how many joules of heat are needed to vaporize 50.0 g water at 100.0 C?
11293J
q = mass H2O x deltaHvap
To calculate the heat needed to vaporize water, we need to consider the heat required for two separate processes: heating the water from its initial temperature to its boiling point, and then vaporizing it.
1. Heating the water:
The heat required to heat water can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat required (in joules),
m is the mass of water (in grams),
c is the specific heat capacity of water (4.18 J/g°C), and
ΔT is the change in temperature (in °C).
Since we need to heat the water from 100.0°C to its boiling point, the change in temperature is:
ΔT = boiling point - initial temperature
= 100.0°C - 0.0°C
= 100.0°C
Substituting the values into the formula, we have:
Q = 50.0 g * 4.18 J/g°C * 100.0°C
= 20900 J
Therefore, the heat required to heat the water to its boiling point is 20900 joules.
2. Vaporizing the water:
The heat required to vaporize the water is given by the formula:
Q = m * Hv
Where:
Q is the heat required (in joules),
m is the mass of water (in grams), and
Hv is the heat of vaporization of water (2260 J/g).
Substituting the values into the formula, we have:
Q = 50.0 g * 2260 J/g
= 113000 J
Therefore, the heat required to vaporize the water is 113000 joules.
Adding both values together, the total heat required to vaporize 50.0 g of water at 100.0°C is:
20900 J + 113000 J = 133900 J
Therefore, 133900 joules of heat are needed to vaporize 50.0 g of water at 100.0°C.
To determine the amount of heat required to vaporize a substance, you need to use the formula:
q = m * ΔH
Where:
q = heat energy
m = mass of the substance
ΔH = enthalpy of vaporization (also known as heat of vaporization)
First, let's find the enthalpy of vaporization for water. The enthalpy of vaporization for water is 40.7 kJ/mol.
Next, we need to convert the mass of water from grams to moles. To do this, we will use the molar mass of water, which is approximately 18.015 g/mol.
50.0 g of water is equal to 50.0 g / 18.015 g/mol ≈ 2.7754 mol.
Now we can calculate the heat energy using the formula:
q = m * ΔH
q = 2.7754 mol * 40.7 kJ/mol
q = 110.57 kJ
Therefore, approximately 110.57 kJ of heat energy is required to vaporize 50.0 g of water at 100.0 °C.