# Question f631a

Apr 9, 2017

${\text{11 g CH}}_{4}$

#### Explanation:

The first thing you need to do here is to calculate the amount of heat needed to get $\text{2 kg}$ of liquid water at ${25}^{\circ} \text{C}$ to $\text{2 kg}$ of liquid water at ${100}^{\circ} \text{C}$.

To do that, use the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

• $q$ is the heat gained by the water
• $m$ is the mass of the water
• $c$ is the specific heat of water, equal to ${\text{4.18 J g"^(-1)""^@"C}}^{- 1}$
• $\Delta T$ is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

In your case, you have a mass of

2 color(red)(cancel(color(black)("kg"))) * (10^3color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = 2 * 10^3 $\text{g}$

and a change in temperature of

$\Delta T = {100}^{\circ} \text{C" - 25^@"C" = 75^@"C}$

which means that you'll have

$q = 2 \cdot {10}^{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 75color(red)(cancel(color(black)(""^@"C}}}}$

q = 6.27 * 10^2 * overbrace(10^3 color(white)(.)"J")^(color(blue)("= 1 kJ"))

$q = 6.27 \cdot {10}^{2}$ $\text{kJ}$

Now, you know that the enthalpy change of combustion for methane is equal to

$\Delta {H}_{\text{comb" = - "890 kJ mol}}^{- 1}$

This means that when $1$ mole of methane undergoes combustion, $\text{890 kJ}$ of heat are being given off, hence the minus sign used for the value of $\Delta {H}_{\text{comb}}$.

You can thus say that in order to give off $6.27 \cdot {10}^{2}$ $\text{kJ}$ of heat, you must burn

6.27 * 10^2 color(red)(cancel(color(black)("kJ"))) * "1 mole CH"_4/(890color(red)(cancel(color(black)("kJ")))) = "0.7045 moles CH"_4#

To convert this to grams, use the molar mass of methane

$0.7045 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CH"_4))) * "16.05 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = color(darkgreen)(ul(color(black)("11 g}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the mass of water and for the final temperature.