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

`T_f=5854.76` °C

Step-by-step explanation:

Given:

mass of hiker, m= 63 kg

height to be climbed, h= 828 m

energy produced by an energy bar,
`E= 1.10* 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* 9.8* 828`

W= 511207.2 J

∵Hike eats 2 energy bars=
`2* 1.1* 10^(6) J`

Energy produced
`= 2.2* 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* (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* 4.184* (T_f-36.6)`

`(T_f-36.6)\approx 5818.16`

`T_f\approx 5854.76` °C