with an elevation gain of 1? Calculate the mass of glucose metabolized by a 80.1 person in climbing a mountain with an elevation gain of 1310 . Assume that the work performed in the climb is four times that required to simply lift 80.1 by 1310 .
The metabolism of glucose, C6H12O6, yields CO2(g) and H20(l) as products. Energy released in the process is converted to useful work, with about 68.0% efficiency. deltaHf of glucose is -1273.3.
I can get as far as calculating the work = 4116 kj and the std enthalpy of formation for the balanced equation C6H12O6(s) + 3 O2(g) ----> 6 CO2(g) + 6 H2O(l) as -2802.5 kj
I am stuck on how to set up the unit conversions from here. I think the heat metabolism of one mole glucose (-2802.5 kj/mol) is needed to convert the 68% some how ...does this get me to moles of glucose? if so, how do I set it up? I'm pretty sure I use the moles of glucose and the molar mass of glucose (180.18 g/mol) to get to grams of glucose...but again how? Please do not leave out any steps so I can see where I keep making my errors. Thanks for your time
The metabolism of glucose, C6H12O6, yields CO2(g) and H20(l) as products. Energy released in the process is converted to useful work, with about 68.0% efficiency. deltaHf of glucose is -1273.3.
I can get as far as calculating the work = 4116 kj and the std enthalpy of formation for the balanced equation C6H12O6(s) + 3 O2(g) ----> 6 CO2(g) + 6 H2O(l) as -2802.5 kj
I am stuck on how to set up the unit conversions from here. I think the heat metabolism of one mole glucose (-2802.5 kj/mol) is needed to convert the 68% some how ...does this get me to moles of glucose? if so, how do I set it up? I'm pretty sure I use the moles of glucose and the molar mass of glucose (180.18 g/mol) to get to grams of glucose...but again how? Please do not leave out any steps so I can see where I keep making my errors. Thanks for your time