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Question

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 kJ35 \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 K2.5 \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 kg51 \mathrm{~kg}. State any additional assumptions.

Solution

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if the reduction in body temperature is 2525 K

 so that

q=mCΔT=51 Kg×4184( J K1 Kg1)×2.5 K=532.95×103 J\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

X×35×103J g1=532.95×103 JX=532.95×103 J35×103J g1=15.23 g\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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