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

Question

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

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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Expert Answer

Carl Swisher

Answered 2022-01-08 Author has 28 answers

The time taht could the student runs before body damage occured

m_{s t u d e n t} = 70 k g P o w e r = 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_{b o d y} = m_{s t u d e n t} \times C \times δ T

= 70 k g \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

T i m e = \frac{Q}{P}

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

= 1400 \sec

= 23 min u t e

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 = H e a t = 1200 W

c = Specific heat of human body = 3480 \frac{J}{k} g K

△ T = Change in temperature = {(44 - 37)}^{\circ} C

t = Time taken

Q = \frac{m c △ T}{t}

\Rightarrow t = \frac{m c △ T}{Q}

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

\Rightarrow t = 1421 s

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

$q = m a s s \times specific heat \times d e l t a T$
$q = 70 k g \times 3480 J / k g \cdot K \times 7 = ? ?$
Then $q \times (\frac{1 s e c}{1200 J}) =$ time in sec.
Check my thinking. 1421 seconds (about 24 minutes)

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