1
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ 3.0 grams or kg of water? (I suspect the difficulty is there or going from kJ to J) $\endgroup$
    – jonsca
    Oct 15, 2012 at 10:40

1 Answer 1

1
$\begingroup$

$$\Delta H = mc_p\Delta T$$

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. $$ m_\text{fuel}= \frac{1\,004\,160\ \mathrm J}{36\,000\ \mathrm{J\cdot g^{-1}}} = 27.89\ \mathrm g $$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.