Respuesta :
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 [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] Â
   [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].
