obrozenecy6

2022-01-07

A 70-kg student produces 1200 W of heat energy per second. For the runner to maintain a constant body temperature of ${37}^{\circ}C$, this energy needs to be released through mechanisms like 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.)

Carl Swisher

Beginner2022-01-08Added 28 answers

The amount of time the pupil has to run before suffering physical harm

$m}_{student}=70kgPower=1200\frac{J}{s$

${T}_{1}=37{C}^{0}{T}_{2}=44{C}^{0}C=3480\frac{J}{k}g.K$

Solution:

Determine the body's amount of heat

${Q}_{body}={m}_{student}\times C\times \delta T$

$=70kg\times 3480\frac{J}{k}g.K\times (44-37){C}^{0}$

$=1.7\times {10}^{6}J$

From the Power (rate) we would find the time where

$Time=\frac{Q}{P}$

$=\frac{1.27\times {10}^{6}J}{1200\frac{J}{s}}$

$=1400\mathrm{sec}$

$=23minute$

So the time before damage occurs to the body is

23 minutes

Barbara Meeker

Beginner2022-01-09Added 38 answers

Answer:

1421 seconds or 23.67 minutes

Explanation:

$Q=Heat=1200W$

$c=\text{Specific heat of human body}=3480\frac{J}{k}gK$

$\mathrm{\u25b3}T=\text{Change in temperature}={(44-37)}^{\circ}C$

$t=\text{Time taken}$

$Q=\frac{mc\mathrm{\u25b3}T}{t}$

$\Rightarrow t=\frac{mc\mathrm{\u25b3}T}{Q}$

$\Rightarrow t=\frac{70\times 3480\times (44-37)}{1200}$

$\Rightarrow t=1421s$

Before permanent bodily harm occurs, a student can jog for 1421 seconds, or 23.67 minutes

karton

Expert2022-01-11Added 613 answers

Then

Check my thinking. 1421 seconds (about 24 minutes)

22+64

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