Question

A gallon of gas contains \(1.21 \cdot 10^{5}\) Btu and weighs 6 pounds [8] . Calculate the energy stored in the gallon of gas in joules, and calculate the specific energy in joules per kilogram.

Short answer

Energy in joules: \(1.28 \times 10^{8}\); Specific energy: \(4.69 \times 10^{7}\) J/kg.

Step-by-step solution

Convert Btu to Joules

To convert energy from British thermal units (Btu) to joules, use the conversion factor: 1 Btu = 1055.06 joules. Calculate the energy in joules as follows: \[ 1.21 \times 10^{5} \text{ Btu} \times 1055.06 \text{ joules/Btu} = 1.277 \times 10^{8} \text{ joules} \] Thus, the energy stored in the gallon of gas is approximately \(1.28 \times 10^{8}\) joules.

Convert Pounds to Kilograms

The gas weighs 6 pounds. To convert this to kilograms, use the conversion factor: 1 pound = 0.453592 kilograms. Calculate the weight in kilograms: \[ 6 \text{ pounds} \times 0.453592 \text{ kg/pound} = 2.721552 \text{ kg} \] So, the gallon of gas weighs approximately 2.721552 kilograms.

Calculate Specific Energy in Joules per Kilogram

The specific energy is the energy per unit mass. Use the energy in joules and the mass in kilograms to find specific energy: \[ \text{Specific Energy} = \frac{1.277 \times 10^{8} \text{ joules}}{2.721552 \text{ kg}} \approx 4.693 \times 10^{7} \text{ joules/kg} \] Thus, the specific energy of the gas is approximately \(4.69 \times 10^{7}\) joules per kilogram.

Key concepts

Btu to Joules Conversion

When working within physics and chemistry, converting energy units is essential to ensure accurate calculations. One common conversion involves transforming energy expressed in British thermal units (Btu) to joules.
Btu is predominantly used in the United States for heating and cooling systems, whereas joules are a standard unit in the International System of Units (SI).
To perform this conversion, one must use the correct conversion factor: 1 Btu is equivalent to 1055.06 joules. Therefore, if a substance has an energy content of \(1.21 \times 10^{5}\) Btu, you multiply by this conversion factor to find the energy in joules:

• Multiply the Btu value by 1055.06 to convert to joules.
• For our example, the calculation is \(1.21 \times 10^{5} \text{ Btu} \times 1055.06 \text{ joules/Btu} = 1.277 \times 10^{8} \text{ joules}\).

Consequently, the energy content in this context is approximately \(1.28 \times 10^{8}\) joules. This conversion provides a clearer understanding on a global scale.

Specific Energy Calculation

Specific energy is a critical concept that illustrates how much energy is stored per unit mass of a substance. Knowing the specific energy helps in comparing the energy density of different fuels or materials.
To calculate specific energy, you need two pieces of information: the total energy and the mass. In this context, we are given the energy in joules and need the mass in kilograms. To find the specific energy:

• Divide the total energy (in joules) by the mass (in kilograms).
• For example, \(\text{Specific Energy} = \frac{1.277 \times 10^{8} \text{ joules}}{2.721552 \text{ kg}}\).

This results in a specific energy of \(4.693 \times 10^{7}\) joules per kilogram. Comparing specific energies helps in selecting materials for energy applications, as higher specific energy indicates more energy storage capacity per kilogram.

Mass Conversion from Pounds to Kilograms

Converting mass units from pounds to kilograms is a routine operation in scientific calculations, especially when working with international data. The kilogram is the base unit of mass in the SI System, and many scientific formulas use kilograms to maintain consistency.
To convert pounds to kilograms, use the conversion factor that 1 pound equals 0.453592 kilograms. Performing this conversion involves simple multiplication:

• Multiply the mass in pounds by 0.453592 to get the mass in kilograms.
• For example, if you have 6 pounds of gas, it converts to \(6 \times 0.453592 = 2.721552\) kilograms.

Understanding this conversion allows seamless communication and computation in scientific work and enables accurate answers when inputs are provided in various units.