/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 97 You want to heat the air in your... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 97

You want to heat the air in your house with natural gas \(\left(\mathrm{CH}_{4}\right) .\) Assume your house has \(275 \mathrm{m}^{2}\) (about \(2800 \mathrm{ft}^{2}\) ) 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 \(29.1 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K} .\) (The number of moles of air in the house can be found by assuming that the average molar mass of air is \(28.9 \mathrm{g} / \mathrm{mol}\) and that the density of air at these temperatures is \(1.22 \mathrm{g} / \mathrm{L} .\). What mass of methane do you have to burn to heat the air from \(15.0^{\circ} \mathrm{C}\) to \(22.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Burn approximately 106.86 grams of methane.

Step by step solution

01

Calculate Air Volume

First, calculate the volume of air in the house. Use the formula for volume: \( \text{Volume} = \text{floor area} \times \text{ceiling height} \). Substitute the given numbers: \[ \text{Volume} = 275 \ \mathrm{m}^2 \times 2.50 \ \mathrm{m} = 687.5 \ \mathrm{m}^3 \]. Convert this volume from cubic meters to liters (since 1 cubic meter = 1000 liters): \[ 687.5 \ \mathrm{m}^3 = 687,500 \ \mathrm{L} \].
02

Calculate Mass of Air

To find the mass of air in the house, use the density of air. Density is mass per volume, so rearrange the formula to solve for mass: \( \text{mass} = \text{density} \times \text{volume} \). Plug in the values: \[ \text{mass} = 1.22 \ \mathrm{g/L} \times 687,500 \ \mathrm{L} = 839,750 \ \mathrm{g} \].
03

Calculate Moles of Air

Convert the mass of air to moles using the average molar mass of air. The formula is: \( \text{moles} = \text{mass} / \text{molar mass} \). Substitute the values to find the moles:\[ \text{moles} = \frac{839,750 \ \mathrm{g}}{28.9 \ \mathrm{g/mol}} \approx 29,062.28 \ \mathrm{mol} \].
04

Calculate Heat Required

Determine the heat energy required to increase the air temperature from 15.0掳C to 22.0掳C. Use the heat capacity formula, \( q = nC_\text{m}(T_f - T_i) \), where \( n \) is the number of moles, \( C_\text{m} \) is the molar heat capacity, and \( T_f \) and \( T_i \) are the final and initial temperatures, respectively. Plug in the values:\[ q = 29,062.28 \ \mathrm{mol} \times 29.1 \ \mathrm{J/mol \cdot K} \times (22.0 - 15.0) \ \mathrm{K} \approx 5,929,894.84 \ \mathrm{J} \].
05

