Question

The heater used in a 4.32 m×3.55 m×2.91 m dorm room uses the combustion of natural gas (primarily methane gas) to produce the heat required to increase the temperature of the air in the dorm room. Assuming that all of the energy produced in the reaction goes towards heating only the air in the dorm room, calculate the mass of methane required to increase the temperature of the air by 6.97 °C. Assume that the specific heat of air is 30.0 J/K·mol and that 1.00 mol of air occupies 22.4 Lat all temperatures. Enthalpy of formation values can be found in this table. Assume gaseous water is produced in the combustion of methane.

Answer 1

Heat combustion of methane ( gaseous water is formed)

= - 802.3 KJ/mole.

Volume of room

= (4.32×3.55× 2.91) m³

= 44.6 m³

= 44.6 × 10³ dm³

= 4.46 × 10⁴ L.

Volume of air = 4.46 ×10⁴ L

Moles of air present in the room

= (4.46×10⁴/22.4)

= 1991.07

Now heat required to increase the temperature by 6.97⁰ c

= Moles of air × specific heat × increase in temperature

= ( 1991.07 × 30.0× 6.97) J

= 416333 J

= 416.33 KJ.

Now moles of methane required to release 416.33 KJ

= (416.33/802.3)

= 0.519

Mass of methane = (mole × molar mass )

= 0.519 × 16

= 8.30 g.