A $60-kg$hiker wishes to climb to the summit of Mt. Ogden, an ascent of$5000$vertical feet$(1500m)$ .
$\left(a\right)$Assuming that she is $25\%$efficient at converting chemical energy from food into mechanical work, and that essentially all the mechanical work is used to climb vertically, roughly how many bowls of corn flakes (standard serving size $1$ounce,$100$ kilocalories) should the hiker eat before setting out?
$\left(b\right)$ As the hiker climbs the mountain, three-quarters of the energy from the corn flakes is converted to thermal energy. If there were no way to dissipate this energy, by how many degrees would her body temperature increase?
$\left(c\right)$In fact, the extra energy does not warm the hiker's body significantly; instead, it goes (mostly) into evaporating water from her skin. How many liters of water should she drink during the hike to replace the lost fluids? (At${25}^{\circ }C$, a reasonable temperature to assume, the latent heat of vaporization of water is $580cal/g$more than atrole="math" localid="1650290792399" ${100}^{\circ }C$).

Let's say a $60kg$ hiker try to climb a $1500m$ high peak. She intends to eat solely corn flakes to get the energy she needs for the ascent. To acquire an estimate, we'll make a number of oversimplifying assumptions.

Assume that the sole effort used during the trek is the energy necessary to ascend $1500metres$. Because the work is equal to the potential energy obtained, the following equation applies:

$W=mgh$

Substitute $m$is mass of hiker and $h$is height of mountain and $g$is acceleration gravity and where$1cal=4.184J$.

$\begin{array}{l}W=\left(60\right)\times \left(9.8\right)\times \left(1500\right)\\ W=8.82\times {10}^5\mathrm{J}.\\ W=8.82\times {10}^5\mathrm{J}\times \frac{1\mathrm{cal}}{4.184J}\\ W=2.108\times {10}^5\mathrm{cal}.\end{array}$

Assuming that can convert $25\%$of the energy in food into work, she will need to ingest four times the amount of energy required to climb the mountain, which means:

$$\begin{gathered} \text{Energy}=4\times 2.108\times {10}^5 \\ \text{Energy}=8.432\times {10}^5\mathrm{cal}. \end{gathered}$$

She'll need to eat: Because the enthalpy change in "burning" corn flakes is 1$100kcalper28g$, she'll need to eat:

$\begin{array}{l}{m}_{\text{flakes}}=\frac{\text{Energy}}{\mathrm{\Delta }H}\times \text{mass of flakes that produce}\mathrm{\Delta }H\\ m_{\text{flakes}}=\frac{8.432\times {10}^5}{100\times {10}^3}\times 28\\ m_{\text{flakes}}=236.1\mathrm{g}.\end{array}$

The hiker's temperature will rise if the remaining $75\%$of the energy is stored as heat. Assuming that the body is mainly water, the specific heat of water is $c=1J\times g^{(-1)}\times C^{(-1)}.$. As a result of her body holding $843.2-210.8=632.4kcal$of heat, her temperature will rise by:

$\begin{array}{l}Q=mc\mathrm{\Delta }T\\ \mathrm{\Delta }T=\frac{Q}{mc}\\ \mathrm{\Delta }T=\frac{632.4\times {10}^3}{60\times {10}^3\times 1}\\ \mathrm{\Delta }T={10.54}^{\circ }\mathrm{C}.\end{array}$

The majority of heat is lost via the skin when perspiration evaporates. Given that water ${25}^{\circ }C$has a latent heat of vaporisation of $580cal{g}^{(-1)}$, the surplus heat will vaporize a quantity of water equal to:

$$\begin{gathered} L=\frac{Q}{m} \\ m=\frac{Q}{L} \end{gathered}$$

$Q$is the leftover heat, which is $632.4kcal$, so:

$\begin{array}{l}m=\frac{Q}{L}\\ m=\frac{632.4\times {10}^3}{580}\\ m=1090.34\mathrm{g}.\end{array}$

She'd have to drink little over a liter of distilled water $1090mL$to replenish it.