-3
$\begingroup$

A room in a house measures 3.7 m × 4.7 m × 4.0 m. Assuming no heat or material losses, how many grams of natural gas (methane, CH4) must be burned to heat the air in this room from 15.0 degrees C to 25.0 degrees C. Assume that air is 78% N2 and 22% O2.

$\endgroup$

closed as off-topic by Ivan Neretin, Loong, Geoff Hutchison, ringo, Wildcat Oct 10 '16 at 16:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Welcome to chemistry.SE! If you have any questions about the policies of our community, please ‎visit the help center. This is a homework question. We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. $\endgroup$ – Loong Oct 10 '16 at 15:42
  • $\begingroup$ Related: How much does a light heat the air around it? $\endgroup$ – Loong Oct 10 '16 at 15:42
1
$\begingroup$

In the first step, we calculate the energy that is needed to heat the room by ten degrees. To do that, you need the heat capacity of air: $C_\text{air} = 1210~ \text{J} \cdot \text{m}^{-3}\cdot \text{K}^{-1} $ This is the energy that is needed per volume unit to increase the temperature by one Kelvin (i.e. one Degree). To increase the temperature by $\Delta T = 10 ~\text{K}$ for a volume of $V_\text{room}=\text{3.7 m}\cdot\text{4.7 m}\cdot\text{4.0 m}=69.56 ~\text{m}^3$ you would thus need an energy of $$ E = C_\text{air}\cdot V_\text{room} \cdot \Delta T = 1210~ \text{J} \cdot \text{m}^{-3}\cdot \text{K}^{-1} \cdot 69.56~\text{m}^3 \cdot 10~\text{K} \approx 842~\text{kJ}$$ We can now calculate the amount of methane that needs to be burned to get that energy. For this, we take the enthalpy of combustion $\Delta H_\text{comb}$: $$\Delta H_\text{comb} = −882.0 ~\text{kJ/mol} \equiv 55.1~\text{kJ/g}$$ Thus, we calculate the mass of methane to be $$ m_\text{methane} = 842/55.1 ~\text{g} = 15.3~\text{g}.$$ So about 15 g of methane would be needed to increase the temperature of that room by 10 degrees. By the way, 15 g methane correspond to a volume of about 20 L.

$\endgroup$
  • $\begingroup$ Is that the upper heating value of methane or the lower heating value? $\endgroup$ – Chet Miller Oct 10 '16 at 17:21
  • $\begingroup$ The upper heating value. $\endgroup$ – logical x 2 Oct 10 '16 at 19:08

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