Question

A lightning flash transfers 4,40 C of charge and 3.82 M) of energy to the Earth. Between what potential difference did it travel? 8.68 x 10^-1 MV How much water could this bring to the boiling point, starting from room temperature (assume room temperature to be 20.0°C)?

Answer 1

The question asks to calculate the potential difference a lightning flash traveled through given the charge and energy, and the mass of water it could boil. We use the formula E = Q × V to find the potential difference and thermal energy concepts to calculate the water boiling aspect.

Step-by-step explanation:

The question involves the physics concept of electrical energy transferred during a lightning strike, the conversion of that energy to thermal energy, and the amount of water that could be boiled by this energy. To find the potential difference the lightning traveled through, we can use the relationship between charge (Q), potential difference (V), and energy (E), which is E = Q × V. Given that 4.40 C of charge transferred 3.82 MJ of energy, the potential difference V can be calculated by rearranging the formula to V = E/Q.

For the second part of the question, we need to calculate the amount of water that can be brought to the boiling point using the energy transferred from the lightning strike. Using the specific heat capacity of water (c) and the heat of vaporization, we can find the mass of water that could be heated and boiled by the provided energy. The specific heat capacity is the energy required to raise 1 gram of water by 1°C, and the heat of vaporization is the energy required to convert water from liquid to gas at the boiling point.

Answer 2

The lightning flash travelled through a potential difference of approximately 8.68 x 10^-1 megavolts. Using the energy transferred (3.82 MJ) and the specific heat capacity and latent heat of vaporization of water, one can calculate the mass of water that could be boiled starting from room temperature.

Step-by-step explanation:

To calculate the potential difference a lightning flash travelled through, we use the relationship between energy (E), charge (Q), and potential difference (V): E = Q * V. The energy transferred is given as 3.82 MJ (megajoules), and the charge transferred is 4.40 C (coulombs).

To find the potential difference V, we rearrange the formula to V = E/Q:

V = 3.82 MJ / 4.40 C = 3.82 x 106 J / 4.40 C ≈ 868,181.82 V, or 8.68 x 10-1 MV (megavolts).

To calculate how much water can be brought to the boiling point, we use the specific heat capacity of water (4.184 J/g°C) to find out the energy required to heat 1 gram of water by 1°C. To boil water starting at 20.0°C, the water needs to be heated to 100°C (boiling point) and then vaporized, requiring additional energy known as latent heat of vaporization (approximately 2260 J/g). Using the total energy provided by the lightning (3.82 MJ), we can calculate the mass (m) of the water that can be heated and vaporized.