91Ó°ÊÓ

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$1cal=4.184J$.

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

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:

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

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

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

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 $c=1J×g^{(-1)}×C^{(-1)}.$. As a result of her body holding $843.2-210.8=632.4kcal$of heat, her temperature will rise by:

$\begin{array}{l}Q=mc\mathrm{Δ}T\\ \mathrm{Δ}T=\frac{Q}{mc}\\ \mathrm{Δ}T=\frac{632.4×{10}^3}{60×{10}^3×1}\\ \mathrm{Δ}T={10.54}^{∘}\mathrm{C}.\end{array}$

The majority of heat is lost via the skin when perspiration evaporates. Given that water ${25}^{∘}C$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:

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

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