Question

Jake weighs 70 kg and has a core body temperature of 37 ° C when he begins steady state exercise at a VO2 of 2.5 L O2/min (12.5 kcal/min). His gross mechanical efficiency is 15%, and his sweat rate is 10 mL/min. Assume that all sweat produced evaporates and that 10% of heat produced is lost through radiative, conductive, and convective heat losses combined. After 45 minutes of exercising at this VO2, how much will his body temperature increase? (5 pts) **Remember that specific heat of human body = 0.83 kcal/kg for every 1° C increase in temperature.

Answer 1

Criven data mass (on) - foreg, of Core I body = 340 (+3+20= 3106) temporar ure Gross mechanical Efficiency = 15% rauc = leat lom2/min Orygen Enfance rare = 2.5402 /min. = 12.5 kcal/min) Time for which = 45 minutes - alick was Gnertising specifica Heak of Human Body = 0.831 (alleg Sulutions to fal onygen faccen = 12.5 18 cal/min x Eneritie a time = 12.5k (allonin xusmin - 562.5kcal There was Heat produced & roy. Energy

was lost Init Energy produced by body via respirat in (Shrough in Haling O2) = 562.5-(562.5x10] 100 -506-25 kcal * Gross mechanical efficiency of fur body is defined as perten fuge ratio of work done. I produce Here GME = 15% (Thus only 15%. is uti- O lized for Enercie) ... Energy utlized for Grercise Q = 506 25 XIS 100 25.9375 kcal . .. by fruatim of specific Hear a mear . 75.9375 i full = 0.85 kd X70kg x (T2-37) J.: T-37=1.31 iza 38.312 Thus body temperature by 1.916 of final would increase temp is 38.3106