91Ó°ÊÓ

An ampere-hour (Ah) is a unit of electric charge, commonly used to describe the capacity of a battery. It tells us how much current a battery can supply over a given period of time. Imagine you have a bucket (battery) and a hose (current). The ampere-hour rating tells you how long you can keep the hose running before the bucket is empty. Here's a simple breakdown:

• 1 ampere-hour is the charge conducted by a steady current of 1 ampere flowing for 1 hour.
• If a battery is rated at 60 Ah, it can deliver a current of 60 A for 1 hour, or 30 A for 2 hours, as long as the voltage remains constant.

Understanding ampere-hours helps to predict how long a battery will last under specific conditions, making it a crucial factor in designing and using electronic devices powered by batteries. In the context of the exercise, knowing that 1 Ah is equivalent to 3600 Coulombs is essential when calculating the total energy the battery can supply.

To find the total energy supplied by a battery, we use the concept of electrical energy, which can be calculated using the formula: \( \text{Energy} = \text{Voltage} \times \text{Current} \times \text{Time} \). This formula derives from the basic principle that energy is work done over time. For a battery rated at 60 A * h and 12 V, we need to convert the ampere-hours into seconds, since energy is calculated in Joules, where 1 Ah is equal to 3600 Coulombs (C). Thus, \( 60 \text{ A} \cdot \text{h} = 60 \times 3600 \text{ C} = 216000 \text{ Coulombs} \) and calculating energy gives: \( E = 12 \times 216000 \text{ C} = 2.592 \times 10^6 \text{ J} \). This calculation tells us how much total energy the battery can deliver when in use. Fully understanding this computation allows practical applications, such as estimating how long electronic devices can operate on a fully charged battery.

The latent heat of combustion refers to the amount of energy released as heat when a substance undergoes complete combustion. In real-world terms, it's the energy you get from fuel, for instance, gasoline, when it's burnt. In this exercise, we're looking to match the energy from a fully charged battery with the energy derived from combustion. The given latent heat of combustion for gasoline is \( 46 \times 10^6 \text{ J/kg} \). To find out how much gasoline produces the same amount of energy as the battery, we set up the equation \( 2.592 \times 10^6 \text{ J} = 46 \times 10^6 \text{ J/kg} \times m \). Solving for mass \( m \), we find: \( m = \frac{2.592 \times 10^6}{46 \times 10^6} \approx 0.0564 \text{ kg} \).Now, converting \( 0.0564 \text{ kg} \) to volume using the known density of gasoline \( 900 \text{ kg/m}^3 \), Volume \( V = \frac{m}{\rho} = \frac{0.0564}{900} \approx 6.27 \times 10^{-5} \text{ m}^{3} \), or roughly, \( 6.27 \times 10^{-2} \text{ L} \).This exercise shows not only how efficient gasoline can be as a source of energy but also helps us compare it with electrical energy to better understand different energy sources.