Question

A healthy human body maintains a central core of about 37.0 C (98.6 F) To do so, the body must shed excess thermal energy that is comverted from chemical energy due to metabolic processes. This is especially challenging during vigorous exercise, when the body's care temperature can increase by as much as 6 Cin only 15 min, which can lead to convulsions or even brain damage Energy transfer is the key to maintaining a consistent temperature. We can analyze how the human body uses heating mechanisms of energy transfer by applying the first law of thermodynamics (Wen S+Q to 8AU) to the body (the system) during a short time interval At Divide the quantities in the first law by the time intenal during which the changes occurred WE AU At At At The second term in the above equation is the heating energy transfer rate (Q to 8/At) fom the environment to the system. We can separate Q into four separate terms, which represent the rates of conduction, convection, radiation, and evaporation (represented by H symbols with subscripts) Ws At AU AU +(Htion +He +Hito +Hgration) At At We also substituted the two types of intenal energy change that can occur in a person's body the rate of thermal energy change and the rate of chemical energy change. This latter term is the metabolic rate, the rate of energy released by the chemical reactions occuming in the body We apply the above equation to a real process: a person running at moderate speed on an indoor air-conditioned track The table below shows quantities that are representative of a real process Work and heating rates (watts-J/ Type of energy change 100 Work done by drag force exerted by air Conduction -300 Thermal energy transfer due to convection Transfer of energy by radiative cooling 325 Cooling due to evaporation of sweat

Answer 1

A) from given er From giren egn e t hot + obiecte les A the end + (Whereken Hemerint Honorint Mercifusion = + die for constant temperature Alle =0 hence 44 40, A -100 +0-300-75 – 325 = 0 + By S cy Il ya 8004 was time to burn Zookeel ť = 2004.103 = 2505 = 25.0 = 417 min ltz 4.2 min very Energy lost in evaporation E = 4*36004 325 energy requise to evaporate wates : ML = E & L=Latent heat m= &- 43400 x 32,5 l vapoarizinhon ma 2kg and we know that density of water = 103 kg/43 m=ve 2 = Vx103 7 V = 2X103 = 2LAm