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$\cdot$ The emissivity of tungsten is $0.35 .$ A tungsten sphere with a radius of 1.50 $\mathrm{cm}$ is suspended within a large evacuated enclo- sure whose walls are at 290 $\mathrm{K}$ . What power input is required to maintain the sphere at a temperature of 3000 $\mathrm{K}$ if heat conduction along the supports is negligible?

$4.54 \times 10^{3} \mathrm{W}$

Physics 101 Mechanics

Chapter 14

Temperature and Heat

Thermal Properties of Matter

The First Law of Thermodynamics

Simon Fraser University

University of Sheffield

University of Winnipeg

Lectures

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Hey, everyone, this is question number 62 from chapter 14. In this problem, we're given Tungsten spear were given an immersive it. Eah radius. And then we're told that the wall temperatures 290 degrees K and we have desired temperature of 3,000 k way. Well, then asked to find the power input to maintain the desired temperature of 3,000 k. So with the immensity and radius which we could turn into surface temp surface area because we know Abel's for pi R. Okay, we can start with our equation our stuff in Bolton equation e sigma a t to the fourth. Okay, so we need to figure out how much input is required to maintain. So in order to do that, we need to find the radiation difference between this 3,000 K and the 290 k, so we can D'oh Yeah, eh, Well, it e excuse my company Sigma e Sigma A TV The fourth And this is that the 3,000 minus e Sigma surface area to you forth with lower 290 k. So you should recognize that we can pull out e sigma and surface area because it's not changing. So then we have the desired out input the necessary input equal Tio a e sigma times t to the fourth. I'm gonna write high for high temperature minus t to the fourth L for low temperature. Okay, so now we can go ahead and plug in her numbers. So we have I'm gonna go to the next page and do this just to make sure have enough room. So h equals, eh? You'll see why I wrote out front Sigma t hae to the fourth minus t load the fourth. Okay, so h equals. We said the a is for pie are so four pi r radius 1.5 centimeters. So we need Thio divide and convert two meters which gives us your 20.15 meters squared meter it meters and NASCAR exits are squared And then a missive it ease your 0.35 and sigma is our constant. 5.67 times 10 to the minus eighth watts per meter squared times came in fourth and then I'm going to continue down here And then we have multiplied by or tea high, which is 3,000 okay to the fourth minus 290 k to the fourth. And if you plug that into your calculator, you get required input of for 0.54 times 10 to the fourth watts.

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