Chapter 9, Problem 9.85PAE

9.91 You want to heat the air in your house with natural gas (CH₄). Assume your house has 275 m²(ahout 2800 ft²) of floor area and that the ceilings are 2.50 m from the floors. The air in the house has a molar heat capacity of 29.1 J mol-l K-l. (The number of moles of air in the house may he found by assuming that the average molar mass of air is 28.9 g/mol and that the density of air at these temperatures is 1.22 g/L.) What mass of methane do you have to burn to heat the air from 15.0 to 22.0°C?

Interpretation Introduction

Interpretation:

The mass of methane required to raise the temperature of air in the house from ${15}^0\text{C}$ to ${22}^0\text{C}$ should be calculated.

Concept Introduction:

• Chemical reactions proceed through absorption (endothermic) or evolution (exothermic) of heat.
• The quantity of heat absorbed or released depends on the amount of the substance, its specific heat capacity and the change in temperature.
• Specific heat capacity is the amount of heat required to raise a unit mass of a given substance through a unit degree increase in temperature.

Answer to Problem 9.85PAE

Solution: Mass of methane required = 4.25 kg

Explanation of Solution

Given: Initial temperature of air (T₁) = ${15}^0\text{C}$

Final temperature of air (T₂) = ${22}^0\text{C}$

Change in temperature ( $\Delta \text{T}$ ) = $T_2-T_1=22-15=7.0^{0}C=(7+273)K=280K$

Dimensions (volume) of the house = $275m^2\times 2.50m=687.5m^3$

Specific heat capacity of air, c_air = 29.1 J/K-mol

Formula:

1. Enthalpy of combustion:

$\Delta {\text{H}}_{\text{rxn}}={\displaystyle \sum {\text{n}}_{\text{products}}\Delta {\text{H}}_{\text{f}}^0{}_{\text{(products)}}\text{ - }{\displaystyle \sum {\text{n}}_{\text{reactants}}\Delta {\text{H}}_{\text{f}}^0{}_{\text{(reactants)}}}}\text{ -------(1)}$

n_products and n_reactants are the number of moles of products and reactants respectively

$\Delta {\text{H}}_{\text{f (products)}}^0\text{ and }\Delta {\text{H}}_{\text{f (reactants)}}^0$ are the standard heats of formation of the products and reactants respectively

2. Heat required to raise the temperature of air:

${\text{q = m}}_{\text{air}}\times {\text{c}}_{\text{air}}\times \Delta \text{T--------(2)}$

m_air = mass of air

c_air = specific heat capacity of air

$\Delta \text{T}$ = change in temperature

Calculation:

Step I: Find enthalpy of combustion of methane

${\text{CH}}_4{\text{(g) + 2O}}_2\text{(g)}\to {\text{CO}}_2{\text{(g) + 2H}}_2\text{O(g)}$

Based on equation 1 we have:

$\Delta {\text{H}}_{\text{rxn}}=[1\times \Delta {\text{H}}_{\text{f}}^0{\text{(CO}}_{\text{2}}\text{) + 2}\times \Delta {\text{H}}_{\text{f}}^0{\text{(H}}_{\text{2}}\text{O)] - }[1\times \Delta {\text{H}}_{\text{f}}^0{\text{(CH}}_{\text{4}}\text{) + 2}\times \Delta {\text{H}}_{\text{f}}^0{\text{(O}}_2\text{)]}$

Substituting for the respective enthalpies formation:

$\Delta {\text{H}}_{\text{rxn}}=[1\times (-393.5)\text{ + 2}\times (-285.8)\text{] - }[1\times (-74.9)\text{ + 2}\times (0)\text{] = -890}\text{.2 kJ/mol}$

Step II: Calculate the heat required to raise the temperature of air

Volume of air = volume of the room = 687.5 m³ = 687500 L

Now, the density of air = 1.22 g/L

Therefore, the mass of air occupied in the room = $1.22\frac{g}{L}\times 687500L=838750g$

The number of moles of air (n_air) = $\frac{Massofair}{Molarmass}\text{ = }\frac{838750\text{ g}}{28.9\text{ g/mol}}=\text{ 29022}\text{.5 moles }$

Based on equation 2:

$\text{q = 29022}\text{.5 moles }\times \text{ 29}\text{.1 }\frac{\text{J}}{\text{K}\text{.mole}}\times \text{280 K = 236475}\text{.3 kJ }$

Step III: Calculate the mass of methane

From Step II, 1 mole of methane releases an energy = 890.2 kJ

The number of moles of methane required corresponding to 236475.3 kJ energy =

= $\frac{236475.3}{890.2}=265.64\text{ moles}$

1 mole of methane (CH₄) = 16 g

The mass corresponding to 265.64 moles = $265.64\times 16=4250.24\text{ g }\simeq \text{ 4}\text{.25 kg}$

Conclusion

The mass of methane required to raise the temperature of air in the house from ${15}^0\text{C}$ to ${22}^0\text{C}$ is 4.25 kg