Question

Part A You have probably seen people jogging in extremely hot weather and wondered Why" As we shall see, there are good reasons nat to do thisl When jogging strenuously, an average runner of mass 69.0 kg ?nd surface aea 1 84 m2 produces energy at a How much heat per second is produced just by the act ot jogging? rate of up to 1310 W 800 % of which is converted to heat The jogger radiates nest but actually absorbs more from the hot air than he radiales away. At such high levels of activity, the skin's temperature can be elevated to around 33 0 C instead of the usual 300We shall neglect conduction which would bring cven more hoat into his body. The only way ter the body to get rid of this extra heat is by evaporating water (sweating) ine Submit Roquest Answer Part B How much net hest per second does the runner gain just from radiation if the air temperature is 4002C(104?F)? (Remember that he radiates out, but the environment radiates back in Submit Request Answer Part C
Thank you so much!

Answer 1

Given that How much head per second is produced Just by the act of Jogging m = 68.0 kg Area = 1.86 m^2: p = 13000: 80 % conversion T = 33 degree c: Pjog = 80 % of p = 0.8 times 1330 = 10464w = 1064 J/s: t_f = 40 degree c implies (273 + 40)k = 313k: t_f = 33 degree c implies (273 + 33)k implies 306k: Delta_p = 6A (tau_f^4 - tau_i^4) = 5.67 times 10^- 8 times 108m^2 times (9.59 times 10^9 - 8.76 times 10^9) = 10.55 times 10^- 8 times 8.3 times 10^8 Delta_p = 87.56 watts The rate of energy gain from radiation is Q/t = (tau^* - tau^t_0) Q/t = (5.67 times 10^- 8) (1.86 m^2) ((273 + 40k)^4 - (273 + 33k)^4) \r\n Q/t = 87.56 W Q/t = 87.6 W The total amount of excess heat this runner's body must gain rid of per second is, Q_rid/t = [Q/t]_radiation + [Q/t]_heat Q_rid/t = 87.6N