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?

To determine the mass of liquid helium that boils away through the venting valve in one hour, we can use the Stefan-Boltzmann Law, which calculates the power radiated by a perfect blackbody radiator.

The Stefan-Boltzmann Law states that the power radiated per unit area by a blackbody is proportional to the fourth power of its absolute temperature. Mathematically, it can be expressed as:
P = σ * A * T^4

Where:
P is the power radiated (in Watts),
σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/(m^2K^4)),
A is the surface area of the blackbody radiator (in square meters),
T is the temperature (in Kelvin).

In this case, the container is a perfect blackbody radiator, so we can use the Stefan-Boltzmann Law to calculate the power radiated by the container, as well as the power absorbed by the shield surrounding the container.

The power absorbed by the shield can be expressed as:
P_absorbed = σ * A_shield * T_shield^4

Since the container is surrounded by a vacuum, the heat exchange occurs only through radiation between the container and the shield. Therefore, the power radiated by the container should be equal to the power absorbed by the shield. Mathematically, we can write:
P_container = P_absorbed

Now, let's calculate the power radiated by the container:
P_container = σ * A_container * T_container^4

To find the mass of liquid helium that boils away, we need to determine the energy required to vaporize a certain amount of helium. This energy is given by the latent heat of vaporization (L_v) multiplied by the mass (m) of the helium that boils away. Mathematically, we can write:
E = L_v * m

The energy E can be calculated by the power P_container, multiplied by the time t (expressed in seconds). Mathematically, we can write:
E = P_container * t

Setting the two expressions for E equal to each other, we can solve for the mass of helium that boils away:
L_v * m = P_container * t

Rearranging the equation to solve for m, we get:
m = (P_container * t) / L_v

Now, let's plug in the values given in the problem and calculate the mass of liquid helium that boils away.

Given:
T_container = 4.2 K
R_container = 0.30 m
T_shield = 77 K
L_v = 2.1 x 10^4 J/kg
t = 1 hour = 3600 seconds

First, we need to calculate the surface areas of the container and the shield. The surface area of a sphere can be calculated using the formula:
A = 4πR^2

Calculating the surface area of the container:
A_container = 4π * R_container^2

Calculating the surface area of the shield:
A_shield = 4π * R_shield^2

Next, we can calculate the power radiated by the container and the shield:
P_container = σ * A_container * T_container^4
P_absorbed = σ * A_shield * T_shield^4

Since the power radiated by the container is equal to the power absorbed by the shield, we have:
P_container = P_absorbed

Finally, we can use the formula to calculate the mass of liquid helium that boils away:
m = (P_container * t) / L_v