## Related questions with answers

A hiker caught in a thunderstorm loses heat when her clothing becomes wet. She is packing emergency rations that if completely metabolized will release $35 \mathrm{~kJ}$ of heat per gram of rations consumed. How much rations must the hiker consume to avoid a reduction in body temperature of $2.5 \mathrm{~K}$ as a result of heat loss? Assume the heat capacity of the body equals that of water and that the hiker weighs $51 \mathrm{~kg}$. State any additional assumptions.

Solution

Verifiedif the reduction in body temperature is $25$ K

```
so that
```

$\begin{align*} q&=mC\Delta T\\ &=51\text{ Kg}\times 4184\left(\text{ J K}^{-1}\text{ Kg}^{-1}\right)\times 2.5 \text{ K}\\ &=\boxed{532.95\times 10^3\text{ J}} \end{align*}$

```
assume that the rations must the hiker consume $X$
so that
```

$\begin{align*} X\times 35 \times 10^3\text{J g}^{-1}&=532.95\times 10^3 \text{ J}\\ X&=\dfrac{532.95\times 10^3 \text{ J}}{35 \times 10^3\text{J g}^{-1}}\\ &=\boxed{15.23\text{ g}} \end{align*}$

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