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A 56-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 the increase in her temperature at the top of the structure.

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The given question is incomplete. The complete question is as follows.

A 56-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 the increase in her temperature at the top of the structure.

Calculate the temperature at the top of the structure. Assume her intitial temperature to be
36.6^(o)C.

Step-by-step explanation:

Energy present in total of two bars is as follows.


2 * 1.2 * 10^(3) kJ


2.4 * 10^(3) kJ


2.4 * 10^(6) J (as 1 kJ = 1000 J)

As 75% of energy releases out. Let us assume that energy releases out of its body be E. Then, energy will be calculated as follows.

E =
(75)/(100) * 2.4 * 10^(6) J

=
180 * 10^(4) J

Given data is as follows.

mass (m) = 56 kg

= 56000 g (1 kg = 1000 g)

Specific heat = 4.18
J/g ^(o)C

As heat is releasing which means that value of E will be negative.

-E =
mC (T_(2) - T_(1))


-(180 * 10^(4) = 56000 g * 4.18 * (T_(2) - 36.6)^(o)C


T_(2) = 28.74^(o)C

Her temperature at the top of the structure is
28.74^(o)C.

Now, change in temperature is calculated as follows.


\Delta T = (36.6 - 28.74)^(o)C

=
7.86^(o)C

Thus, we can conclude that increase in her temperature at the top of the structure is
7.86^(o)C.

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