Question

A 63-kg hiker is climbing the 828-m-tall Burj Khalifa in Dubai. If the efficiency of converting the energy content of the bars into the work of climbing is 25%, the remaining 75% of the energy released through metabolism is heat released to her body. She eats two energy bars, one of which produces 1.10×103kJ of energy upon metabolizing. Assume that the heat capacity of her body is equal to that for water (75.3 Jmol-1 .K-1.) Calculate her temperature at the top of the structure. Assume her initial temperature to be 36.6 °C. Express your answer in degrees Celsius to three significant figures. oC

Answer 1

Answer:

$T_f=5854.76$ °C

Explanation:

Given:

mass of hiker, m= 63 kg

height to be climbed, h= 828 m

energy produced by an energy bar, $E= 1.10\times 10^6 J$

heat capacity of the hiker, $c=75.3 J.mol^{-1}.K^{-1}= 4.184 J.kg^{-1}.K^{-1}$

initial body temperature of hiker, $T_i=36.6 \degree C$

The efficiency of converting the energy content of the bars into the work of climbing is 25%, the remaining 75% of the energy released through metabolism is heat released to her body.

We find the energy required for climbing 828 m height:

W=m.g.h

$W=63\times 9.8\times 828$

W= 511207.2 J

∵Hike eats 2 energy bars= $2\times 1.1\times 10^{6} J$

Energy produced$= 2.2\times 10^{6} J$

Now, according to her efficiency:

Total energy required for producing the work of W= 511207.2 J which is required to climb the given height will be (say, E):

$25\% of E= 511207.2$

$\Rightarrow E= 511207.2\times \frac{100}{25}$

$E=2044828.8 J$

&

Amount of total energy (E) converted into heat(Q) is:

$Q=2044828.8-511207.2\\Q=1533621.6J$

As we know that:

heat, $Q=m.c. (T_f-T_i)$.................(1)

where:

$T_f$ is the final temperature

Putting respective values in the eq. (1)

$1533621.6= 63\times 4.184\times (T_f-36.6)$

$(T_f-36.6)\approx 5818.16$

$T_f\approx 5854.76$ °C