Solution of question Chapter 5
Q5.5 a(a) What is the energy required to walk 20 km at 5 km/hr ? From the table , the energy rate of walking activity at 5km/hr is 3.8 kcal /min . The energy required to walk 20 km =3.8 kcal/min x 20km x 60 min/hr 5km/hr Energy = 912 kcal
Q5.5 b
(b) Assuming 5 kcal /g of food ,calculate the grams of food needed for walk . The amount of food needed for walk =[ Energy ] [ Energy /gm ] = 912 kcal 5 kcal/gm = 182 gramQ 5.7Suppose that the elevator is broken in the building in which you work and you have to climb 9 stories – a height of 45 m above ground level .How many extra calories will this external work cost you if your mass is 70 kg and your body at 15% efficiency ?
5.9 A 70 Kg hiker climbed a mountain 1000 m high . He reached the peak in 3 hr . a . calculate the external work done by the climber . External work = m g h = 70 x 9.8 x 10і = 6.9 x 10 J
b. Assuming the work was done at a steady rate during the 3 hr period ,calculate the power generated during climb. Power = Work Time Power = 6.9 x 10 J = 64 watts 3x 3600 sec
c. Assuming the average O2 consumption during the climb was 2 liter /min (corresponding to 9.6 Kcal /min ), find the efficiency of the hiker’s body . Energy consumed =( 9.6 Kcal /min )(180 min )(4.2 x 10іJ/kcal) 6 = 7.3 x 10 J Efficiency = work done Energy consumed € = 6.9 x 10 6 7.3 x 10 € = 0.094 € = 9.4 %
d . How much energy appeared as heat in the body ? Δ U =Δ Q + Δ W 6 7.3 x 10 J = Δ Q + 6.9 x 10 6 Δ Q = 6.6x 10 J
Hr = Kr Ar e (Ts –Tw) Since Kr = 5Kcal/ mІ. hr .°C e=1 Hr = 5x 0.9 (32 -30) =9 Kcal/hr
b . Hc = Kc Ac (Ts – T α) Kc =10.45 – v + 10 √ v Kc = 10.45 - 4 + 10 √4 Kc = 6.45 +20 Kc = 26.5 Kcal /hr.mІ .°C Hc = 26.5 x 0.9 (32 – 30 ) Hc = 48 Kcal /hr
c. Heat lost = Heat gain Heat lost by radiation +evaporation +convection +respiration Heat lost = 9 +evaporation +48 +10 Heat lost =67 + evaporation Heat gain = 80 +30 = 110 110 =67 + evaporation Heat lost by evaporation =110 -67 Heat lost by evaporation = 43 Kcal /hr