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 [tex]36.6^{o}C[/tex].

Explanation:

Energy present in total of two bars is as follows.

[tex]2 \times 1.2 \times 10^{3} kJ[/tex]

[tex]2.4 \times 10^{3}[/tex] kJ

[tex]2.4 \times 10^{6} J[/tex] (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 = [tex]\frac{75}{100} \times 2.4 \times 10^{6} J[/tex]

= [tex]180 \times 10^{4}[/tex] J

Given data is as follows.

mass (m) = 56 kg

= 56000 g (1 kg = 1000 g)

Specific heat = 4.18 [tex]J/g ^{o}C[/tex]

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

-E = [tex]mC (T_{2} - T_{1})[/tex]

[tex]-(180 \times 10^{4} = 56000 g \times 4.18 \times (T_{2} - 36.6)^{o}C[/tex]

[tex]T_{2} = 28.74^{o}C[/tex]

Her temperature at the top of the structure is [tex]28.74^{o}C[/tex].

Now, change in temperature is calculated as follows.

[tex]\Delta T = (36.6 - 28.74)^{o}C[/tex]

= [tex]7.86^{o}C[/tex]

Thus, we can conclude that increase in her temperature at the top of the structure is [tex]7.86^{o}C[/tex].