Calculate Mass of Methane Required

To find the amount of methane needed, use the enthalpy of combustion for methane, \( \Delta H_c \), which is \(-890.3 \ \mathrm{kJ/mol} \). Convert the required energy from joules to kilojoules: \[ 5,929,894.84 \ \mathrm{J} = 5,929.89 \ \mathrm{kJ} \].Next, solve for the moles of methane needed: \( n_\text{CH4} = \frac{q}{|\Delta H_c|} \):\[ n_\text{CH4} = \frac{5,929.89 \ \mathrm{kJ}}{890.3 \ \mathrm{kJ/mol}} \approx 6.66 \ \mathrm{mol} \].Finally, convert the moles of methane to mass using its molar mass \(16.04 \ \mathrm{g/mol}\):\[ \text{mass} = 6.66 \ \mathrm{mol} \times 16.04 \ \mathrm{g/mol} \approx 106.86 \ \mathrm{g} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Heat Capacity
The molar heat capacity is a significant property in thermodynamics. It represents the amount of heat required to raise the temperature of one mole of a substance by one Kelvin (or degree Celsius). This concept is vital when determining the energy needed to alter the temperature of a substance.
In the given problem, the molar heat capacity of air is provided as 29.1 J/mol路K. This value helps us calculate how much heat is necessary to increase the air's temperature in the house from 15掳C to 22掳C. The formula used is:
  • q = nCm(Tf - Ti)
where:
  • q is the heat energy
  • n is the number of moles
  • Cm is the molar heat capacity
  • Tf and Ti are the final and initial temperatures, respectively
This equation showcases the direct relationship between the heat capacity, temperature change, and number of moles, which ties directly into how much energy is required.
Enthalpy of Combustion
Enthalpy of combustion is a crucial concept in thermodynamics that refers to the heat released when one mole of a substance fully burns in oxygen. For methane ( CH4), it is provided as -890.3 kJ/mol. This value indicates how much energy is produced from the combustion of a mole of methane.
This quantity is essential for the problem because it tells us how much methane we need to burn to generate the exact amount of energy needed to heat the air in the house. By understanding the enthalpy of combustion, we can efficiently compute energy transformations and conversions.
The overall energy required for heating (q) is matched against this energy released per mole combustion:
  • nCH4 = q / | 螖Hc|
This equation helps in figuring out the number of moles and eventually the mass of methane needed, illustrating how understanding combustion plays a key role in practical heating applications.
Moles Calculation
Mole calculation is a fundamental aspect of chemistry used to connect the macroscopic quantities with the number of entities at the molecular level. In this exercise, several calculations involve moles:
  • Moles of air
  • Moles of methane
The moles of air are calculated using the mass of air and its average molar mass. The formula is:
  • moles = mass / molar mass
Here, we convert the mass of air (obtained using its volume and density) into moles, enabling us to perform heat capacity calculations.
Similarly, calculating the moles of methane involves the energy required and the enthalpy of combustion. The formula is beneficial for understanding the magnitude of chemical reactions and energy needs. It allows one to bridge the gap between mass and energy and is a crucial tool in analyzing a wide range of thermodynamic and chemical processes.

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Most popular questions from this chapter

A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\ell) $$ The bomb had a heat capacity of \(550 \mathrm{J} / \mathrm{K},\) and the calorimeter contained \(650 \mathrm{g}\) of water. Burning \(4.20 \mathrm{g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)\) resulted in a rise in temperature from \(18.5^{\circ} \mathrm{C}\) to \(22.3^{\circ} \mathrm{C} .\) Calculate the enthalpy of combustion of ethanol, in \(\mathrm{kJ} / \mathrm{mol}\).

The energy required to melt \(1.00 \mathrm{g}\) of ice at \(0^{\circ} \mathrm{C}\) is 333 J. If one ice cube has a mass of \(62.0 \mathrm{g}\) and a tray contains 16 ice cubes, what quantity of energy is required to melt a tray of ice cubes to form liquid water at \(0^{\circ} \mathrm{C} ?\)

An "ice calorimeter" can be used to determine the specific heat capacity of a metal. A piece of hot metal is dropped onto a weighed quantity of ice. The energy transferred from the metal to the ice can be determined from the amount of ice melted. Suppose you heated a 50.0 -g piece of silver to \(99.8^{\circ} \mathrm{C}\) and then dropped it onto ice. When the metal's temperature had dropped to \(0.0^{\circ} \mathrm{C},\) it is found that \(3.54 \mathrm{g}\) of ice had melted. What is the specific heat capacity of silver?

You add \(100.0 \mathrm{g}\) of water at \(60.0^{\circ} \mathrm{C}\) to \(100.0 \mathrm{g}\) of ice at \(0.00^{\circ} \mathrm{C} .\) Some of the ice melts and cools the water to \(0.00^{\circ} \mathrm{C} .\) When the ice and water mixture reaches thermal equilibrium at \(0^{\circ} \mathrm{C},\) how much ice has melted?

How much energy as heat is required to raise the temperature of \(50.00 \mathrm{mL}\) of water from \(25.52^{\circ} \mathrm{C}\) to \(28.75^{\circ} \mathrm{C} ?\) (Density of water at this temperature = \(0.997 \mathrm{g} / \mathrm{mL} .)\)
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