Bicycle riders in the Tour de France consume on average ~8000 kcal = 33.4 MJ of

Question

Bicycle riders in the Tour de France consume on average ~8000 kcal = 33.4 MJ of food per day! About 10.0 MJ per day is lost to air drag. (19 points total)
i) Part of the remaining energy goes to fighting gravity. The brutal stage 6 of the 2005 tour climbed from Fontenay-les-Briis at 50m to Carrefour at 150m. How much energy did Lance Armstrong expend in pushing himself (75kg) and his bike (8kg) up the hill?
iii) Now consider a flat stage from Paris to Pont de l’Alma. The tour is held in July, which is hot. Bicyclists drink water, which evaporates as sweat at (37°C). If all of the remaining energy goes to water evaporation, what volume of water does a bicyclist drink on the flat stage?
v) What is the total change in entropy for the evaporation of the water in part iii?

Explanation / Answer

i) it is (75+8)*9.8*100=81340(J)=0.08(MJ) ii) energy left. 33.4-10-0.08=23.32e6(J) assume that mass of m water be evaporated. so m*4200*63+m*2260e3=23.32e6 so m=9.23(kg) ii). it is E=ln(T'/T)*m*C+m*muy/T'=63100

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