Liquid helium, with a boiling point of 4.2 K, is used in ultralow-temperature experiments and also for cooling the superconducting magnets used in MRI imaging in medicine. Storing liquid helium so far below room temperature is a challenge because even a small 'heat leak' will boil the helium away. A standard helium dewar, shown in FIGURE P19.67, has an inner stainless-steel cylinder filled with liquid helium surrounded by an outer cylindrical shell filled with liquid nitrogen at –196°C. The space between is a vacuum. The small structural supports have very low thermal conductivity, so you can assume that radiation is the only heat transfer between the helium and its surroundings. Suppose the helium cylinder is 16 cm in diameter and 30 cm tall and that all walls have an emissivity of 0.25. The density of liquid helium is 125 kg/m^3 and its heat of vaporization is 2.1×10^4 J/kg. a. What is the mass of helium in the filled cylinder?

Video transcript

Hey, everyone. So this problem is dealing with material properties. Let's see what it's asking us and all of the information that it's giving us in this problem. So we're dealing with liquid helium that's super cooled to negative 269 degrees Celsius. We have a double walled stainless steel vessel. It's equipped with a pressure relief valve. The container has a cross sectional area that's given to us as 0.15 m squared, a length of 0.25 m. The affected emi emissivity is given to us as well as 0.2. And then finally, they tell us that the liquid helium has a density of 125 kg per meter cubed and a heat of vaporization of 2.1 times 10 to the four tools per kilogram. And we're asked to determine the mass of the liquid helium inside the cylinder. It's completely failed. Our multiple choice answers here are a 0. kg. B 0.47 kg C 1.25 kg or D 2. kg. So this is a great example of a problem that is testing your basic understanding of physics. So they're asking they're asking us to determine the mass and they have given us density. And so this is a very simple mass equals density multiplied by a volume problem, right? This is they are giving all of this other information to throw you off your tracks, make you double guess um you know, the basics, the things that, that we know. So as long as we can recall that mass is equal to density multiplied by volume, we can ignore everything else in this problem. So our volume of the cylinder, that's the only piece that they didn't outright give us in. This problem is going to be the cross sectional area multiplied by the height or the length of the cylinder. So that's gonna be 0.15 m squared multiplied by 0.25 m. So that comes out to 3.75 times 10 to the negative three m cubed. And from there, we're just gonna, we are just going to multiply that by the density that was given to us in the problem 125 kg per meter cubed 3.75 times 10 to the negative three m cubed in our mass is 0. kg that aligns with answer choice B. So that's the correct answer for this problem. So that's all we have for this one. We'll see you in the next video