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

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 ${37}^{\circ }C$, 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 ${44}^{\circ }C$ or higher. The specific heat of a typical human body is $3480\frac{J}{k}g\cdot K$, 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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Carl Swisher
The time taht could the student runs before body damage occured
${m}_{student}=70kgPower=1200\frac{J}{s}$
${T}_{1}=37{C}^{0}{T}_{2}=44{C}^{0}C=3480\frac{J}{k}g.K$
Solution:
Calculate the quantity heat of the body
${Q}_{body}={m}_{student}×C×\delta T$
$=70kg×3480\frac{J}{k}g.K×\left(44-37\right){C}^{0}$
$=1.7×{10}^{6}J$
From the Power (rate) we would find the time where
$Time=\frac{Q}{P}$
$=\frac{1.27×{10}^{6}J}{1200\frac{J}{s}}$
$=1400\mathrm{sec}$
$=23minute$
So the time before damage occurs to the body is
23 minutes
###### Not exactly what you’re looking for?
Barbara Meeker
1421 seconds or 23.67 minutes
Explanation:
$Q=Heat=1200W$
$c=\text{Specific heat of human body}=3480\frac{J}{k}gK$
$\mathrm{△}T=\text{Change in temperature}={\left(44-37\right)}^{\circ }C$
$t=\text{Time taken}$
$Q=\frac{mc\mathrm{△}T}{t}$
$⇒t=\frac{mc\mathrm{△}T}{Q}$
$⇒t=\frac{70×3480×\left(44-37\right)}{1200}$
$⇒t=1421s$
The time student jogs before irreversible body damage occurs is 1421 seconds or 23.67 minutes.
###### Not exactly what you’re looking for?
karton

$q=mass×\text{specific heat}×deltaT$
$q=70kg×3480J/kg\cdot K×7=??$
Then $q×\left(\frac{1sec}{1200J}\right)=$ time in sec.
Check my thinking. 1421 seconds (about 24 minutes)