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I have 126.39 L of water. What volume of methane gas would I need to combust to heat it (at room temp, 21 degrees Celsius) to 38 degrees Celsius? How much $\ce{CO2}$ gas would be produced?

I calculated the molar enthalpy of combustion of methane and got −890.27 kJ/mol, and I know that factors into my calculation. I'm just confused as to what I actually have to do to determine exactly how much methane I need to heat the water.

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A great place to start would be the heat capacity $C$ of water. It is defined to be the "amount" of heat change divided by the resulting temperature change. In symbols,

$$C=\frac{q}{\Delta T}.$$

Heat capacity is an extensive property; the size of $C$ depends on how much matter there is. We take this into account by dividing the equation by mass $m$.

$$C_s=\frac{q}{m\Delta T}$$

$C_s$ is called specific heat (capacity). Taking $\rho_{\ce{(H2O)}}\approx1$,

$$C_s \approx \frac{q}{V\Delta T}.$$

Can you take it from here? (Be careful with units.)


Note: It is probably assumed that

  • the combustion of $\ce{CH4}$ is complete, yielding only $\ce{CO2}$ and $\ce{H2O}$;
  • density of water does not change over the temperature range;
  • heat capacity is constant as $\Delta T$ is rather small;
  • vaporisation is neglible;
  • no heat lost to surroundings;
  • decreasing solubility of $\ce{CO2}$ does not contribute to the overall amount of gaseous carbon dioxide (joking...)
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