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
.
Explanation:
Energy present in total of two bars is as follows.
![2 \times 1.2 \times 10^{3} kJ](https://tex.z-dn.net/?f=2%20%5Ctimes%201.2%20%5Ctimes%2010%5E%7B3%7D%20kJ)
kJ
(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 = ![\frac{75}{100} \times 2.4 \times 10^{6} J](https://tex.z-dn.net/?f=%5Cfrac%7B75%7D%7B100%7D%20%5Ctimes%202.4%20%5Ctimes%2010%5E%7B6%7D%20J)
=
J
Given data is as follows.
mass (m) = 56 kg
= 56000 g (1 kg = 1000 g)
Specific heat = 4.18 ![J/g ^{o}C](https://tex.z-dn.net/?f=J%2Fg%20%5E%7Bo%7DC)
As heat is releasing which means that value of E will be negative.
-E =
![-(180 \times 10^{4} = 56000 g \times 4.18 \times (T_{2} - 36.6)^{o}C](https://tex.z-dn.net/?f=-%28180%20%5Ctimes%2010%5E%7B4%7D%20%3D%2056000%20g%20%5Ctimes%204.18%20%5Ctimes%20%28T_%7B2%7D%20-%2036.6%29%5E%7Bo%7DC)
![T_{2} = 28.74^{o}C](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%2028.74%5E%7Bo%7DC)
Her temperature at the top of the structure is
.
Now, change in temperature is calculated as follows.
![\Delta T = (36.6 - 28.74)^{o}C](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%20%2836.6%20-%2028.74%29%5E%7Bo%7DC)
= ![7.86^{o}C](https://tex.z-dn.net/?f=7.86%5E%7Bo%7DC)
Thus, we can conclude that increase in her temperature at the top of the structure is
.