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Read cumilirie fat move and aEsWer the foll: Whial Cuninnc? Cinc QuaCTIMn- jlmune 4ouD cannlnc Where carnitinc S}thesired? Ma sonie that mmint calnunntnn fltitinc weight loss supplewent WIM /3s childrew produces Mctubolic decompensation Whut Incabdlic dcccmnnsation Delicien camicn Wuham [Ic #AAO1S deramnensition MILA H au caloli decompensation lard"

Read cumilirie fat move and aEsWer the foll: Whial Cuninnc? Cinc QuaCTIMn- jlmune 4ouD cannlnc Where carnitinc S}thesired? Ma sonie that mmint calnunntnn fltitinc weight loss supplewent WIM /3s childrew produces Mctubolic decompensation Whut Incabdlic dcccmnnsation Delicien camicn Wuham [Ic #AAO1S deramnensition MILA H au caloli decompensation lard"



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$\bullet$ Burning fat by exercise. Each pound of fat contains 3500 food calories. When the body metabolizes food, 80$\%$ of this energy goes to heat. Suppose you decide to run without stopping, an activity that produces 1290 $\mathrm{W}$ of metabolic power for a typical person. (a) For how many hours must you run to burn up 1 lb of fat? Is this a realistic exercise plan? (b) If you followed your planned exercise program, how much heat would your body produce when you burn up a pound of fat (c) If you needed to get rid of all of this excess heat by evaporating water (i.e., sweating), how many liters would you need to evaporate? The heat of vaporization of water at body temperature is $2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}$

So the fats in our body are generally very large molecules. We're gonna go ahead and metabolize a typical fat. Which just means we're going to burn it. Okay? So we're going to take C57 H104 oh six and we're gonna react it with oxygen. Okay? And when we burned Things with hydrogen, carbon and oxygen of course we get co two and water. Okay? So to bounce us looks like we'll need a 57 here And a 52 here. Which means we're gonna need an 83 in front of our oxygen. Okay? So we have a balanced equation. Let's go ahead and turn this into a thermal chemical equation by adding the heat term which was given mm this much this many killing joules per one mole of our fat. So there's our thermo chemical equation. All right now let's see if we can figure out how many killer jewels of energy will be given off If I kind of burned five lbs of fat. All right. So, I'm gonna start with £5 and I'm gonna go ahead and change a pound to a kilogram. All right, so, there's 5454 kg and a pound. Then I'm gonna change kg two g. So, we've got 1000 g in a kilogram and then we'll go ahead and change grams to moles using the molar mass, which is quite large. 8 85.61. And then the last step is we'll just change moles to kill the jewels using our delta H. Here and to the fourth. So, this is going to be negative 77 460 killer jewels. Mhm. Or if we want to use proper sig figs, we'll go ahead and negative 7.75 Times 10 to the 4th killed yours. Mhm. Okay. So, then they've asked us to change this to nutritional calories. Well, a nutritional calorie. All right. It's just a killer calorie. Okay, it's 1000 calories or one kilocalories. So all we're really doing is changing killer jewels, two kilocalories. Okay, so we'll take our negative 77 for 60, 60 killer jewels and we'll just change killer jewels. Two kilocalories. Okay? And there's 4.184 killer jewels in one killer calorie. So that will give us -18 513 kilocalories. Okay, 4 -1.85 Times 10 to the 4th. Killer croc calories or nutritional calories. Just showing the proper sick pigs here and there's your answer.

So the balanced thermo chemical equation for the metabolism of fat in the body is as follows. C. 57 H. 104 oh six at 80 oh two Gives us 57 c. 02 at 52 H 20. Adults H. is negative 3.0-2 times 10 to the four kg jewels. So the molar mass of C. 57 81 oh four oh six is 884.57 g per mole. And so what we need to do is convert the pounds into g. So we have 226 7.95 g from £5. And so we can calculate the amount of heat evolved and we have this amount of grams metabolized by the body. So if you have 3.0 to 2 times 10 to the four kg joules of heat released per 884.57 g. This is equal to X kilo jewels, divided by the new mass 2267.95 g. Where we solve for X, which is equal to 7.748 times 10 to the four killed jules. And so what we can do is convert kilo jewels into calories, and so what we get is 1.852 times turned to the four calories.

