A 70-kg student produces 1200 W of heat energy per second. For the runner to maintain a constant body temperature of&nbsp;37∘C, this energy needs to be released through mechanisms like &nbsp;perspiration or other mechanisms. If these mechanisms failed and the energy could not flow out of the student’s body, for what amount of time could a student run before irreversible body damage occurred? (Protein structures in the body are irreversibly damaged if body temperature rises to&nbsp;44∘C&nbsp;or higher. The specific heat of a typical human body is&nbsp;3480Jkg⋅K, slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

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A 70-kg student produces 1200 W of heat energy per second. For the runner to maintain a constant body temperature of&amp;nbsp;37∘C, this energy needs to be released through mechanisms like &amp;nbsp;perspiration or other mechanisms. If these mechanisms failed and the energy could not flow out of the student’s body, for what amount of time could a student run before irreversible body damage occurred? (Protein structures in the body are irreversibly damaged if body temperature rises to&amp;nbsp;44∘C&amp;nbsp;or higher. The specific heat of a typical human body is&amp;nbsp;3480Jkg⋅K, slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

Carl Swisher · Accepted Answer

The amount of time the pupil has to run before suffering physical harmmstudent=70kg&amp;nbsp;Power&amp;nbsp;=1200Js&amp;nbsp;T1=37C0&amp;nbsp;T2=44C0&amp;nbsp;C=3480Jkg.K&amp;nbsp;Solution:&amp;nbsp;Determine the body&#039;s amount of heatQbo&amp;nbsp;dy&amp;nbsp;=mstudent×C×δT&amp;nbsp;=70kg×3480Jkg.K×(44−37)C0&amp;nbsp;=1.7×106J&amp;nbsp;From the Power (rate) we would find the time where&amp;nbsp;Time=QP&amp;nbsp;=1.27×106J1200Js&amp;nbsp;=1400sec⁡&amp;nbsp;&amp;nbsp;=23minute&amp;nbsp;So the time before damage occurs to the body is&amp;nbsp;23 minutes

Barbara Meeker · Accepted Answer

Answer:&amp;nbsp;1421 seconds or 23.67 minutes&amp;nbsp;Explanation:&amp;nbsp;Q=Heat=1200W&amp;nbsp;c=Specific heat of human body=3480JkgK&amp;nbsp;△T=Change in temperature=(44−37)∘C&amp;nbsp;t=Time taken&amp;nbsp;Q=mc△Tt&amp;nbsp;⇒t=mc△TQ&amp;nbsp;⇒t=70×3480×(44−37)1200&amp;nbsp;⇒t=1421s&amp;nbsp;Before permanent bodily harm occurs, a student can jog for 1421 seconds, or 23.67 minutes

karton · Accepted Answer

q=mass×specific heat×deltaT q=70kg×3480J/kg⋅K×7=?? Then q×(1sec1200J)= time in sec. Check my thinking. 1421 seconds (about 24 minutes)

While running, a 70-kg student generates thermal energy at a rate of 1

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