# Fuel mass required to heat a room [closed]

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.

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– user7951
Oct 10 '16 at 15:42
• – user7951
Oct 10 '16 at 15:42

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.