Assume that a 65-kg hiker needs to eat 4.0 x 10³ kcal of energ...

Question

Assume that a 65-kg hiker needs to eat 4.0 x 10³ kcal of energy to supply a day’s worth of metabolism ( = Q_H). Estimate the elevation change the person can climb in one day, using only this amount of energy. As a fun and rough prediction, treat the person as an isolated heat engine, operating between the internal temperature of 37°C (98.6°F) and the ambient air temperature of 20°C.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice from together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem, a mountaineer weighing around 75 kg consumes about 5.0 multiplied by 10 to the power of three kilocalories of food energy per day in order to maintain his metabolism, which is equal to Q subscript capital H Imagine he acts like an insulated heat engine that operates between his body temperature of approximately 37 °C and an ambient temperature of negative 5.0 °C. What would be his potential altitude gain in one day using only this particular amount of energy? So that's our goal is we're trying to figure out what the potential gain in one day using only this particular amount of energy. So what is the altitude gain? Awesome. So now that we know that we're ultimately trying to figure out what the potential altitude gain value is for this particular problem. Let's read off the multiple choice answers to see what our final answer might be. And let us note that they're all in the same units of meters. So A is 3.5 multiplied by 10 to the power of three, B is 3.9 multiplied by 10 to the power of three C is 4.1 multiplied by 10 to the power of three and D is 4.5 multiplied by 10 to the power of three. OK. So first off, let us write down all of our known variables. So we know the mass of the mountaineer, which we're gonna call this value M is equal to 75 kg. We know the energy available, which is Qh and Qh is equal to 5.0 multiplied by 10 to the power of three and its units are gonna be kilocalories. So 10 to the power of three and then killer calories. Awesome. And then the internal temperature which we're gonna call T subscript capital H. So th is equal to 37 °C, which I'm gonna go ahead and give you the conversion, but this is also equal to drum roll 310 Kelvin. So I've gone ahead and given you the conversion, but you can use the conversion factor on your own. All you have to do is just add 273 to 37 to get 310,000. And we know the ambient temperature which we're gonna call T subscript capital LTL which is equal to negative 5.0 °C, which is also equal to 268 Kelvin. So the reason why we're converting degrees Celsius to Kelvin is we need to have everything in the proper units in order to solve for our final answer in meters. Awesome. And we also know that the gravitational acceleration or the acceleration due to gravity is equal to 9.81 m per second. We're gonna call g the acceleration due to gravity. And we should now recall or we'll quickly look up. We need to note that the conversion factor for 1 kg calorie. So 1 kg calorie is also equal to 4186 jewels. And we're gonna need this value in order to solve for our final answer using the proper unit. So we'll hold on to that piece of information for a little bit. So moving right along, we now need to assume that the efficiency of the mountaineer's metabolism is that of a reversible engine. So therefore, with that in mind, we can recall and note that we can take the work from the engine. Engine quotation marks around engine Macy is technically not an engine. And we need to assume that it is all used to increase the mountaineer's potential energy by climbing higher. So with that in mind, we can write that lowercase E is equal to one minus TL divided by th where lowercase E is the energy which TL divided by th this is equal to W divided by QH, which is equal to M multiplied by G multiplied by H all divided by QH. And we need to note and recall that H the height is equal to E multiplied by Q H all divided by M multiplied by G. So I said E is the energy but specifically, I should technically this would be the efficiency E of a Carno engine. So focusing just on E right now and plugging in our known variables, we will find that E is equal to one minus TL divided by th which we know TL and using our Kelvin units as equal to 268 Calvin divided by 310,000. So when we plug that into our calculator, we will get when we round to four decimal places, 0.1355. So now we need to, this is where our conversion factor is gonna be really helpful. We need to take our qh value and convert it from kilocalories to jewels. So with that in mind, we will take 5.0 multiplied by 10 to the power of three kilo calories. And we will need to multiply it by 4186 tools in 1 kg calorie. Since we're using dimensional analysis to do this conversion, which as you can see our Koc calorie units cancel out leaving us with just tools, which is perfect. That's what we want our answer. To be in and we will get 2093 zero, 000. So it'll be 20,000, so 20,930,000 jewels. Ok. So now we can solve for H which is the height that the or the altitude that the climber will obtain. So all we have to do is take our H equation and plug in our known variable. So we know that the value of E we determine this to be 0.13 55. And then we need to multiply this by our qh value which we determined to be 20,930,000 tools. And don't forget your parenthesis, all divided by the mass of the climer, which is 75 kg multiplied by acceleration due to gravity which is 9.81 m per second squared. So let me plug that into our calculator and we round to one decimal place, writing our final answer in scientific notation, 3.9 multiplied by 10 to the power of 3 m. Awesome. And that's it. That's our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter B 3.9 multiplied by 10 to the power of 3 m. Thank you so much for watching. Hopefully, that helped and I can't wait to see in the next video. Bye.

Key Concepts

Heat Engine Efficiency

Work Done Against Gravity

Conversion of Energy Units

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