Question

what is the mass of a liquid water that vaporizes if 7700 joules of energy is absorbed by the h2o at 100 celsious

Answer 1

The mass of the liquid water that vaporizes if 7700 joules of energy is absorbed by the water at 100 degrees Celsius is approximately 3.41 kg.

Step-by-step explanation:

To calculate the mass of the water that vaporizes, we can use the formula Qv = mLv, where Qv represents the amount of heat absorbed by the water, m represents the mass of the water, and Lv represents the latent heat of vaporization.

Given that 7700 joules of energy is absorbed by the water at 100 degrees Celsius, we can rearrange the formula to solve for the mass, m. Since Lv is equal to 2,256 kJ/kg, we have:

m = Qv / Lv

Substituting the values, we have m = 7700 J / 2256 J/kg = 3.41 kg. Therefore, the mass of the liquid water that vaporizes is approximately 3.41 kg.

Answer 2

Approximately 3.41 grams of liquid water would vaporize when 7700 joules of energy is absorbed at 100°C, using the latent heat of vaporization for water which is 2256 kJ/kg.

Step-by-step explanation:

We can determine the mass of water that vaporizes using the supplied energy and the latent heat of vaporization for water. The latent heat of vaporization for water is about 2256 kJ/kg, which implies that a kilogram of water requires 2256 kJ of energy to vaporize at 100°C.

First, convert the energy from joules to kilojoules:

Energy (kJ) = 7700 joules × (1 kJ / 1000 J)
Energy (kJ) = 7.7 kJ

Next, use the formula Q = m × Lv, where Lv is the latent heat of vaporization, Q is the amount of heat, and m is the mass, to find the mass (m):

m = Q / Lv
m = 7.7 kJ / 2256 kJ/kg
m ≈ 0.00341 kg

To convert to grams, as most laboratory measurements are in grams:

m ≈ 3.41 grams

Therefore, approximately 3.41 grams of liquid water would vaporize when 7700 joules of energy is absorbed at 100°C.