Calculate the heat (in kJ) required to change 40.0g of ice at -30.0 deg C to steam at 150.0 deg C?
Two formulas you need.
Within a phase is
mass x specific heat x (Tfinal-Tinitial)
Use that formula for ice at -30 to ice at zero, for water at zero to water at 100, for steam at 100 to steam at 150.
At the phase change.
mass x heat fusion for melting.
mass x heat vaporization for boiling.
Then add each part to find the total.
-760kJ cal
To calculate the heat required to change the state of a substance, we need to consider the following steps:
1. Heat required to raise the temperature of ice from -30.0°C to 0°C.
2. Heat required to melt the ice at 0°C.
3. Heat required to raise the temperature of water from 0°C to 100°C.
4. Heat required to convert the water at 100°C to steam at 100°C.
5. Heat required to raise the temperature of steam from 100°C to 150°C.
Now, let's calculate each step one by one:
Step 1:
The specific heat capacity of ice is 2.09 J/g°C.
The mass of ice is 40.0 g.
The temperature change is 0 - (-30) = 30°C.
Using the formula Q = mcΔT (where Q is the heat, m is mass, c is specific heat capacity, and ΔT is the temperature change), let's calculate the heat required to raise the temperature of ice from -30.0°C to 0°C:
Q1 = (40.0 g) * (2.09 J/g°C) * (30°C)
Q1 = 2514 J
Step 2:
The heat of fusion for water is 334 J/g.
Using the same mass of ice (40.0 g), let's calculate the heat required to melt the ice at 0°C:
Q2 = (40.0 g) * (334 J/g)
Q2 = 13360 J
Step 3:
The specific heat capacity of water is 4.18 J/g°C.
The mass of water is also 40.0 g.
The temperature change is 100°C - 0°C = 100°C.
Using the same formula, let's calculate the heat required to raise the temperature of water from 0°C to 100°C:
Q3 = (40.0 g) * (4.18 J/g°C) * (100°C)
Q3 = 16720 J
Step 4:
The heat of vaporization for water is 2260 J/g.
Using the same mass of water (40.0 g), let's calculate the heat required to convert the water at 100°C to steam at 100°C:
Q4 = (40.0 g) * (2260 J/g)
Q4 = 90400 J
Step 5:
The specific heat capacity of steam is 2.03 J/g°C.
The mass of steam can be calculated using the principle of conservation of mass: mass of steam = mass of ice = 40.0 g.
The temperature change is 150°C - 100°C = 50°C.
Using the same formula, let's calculate the heat required to raise the temperature of steam from 100°C to 150°C:
Q5 = (40.0 g) * (2.03 J/g°C) * (50°C)
Q5 = 4060 J
Finally, let's add up all the heat values to get the total heat required:
Total Heat = Q1 + Q2 + Q3 + Q4 + Q5
Total Heat = 2514 J + 13360 J + 16720 J + 90400 J + 4060 J
Total Heat = 124054 J
To convert the answer to kJ:
Total Heat = 124054 J / 1000
Total Heat ≈ 124.1 kJ
Therefore, the heat required to change 40.0g of ice at -30.0°C to steam at 150.0°C is approximately 124.1 kJ.
To calculate the heat required for this problem, we need to consider three different processes: heating the ice to its melting point, melting the ice into water, and heating the water to its boiling point, and then converting the water into steam.
First, let's calculate the heat required to warm the ice from its initial temperature (-30.0°C) to its melting point (0°C).
The heat required to warm a substance can be calculated using the formula:
Q = m × C × ΔT
Where:
Q = heat (in joules)
m = mass of the substance (in grams)
C = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
The specific heat capacity of ice is approximately 2.09 J/g°C.
So, the heat required to warm the ice to its melting point can be calculated as:
Q1 = 40.0g × 2.09 J/g°C × (0°C - (-30.0°C))
Next, let's calculate the heat required to melt the ice. The heat of fusion (also known as the heat of melting) is the amount of heat required to convert a substance from a solid to a liquid at its melting point. For water, the heat of fusion is approximately 334 J/g.
The heat required to melt the ice can be calculated as:
Q2 = 40.0g × 334 J/g
Finally, let's calculate the heat required to heat the water from its melting point (0°C) to its boiling point (100°C).
The specific heat capacity of water is approximately 4.18 J/g°C.
So, the heat required to heat the water can be calculated as:
Q3 = 40.0g × 4.18 J/g°C × (100.0°C - 0°C)
Now, let's sum up all the calculated heat values:
Total Heat = Q1 + Q2 + Q3
To convert the total heat from joules to kilojoules, divide the result by 1000:
Total Heat (in kJ) = (Q1 + Q2 + Q3) / 1000
Now, you can substitute the given values into the equations above and calculate the answer.