Final answer:
To boil away one gallon of water, which is 3,785 grams, it requires 8,516,250 joules of energy. This amount of energy is equivalent to around 8.52 sticks of dynamite, given that each stick of dynamite contains roughly 1 megajoule (1,000,000 joules) of energy.
Step-by-step explanation:
To calculate the energy required to boil away one gallon of water, we first need to know the specific heat of vaporization for water. It's given that it takes 2,250 J to evaporate each gram of water. One gallon of water is approximately 3,785 grams (since 1 gallon ≈ 3.785 liters and 1 liter of water has a mass of about 1 kilogram or 1,000 grams).
Therefore, the total energy (E) needed to evaporate one gallon of water is:
E = mass (m) × heat of vaporization (Hv)
Where:
- m = 3,785 grams
- Hv = 2,250 J/g
E = 3,785 g × 2,250 J/g = 8,516,250 J
We can compare this to the energy of a stick of dynamite. A typical stick of dynamite contains about 1 megajoule (1,000,000 J) of energy. Thus, the energy needed to boil one gallon of water is equivalent to approximately:
Energy in sticks of dynamite = 8,516,250 J / 1,000,000 J/stick ≈ 8.52 sticks of dynamite.