Question

The kitchen, bath, and other ventilation fans in a house should be used sparingly since these fans can discharge a houseful of warmed or cooled air in just one hour. Consider a \(200-\mathrm{m}^{2}\) house whose ceiling height is \(2.8 \mathrm{m} .\) The house is heated by a 96 percent efficient gas heater and is maintained at \(22^{\circ} \mathrm{C}\) and \(92 \mathrm{kPa}\). If the unit cost of natural gas is \(\$ 1.20 /\) therm \((1 \text { therm }=105,500 \mathrm{kJ})\) determine the cost of energy "vented out" by the fans in 1 h. Assume the average outdoor temperature during the heating season to be \(5^{\circ} \mathrm{C}\)

Short answer

Answer: The cost of energy vented out by the fans in 1 hour is approximately $0.68.

Step-by-step solution

Calculate the volume of the house

Given the dimensions of the house as \(200\,\mathrm{m}^2\) and the ceiling height as \(2.8\,\mathrm{m}\), the volume of the house can be calculated as follows: $$ V = area \times height = 200\text{ m}^2 \times 2.8\text{ m} = 560\,\mathrm{m}^3 $$

Determine the mass of the air using ideal gas law

The ideal gas law can be expressed as: $$ PV = nRT $$ Where \(P\) is the pressure, \(V\) is the volume, \(n\) is the amount of gas in moles, \(R\) is the ideal gas constant, and \(T\) is the temperature. The mass of the air can be calculated using the molar mass of air, which is approximately \(29\,\mathrm{g/mol}\) or \(0.029\,\mathrm{kg/mol}\). We can rearrange the equation to solve for the mass of the air (\(m\)): $$ m =\frac{PV}{RT} \times \text{M}_\text{air} $$ With the given pressure (\(P=92\,\mathrm{kPa}\)), volume (\(V=560\,\mathrm{m}^3\)), temperature (\(T=22^{\circ}\mathrm{C} = 295.15\,\mathrm{K}\)) and the gas constant (\(R=8.314\,\mathrm{J/(mol\cdot K)}\)), we can determine the mass of the air inside the house: $$ m = \frac{(92\times10^3\,\mathrm{Pa})(560\,\mathrm{m}^3)}{(8.314\,\mathrm{J/(mol\cdot K)})(295.15\,\mathrm{K})} \times 0.029\,\mathrm{kg/mol} \approx 3366\,\mathrm{kg} $$

Calculate the energy required to heat the air inside the house

To calculate the energy required to heat the air inside the house, we need to consider the difference in temperature between the indoor (\(22^{\circ}\mathrm{C}\)) and outdoor (\(5^{\circ}\mathrm{C}\)) temperature: $$ \Delta T=22^{\circ}\mathrm{C} - 5^{\circ}\mathrm{C} = 17\,\mathrm{K} $$ The energy required can be determined using the specific heat capacity of air (\(c_p = 1005\,\mathrm{J/(kg\cdot K)})\) and the mass of the air calculated earlier: $$ Q = mc_p\Delta T = (3366\,\mathrm{kg})(1005\,\mathrm{J/(kg\cdot K)})(17\,\mathrm{K}) \approx 57,440,000\,\mathrm{J} $$

Calculate the cost of energy vented out by the fans in 1 hour

We are given the efficiency of the gas heater as 96%, so the actual energy used to heat the air is: $$ Q_\text{actual} = \frac{Q}{\text{efficiency}} = \frac{57,440,000\,\mathrm{J}}{0.96} \approx 59,833,333\,\mathrm{J} $$ To find the cost of energy vented out by the fans in 1 hour, we need to convert the energy used in Joules to therms: $$ \text{therms} = \frac{59,833,333\,\mathrm{J}}{105,500\,\mathrm{J/therm}} \approx 0.568\,\mathrm{therms} $$ Finally, multiplying the therms with the unit cost of natural gas (\(\$1.20/\mathrm{therm}\)) gives us the cost of energy vented out by the fans in 1 hour: $$ \text{Cost} = 0.568\,\mathrm{therms} \times \$1.20/\mathrm{therm} \approx \$0.68 $$ Therefore, the cost of energy vented out by the fans in 1 hour is approximately \( \$0.68\).