00:01
In this particular question, we are assuming we have a very large tank and we have a black body object that is submerged inside of a pool of tank of water.
00:21
We want to find how much water is being evaporated due to only the black body radiation.
00:33
That is actually in emitting from this body.
00:38
Now we're going to ignore everything that is actually transferred by conduction or convection.
00:46
So only radiation transfer of energy.
00:51
So kind of like they are not really touching, right, except that they are, but we ignore the conduction part.
01:04
So what is the amount of energy being transferred per second? we can use the stefan bozeman's law for the power radiated.
01:19
There is the power that's the power is given by the stiffel bozeman's constant, multiplied by the surface area, what about by temperature to power 4.
01:33
Now the thing is, the water itself gains this amount of energy from the sphere but it also loses a certain amount of energy based on its own temperature right so this is the water's area surface area and water's temperature now we're going to assume that it only radiates its power in this area right it radiates back to the sphere so they will have the same surface area, a, although it doesn't really make sense, right, but we don't know what is this surface area of this.
02:24
Now we're going to assume maybe it's perfectly insulated, but this is not given.
02:31
So we cannot use that surface area, we only can use the surface area of the sphere.
02:37
Right, and of course the temperature of the water.
02:45
So now we can equate this, right, this would be the power of the water, right, so the energy being absorbed by the water must be equals to the rates of change of our mass, multiplied by the latent heat of vaporization.
03:24
Right since this is what causes the water to vaporize alright and if we multiply the mass change in mass times the vaporization this will give us the energy right total energy absorbed and over time will give us power so actually we can equate these two together.
04:08
So we have sigma a times temperature of the sphere to power minus temperature of the water.
04:20
We are assuming that the water is at boiling point already, which is basically 100 degrees celsius.
04:45
So we can rearrange this equation to find the rate of change of our mass, or water just have to divide the latent heat of vaporization over and we can replace the surface area a for pi r square where r would be half of the diameter that's given so that's 0 .12 meters sigma the constant 5 .6 7 temperature of the sphere is 225 degrees.
05:46
Just 498 kelvin's.
05:47
We take 273 plus 225...