Pretty much on my 40th birthday, I could, all of a sudden, gain weight just by eating. \u00a0I’d always been a bone rack; in my 30s I could gain weight by exercise, but my body wouldn’t put on fat. \u00a0My, oh my: \u00a0what a different world I live in now, at age 4_.<\/p>\n
Over the holidays, like so many people, I put on about 10 lbs. \u00a0As of this morning, five of those 10 lbs were already gone. \u00a0Then, I went for a 45 min run (which, by the way, turned out to be much cooler and drier than I’d expected).<\/p>\n
Here’s the topical question: \u00a0how many calories was that run worth?<\/p>\n
There are many ways to calculate the answer to that question. \u00a0Today we’ll do it chemically.<\/p>\n
To start with, according to the bbc<\/a>, inhaled air has 0.04% CO2<\/sub>, while exhaled air has 4% CO2<\/sub>. We’ll take that to mean you breath in virtually no CO2<\/sub>, while a breath out is 4% CO2<\/sub>. We are going to track the consumption of calories via the production of CO2<\/sub>.<\/p>\n I’d estimate, during that run, I exhaled a litre of air every 2 seconds. Therefore, I exhaled 4% of that, or 0.04L, of CO2<\/sub> every 2 seconds. In one minute, that’s 30(0.04)=1.2L; in 45 min, it’s 45(1.2L)=54L.<\/p>\n At STP (standard temperature and pressure: 0\u00b0C, 1atm pressure), there is 22.4L of gas in a mole (for what a mole is, see my post here<\/a>). The conditions of my run were virtually STP (it’s January). 54L of CO2<\/sub> is 54\/22.4 = 2.4 moles CO2<\/sub>. The molar mass (see my post here<\/a> for more about the molar mass) of CO2<\/sub> is 12 +16+16=44g. Therefore, on the run, I produced 2.4x44g=105.6g CO2<\/sub>.<\/p>\n The chemical equation for burning glucose is<\/p>\n C6<\/sub>H12<\/sub>O6<\/sub> + 6O2<\/sub> \u2192 6CO2<\/sub> + 6H2<\/sub>O<\/p>\n Translating to molar masses, we could write the equation as<\/p>\n 180g glucose + 192g oxygen \u2192 264g carbon dioxide + 108g water<\/p>\n The equation indicates a consumption ratio of 180g glucose for every 264g CO2<\/sub> produced. Since I produced 105.6g CO2<\/sub>, my glucose consumption would have been 105.6(180\/264)=72g. Apparently, I consumed 72g glucose on that run.<\/p>\n Calorie King<\/a> tells me that glucose is 4 calories per gram. Therefore, the 72g glucose consumed on that run is worth 72(4)=288 calories. <\/p>\n Apparently, I burned 288 calories during my 45 minute run. We’ll look into how else one could calculate the calorie burn, how similar those findings are, and much more about this topic in future posts:)<\/p>\n Source:<\/p>\n Mortimer, Charles E. Chemistry<\/em>, 6th Ed. Belmont: Wadsworth, 1986.<\/p>\n