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Question:

A backpacker wants to carry enough fuel to heat $\pu{3.0 kg}$ of water from $\pu{20 ^\circ C}$ to $\pu{100.0 ^\circ C}$. If the fuel he carries produces $\pu{36 kJ}$ of heat per gram when it burns, how much fuel should he carry? (For the sake of simplicity, assume that the transfer of heat is $100\%$ efficient.)

My hunch is that it involves the specific heat equation $Q_\mathrm p = m c_\mathrm p \Delta T$ where $Q$ is heat, $m$ is mass, $c_\mathrm p$ is the specific heat capacity (in this case $\pu{4.184 J/(g ^\circ C)}$) under constant pressure and $\Delta T$, but twice I calculated the wrong answer, and would like a more intuitive understanding of the problem.

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    $\begingroup$ 3.0 grams or kg of water? (I suspect the difficulty is there or going from kJ to J) $\endgroup$ – jonsca Oct 15 '12 at 10:40
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$$\Delta H = mc_p\Delta T$$

Where $\Delta H$ is either energy absorbed or released at constant pressure. Assuming this specific problem refers to $3\ \mathrm{kg}$ of water: $$ \Delta H = \left(3000\ \mathrm g\right)\left(4.184\ \mathrm{J\cdot g^{-1}\cdot K^{-1}}\right)\left(80\ \mathrm K\right) $$

That means that the energy required is: $1\,004\,160$ joules.

Now, simply divide those 1 004 160 joules by the 36 000 joules your fuel releases per gram. $$ m_\text{fuel}= \frac{1\,004\,160\ \mathrm J}{36\,000\ \mathrm{J\cdot g^{-1}}} = 27.89\ \mathrm g $$

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