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Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to $\mathbf{34}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}\mathit{C}$overnight and rise to $\mathbf{40}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}\mathit{C}$during the day. To see how effective this mechanism is for saving water, calculate how many litres of water a 400kg camel would have to drink if it attempted to keep its body temperature at a constant by evaporation of sweat during the day ( 12hours ) instead of letting it rise to $\mathbf{40}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}\mathit{C}$. (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, 3480J/kgK . The heat of vaporization of water at $\mathbf{34}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}\mathit{C}$ is $\mathbf{2}\mathbf{.}\mathbf{42}\mathbf{×}{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathit{J}\mathbf{/}\mathit{k}\mathit{g}$ .)

Step by step solution

01

Equate the heat flows from Camel and Water

Given that $m_{Camel}=400kg,c_{Camel}=3480J/kgKand{L}_v=242×{10}^{4}J/kg$.

To solve this problem, we should know there are two heat flows from the Camel and the Water.

For the Camel, $Q=m_{Camel}×c_{Camel}×\mathrm{δ}T$

Where $\mathrm{δ}T={(40-34)}^{0}C$.

For water, $Q=m_{Water}×L_f$.

Equate both the heat flow equations.

$$\begin{gathered} m_{Water}×L_f=m_{Camel}×c_{Camel}×\mathrm{δ}T \\ {m}_w=\frac{m_{Camel}×c_{Camel}×\mathrm{δ}T}{L_v}{L}_v \\ {m}_w=\frac{400×3480×6}{242×{10}^4} \\ {m}_w=3.45kg \end{gathered}$$

Hence, for saving the camel’s body for 12 hours , it should have 3.45L from water.

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