What are the Kilojoules needed to melt 24.0g of ice at O degrees C, warm the liquid to 100 degree C and change it to steam at 100 degree C?
To calculate the total amount of energy required to melt the ice, warm the liquid water, and convert it to steam, we need to consider the different phases of water and their respective energy requirements.
First, we need to calculate the energy required to melt the ice. This can be done using the equation:
Q = m * ΔH_fusion
Where:
Q is the energy in joules
m is the mass of the ice in grams
ΔH_fusion is the heat of fusion, which is the amount of energy required to change a substance from a solid to a liquid phase.
The heat of fusion for water is approximately 334 J/g.
Plugging in the values, we get:
Q1 = 24.0 g * 334 J/g
Q1 ≈ 8016 J
Next, we need to calculate the energy required to warm the liquid water from 0°C to 100°C. This can be done using the equation:
Q = m * c * ΔT
Where:
Q is the energy in joules
m is the mass of the liquid water in grams
c is the specific heat capacity of water, which is approximately 4.18 J/g°C
ΔT is the change in temperature in degrees Celsius
The change in temperature (ΔT) is calculated as:
ΔT = final temperature - initial temperature
ΔT = 100°C - 0°C
ΔT = 100°C
Therefore, the energy required to warm the liquid water can be calculated as:
Q2 = 24.0 g * 4.18 J/g°C * 100°C
Q2 ≈ 10032 J
Lastly, we need to calculate the energy required to convert the liquid water to steam at 100°C. This can be done using the equation:
Q = m * ΔH_vaporization
Where:
Q is the energy in joules
m is the mass of the liquid water in grams
ΔH_vaporization is the heat of vaporization, which is the amount of energy required to change a substance from a liquid to a gas phase.
The heat of vaporization for water is approximately 2260 J/g.
Plugging in the values, we get:
Q3 = 24.0 g * 2260 J/g
Q3 ≈ 54240 J
To find the total energy required, we sum up the three calculated values:
Total energy = Q1 + Q2 + Q3
Total energy ≈ 8016 J + 10032 J + 54240 J
Total energy ≈ 72288 J
Therefore, the total energy required to complete the described process is approximately 72288 joules or 72.29 kilojoules (rounded to two decimal places).
To find the total energy required to melt the ice, warm the liquid, and change it to steam, we need to calculate the energy required for each step and then add them together.
Step 1: Melting the ice
The energy required to melt ice is given by the equation:
Q1 = mass × heat of fusion
Given:
mass of ice (m) = 24.0g
heat of fusion for ice (ΔH1) = 334 J/g
Converting mass from grams to kilograms:
mass = 24.0g ÷ 1000g/kg = 0.024 kg
Calculating the energy required to melt the ice:
Q1 = 0.024 kg × 334 J/g = 7.616 J
Step 2: Warming the liquid
The energy required to warm the liquid from 0 degrees Celsius to 100 degrees Celsius can be calculated using the equation:
Q2 = mass × specific heat capacity × change in temperature
Given:
mass of liquid (m) = mass of ice = 24.0g = 0.024 kg
specific heat capacity for water (c) = 4.184 J/g°C
change in temperature (ΔT) = 100°C
Calculating the energy required to warm the liquid:
Q2 = 0.024 kg × 4.184 J/g°C × 100°C = 10.036 J
Step 3: Changing the liquid to steam
The energy required to change the liquid to steam is given by the equation:
Q3 = mass × heat of vaporization
Given:
mass of liquid = mass of ice = 24.0g = 0.024 kg
heat of vaporization for water (ΔH2) = 2260 J/g
Calculating the energy required to change the liquid to steam:
Q3 = 0.024 kg × 2260 J/g = 54.24 J
Finally, adding all the energies together:
Total energy (Q total) = Q1 + Q2 + Q3
Q total = 7.616 J + 10.036 J + 54.24 J = 71.892 J
Therefore, the total energy required to melt 24.0g of ice, warm the liquid to 100 degrees Celsius, and change it to steam at 100 degrees Celsius is approximately 71.892 Joules (J).