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

obrozenecy6 2022-01-07 Answered
While running, a 70-kg student generates thermal energy at a rate of 1200 W. For the runner to maintain a constant body temperature of 37C, this energy must be removed by 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 44C or higher. The specific heat of a typical human body is 3480JkgK, 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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Expert Answer

Carl Swisher
Answered 2022-01-08 Author has 28 answers
The time taht could the student runs before body damage occured
mstudent=70kgPower=1200Js
T1=37C0T2=44C0C=3480Jkg.K
Solution:
Calculate the quantity heat of the body
Qbody=mstudent×C×δT
=70kg×3480Jkg.K×(4437)C0
=1.7×106J
From the Power (rate) we would find the time where
Time=QP
=1.27×106J1200Js
=1400sec
=23minute
So the time before damage occurs to the body is
23 minutes
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Barbara Meeker
Answered 2022-01-09 Author has 38 answers
Answer:
1421 seconds or 23.67 minutes
Explanation:
Q=Heat=1200W
c=Specific heat of human body=3480JkgK
T=Change in temperature=(4437)C
t=Time taken
Q=mcTt
t=mcTQ
t=70×3480×(4437)1200
t=1421s
The time student jogs before irreversible body damage occurs is 1421 seconds or 23.67 minutes.
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karton
Answered 2022-01-11 Author has 368 answers

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

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