Here. Well, we will determine how long you must run in order to burn off £1 of fat, which is assumed to be 3500 food calories. What many people don't realize is that 3500 is actually in kilo calories. It's a little bit confusing because that is usually written as a calorie with a capital C. But it really is 3500 times 1000 times more calories. And in order to convert that to jules, There are four points 184 jules per calorie. Little calorie. Okay, so the idea is that we know the power at which the energy is getting expanded and we will assume that coast right into the fat, which may not be perfectly true, but we have basically Q dot times delta T in seconds is equal to that for amount of calories. Because we have the power in watts. We do want the uh energy in jules because a lot is the same as actual per second. Yes. And that's a lot of jewels, 1.4, 6 Times 10 to the 7th jules. A jewel, by the way is a very small unit. Um so therefore we can solve for the time to burn that off, simply divide that by the rate in jewels per second. And we get of the order of 11,000 seconds which if we convert that two hours, we see that it's about three hours, a little bit over. So 3.15 hours. And the question is this a reasonable plan. Well, um, most people would have a hard time running at that metabolic rate for three hours if you're a trained runner doing a marathon. Yes, you can expect to be doing that. But usually people do not run to burn off fat. They're doing it for other reasons. So that does not seem very reasonable. uh the next thing that's really interesting is that 80 of the heat that is generated over that three hours Um sorry, 80% of the energy that comes out during that three hours is in the form of heat. So the question is, where does that heat go? And of course this is why you sweat the heat generated. Uh huh. Goes to evaporation. And so we want to pretend that that sweat is water and how much evaporates During the three hour run. Okay, so we're going to take the metabolic energy. And the rule of thumb is that 80% of that goes into heat. And we're running for 11,400 seconds. Yeah. And that will give us the energy and jewels uh Using Q equals m latent heat of vaporization during a phase change. We are going to assume that the temperature of the water is of course, raise to the evaporation point. So we won't use any of it um to raise it up. But we'll just go ahead and use the latent heat of vaporization. That's a little bit over simplification. We could make this more complicated by, first of all raising the temperature From body temperature 200°C. Um Oh, we will use as the latent heat of vaporization for water. Yeah, let's see. That is a well known quantity And it is two for two Times 10 to the six jules per kilogram, roughly. Okay, so using that we can determine the mass that evaporates again. This is a little bit simplistic but will give us a good estimate. Um and that is roughly 4.8 We'll say 4.78 kilograms. Okay, And we can do some things with that to make it feel a little bit better or understand it a little bit better. 4.78 kg has a weight Roughly 2.2 lb per kilogram, So that's a little bit over 10 lbs. Almost 11 lbs. Of course you're going to replenish that by drinking water as you're running. But most athletes will tell you that the weight loss that they experienced during a heavy athletic activity, basketball game, football game, etcetera, that most of that weight loss is due to the fluids rather than the fat that goes away. Um We can convert this into leaders by using the density of water. Um So what is the volume? Um Mass per unit volume? Uh huh. Okay. Is equal to the density which is 1000 kilograms per cubic meters. And so we can solve for the volume is mass Divided by 1000 than cubic meters. Mhm. Okay, so this is roughly um mhm. Mhm is 4.78 Times 10 to the -3 m3. And then we can convert that into leaders by multiplying By one leader is 10 to the -3 cubic meters. So that is almost five L which again seems a little bit mhm. Yeah, overkill. And again most of it gets replenished as the person is running. But yes, it does point out that most of the mass that you lose in doing these activities is the fluids in your body.

Things, question asks. How might an overweight person who does not exercise increase his or her Vo two Peak and therefore his or her lactate threshold? So we know that the video to peak is going to be related to now We're lactate threshold because the higher the video to peak percentage, the higher the lactate threshold, meaning the higher the rate that weaken burn fat. So in order to burn fat at a faster rate for an overweight person, we would want to see their VO two peak increase in the way to do this is via exercise. So my answer here would be exercising will increase the video to peak and thus will increase the lactate threshold. And this makes sense because if we start exercising, we are basically going to increase our view to peak in that we have now a next ra need for oxygen consumption, and thus our consumption of oxygen will go up when we exercise. So this makes sense that since we're exercising now, we will have an increased go to peak and thus will have an increased lactic threshold, and thus, finally, we will have an increased rate of burning fat


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