🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Numerade Educator ### Problem 98 Hard Difficulty # Animals in cold climates often depend on two layers of insulation: a layer of body fat (of thermal conductivity 0.20 W/m$\cdot$K) surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere 1.5 m in diameter having a layer of fat 4.0 cm thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at 2.7$^\circ$C during hibernation, while the inner surface of the fat layer is at 31.0$^\circ$C. (a) What is the temperature at the fat-inner fur boundary so that the bear loses heat at a rate of 50.0 W? (b) How thick should the air layer (contained within the fur) be? ### Answer ## a.$29.6^{\circ} \mathrm{C}$b.$9.1 \mathrm{cm}\$

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Temperature and Heat

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##### Andy C.

University of Michigan - Ann Arbor

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University of Winnipeg

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### Video Transcript

okay. Today's question states that animals in cold climates often depend upon two layers of installation. A layer of body fat with a thermal conductivity of 0.2 watt per meter times Kelvin. I denote that layer thermal body fat as case of F here, and it's surrounded by a layer of air trapped in sired of fur or down we can model. A black bear was as a sphere of 1.5 meters in diameter, having a layer of fat of four centimetres thick. So I draw a model of the black bear here where the red line labeled as our is our radius, which I go ahead and calculate from the diameter that they give by dividing it by two, which gives us a value of 0.75 meters F. Here is the thickness of the fact layer, which is four centimeters or converted two meters is four times 10 to the minus two meters, since one meter is equal to 100 centimeters. And then I do know the thickness of the hare by H okay, and then it then goes on to say, in studies of bear hibernation, it was found that the outer surface layer of the fur is 2.7 degrees Celsius during hibernation, and the inner surface of the fat is 31 degrees Celsius. So I labelled the inner surface temperature of the fat as T's F F and the outer surface layer of the hair, or for as T's of H. Then the question is, what is the temperature of the fat inter for boundary so that the bear loses heat at a rate of 50 watts. So I label the fat inter for hair boundary temperature as T's of I. And then we want to know what that temperature is for part A. If the they're loses the heat at a rate of 50 watts, which I know as Capital H and then for Part B, it wants to know how thick the layer of thie for you should be. In order for this, he lost to be true with the temperature that we're going to find from part, eh? So let's go ahead, start on part A. In order to do party, we're going to make use of the heat current formula, which says that the rate of heat loss, which is capital H is equal to the the thermal conductivity constant for us. That's case of F, in this case, times the area that's going to be the area of the bear, which we're going to calculate shortly. Divided by the thickness of the medium were desire ng or the medium that were calling into account, which the thickness here is the thickness of the fat multiplied by the difference between the temperature of the fat minus temperature of that inner inner layer between the fat and the fur. So we're going to go ahead and solve for tea supply, which is what we're wanting to find anyway, since we know all the other boundaries and that is going to give us an equation, if we solve for Tisa by of tea Sabbeth minus H capital h, times f and then we're going to divide that by case of F time, sea area. Okay, since we have everything in this equation calculated except for the area, let's go ahead and calculate that. So the equation for the area is the equation for the surface area of a sphere, which is four pi r squared, which, if you recall, are a 0.75 meters if we plugged that in 0.75 meters squared times four. Pi gives us 7.1 square meters. Okay, so now that we have everything we need to calculate T's of I, let's go ahead and plug those in T's of F. If you recall, this would be a minus. Sign here. T's of F was 31 degrees Celsius. We're going to subtract that 31. They were goingto subtract from that 31 degrees Celsius. It's 50 wants and that's gonna be multiplied by F, which we found to be four times 10 to the minus two meters and then all that is going to be divided by case of F, which was 0.2 watts per meter squared. I'm sorry. Watch per meter times, Kelvin. And then that's gonna be multiplied by the area which we found to be 7.1 7.1 square meters. Okay, if you plug all that into your calculator, you will find that this comes out to be 29.6 degrees Celsius. So we found that the temperature for the fat enter for boundary is going to be 29.6 degrees Celsius if the rate of he lost his 50 watts. Okay, Now, for part B, If you recall, it wants to. Now it wants us to now find the thickness of the hair layer that needs to be in order for this all to be true. Okay, well, again, since the heat loss is going to be the same for the bear, no matter what, we can use the same equation. H is going to be the same again. Except now our difference is gonna be the difference between the fat enter for boundary temperature and the temperature of the fur. And then that's gonna be multiplied by the A constant now of case of a for the yeah, conductivity of air or which is the same as the fur multiplied by the same area divided by H. We can go ahead and h now being the thinking of the first. So we want to find the thickness of the fur. So we're going toe to rearrange the equation. So now we know that h we just simply swap big age for little. H h is going to be a little H is going to be equal to a k a times that area divided by, um, Big H. And then again, we'll deployed by the difference between these two temperatures. Tisa by minus T said H. Okay, so let's go ahead and plug everything in. Okay, So case of a was 0.2 for air, and that is watts per meter. Killed in okay area from the previous page was 7.1 meters squared. All that's gonna be divided by H, which was, uh, something something good five, which was 50 once and then multiplied by the difference from what we found on the previous page to be 29.6 degrees Celsius. Come minus the given temperature of the hair, which was 2.7 degree Celsius. And once again playing all of that into a calculator, we get 9.1 centimeters. Okay, So guys in box that in because that is our answer for part B.

University of Kansas

#### Topics

Temperature and Heat

##### Andy C.

University of Michigan - Ann Arbor

LB
##### Jared E.

University of Winnipeg

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