Q: Suppose a hiker of mass (m=56.8kg) climbs a mountain of height (h=212.13m) wi

Question

Q: Suppose a hiker of mass (m=56.8kg) climbs a mountain of height (h=212.13m) with an efficiency of (N=.20) on a cool dy with an ambient temperature (T=7.2C=280K): a. How many food calories are required? b. How much thermal energry must the hiker shed? (thermal energy shed by perspiration/evaporation, radiation, and work) c. How much thermal energy will the hiker radiate, assuming an emissivity (e=.6886) if the time takes (t=3600s)? d. How much water must the hiker perspire to shed the remainder of the thermal energy that was not radiated?

Explanation / Answer

a) E = m g h/e = 56.8*9.81*212.13/.2=591003 J = 141 Calories

b) thermal = E - mgh =
56.8*9.81*212.13*(1/.2-1)=472802 J

c) didnt give surface area supposing A =1

E = e sigma A (T^4 - t0^4) = 0.6886*5.67E-8*1*(310.15^4 - 280^4)=121.3 J

d) Q = m L

(472802-121.3) = m*334E3

m=1.42 kg

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