔT is the temperature difference between the blackbody and the surroundings (in Kelvin)
The power radiated by the blackbody is equal to the energy required to vaporize the liquid helium, which can be calculated using the latent heat of vaporization. The formula for energy is:
E = m * L
Where:
E is the energy (in Joules)
m is the mass of the liquid helium (in kilograms)
L is the latent heat of vaporization (in Joules per kilogram)
Since we want to find the mass m, we can rearrange the equation as:
m = E / L
Given that the boiling point temperature of liquid helium is 4.2 K and the temperature of the shield is 77 K, the temperature difference is:
ΔT = 77 K - 4.2 K = 72.8 K
To calculate the surface area of the spherical container, we use the formula:
A = 4πr^2
Where:
A is the surface area (in square meters)
r is the radius of the container (in meters)
Substituting the given values:
A = 4π(0.30 m)^2
Now, we can substitute all the known values into the equations to calculate the mass of liquid helium:
P = ε * σ * A * ΔT^4
Simplifying, we can determine the power radiated by the container.
Next, use the equation:
E = P * t
Where:
E is the energy (in Joules)
P is the power radiated (in Watts)
t is the time (in seconds)
Since we have the time in one hour, we need to convert it to seconds.
Finally, substitute the known values into the equation:
m = E / L
Solve for mass to find the mass of liquid helium that boils away through the venting valve in one hour.