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?

Question

Kty · Answer

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. :)

noah · Answer

oo thx dude