How much heat in kilojoules is evolved in converting 1.00 mol of steam at 145 Celsius to ice at -50 Celsius?
Please, all particular steps and conversions between mole and grams of steam, H2O and ice. Thanks!
q1 = heat evolved in moving steam from 145 to 100.
q1 = mass steam x specific heat steam x (Tfinal-Tinital) where Tfinal is 100 and Tinitial is 145.
q2 = heat evolved when steam at 100 C condenses to liquid at 100 C.
q2 = mass x heat vaporization
q3 = heat evolved in moving water from 100 C to zero.
q3 = mass water x specific heat water x (Tfinal-Tinitial) where T final is zero and Tinitial is 100.
q4 = heat evolved on freezing water at zero to ice at zero.
q4 = mass water x heat fusion.
q5 = heat evolved in moving ice from zero to -50.
q5 = mass ice x specific heat ice x (Tfinal-Tinitial) where Tfinal is -50 and Tinitial is zero.
Total Q = q1+q2+q3+q4+q5
57.8kJ
To calculate the heat evolved in converting 1.00 mol of steam at 145 Celsius to ice at -50 Celsius, we need to consider the various steps involved.
Step 1: Convert moles of steam to grams of steam.
Given:
- 1 mol of steam
We can use the molar mass of water to convert moles to grams.
The molar mass of water (H2O) is approximately 18.015 g/mol.
So, 1 mol of steam is equal to 18.015 grams of steam.
Step 2: Calculate the heat evolved in cooling steam from 145 Celsius to 100 Celsius.
Given:
- Specific heat capacity of steam (H2O): 2.03 J/g·°C
- Initial temperature (T1): 145°C
- Final temperature (T2): 100°C
We can use the formula:
Q = m × c × ΔT
Where:
- Q is the heat transferred (in Joules)
- m is the mass of the substance (in grams)
- c is the specific heat capacity (in J/g·°C)
- ΔT is the change in temperature (in °C)
First, we need to find the mass of steam. We know that 1 mol of steam is equal to 18.015 grams of steam from Step 1. Therefore, 1.00 mol of steam is equal to 18.015 grams as well.
Next, we can calculate the heat evolved in cooling steam from 145°C to 100°C using the formula:
Q1 = m × c × ΔT1
Q1 = 18.015 g × 2.03 J/g·°C × (100°C - 145°C)
Step 3: Calculate the heat evolved in converting steam from 100°C to ice at 0°C (phase change).
Given:
- Heat of vaporization of water: 40.7 kJ/mol
We can use the molar heat of vaporization to calculate the heat evolved during the phase change.
We already know that 1.00 mol of steam is equal to 18.015 grams (Step 1).
Now we need to convert grams to moles:
1.00 mol = 18.015 g of steam
Using the molar heat of vaporization, we can calculate the heat evolved during the phase change:
Q2 = (18.015 g ÷ 18.015 g/mol) × 40.7 kJ/mol
Step 4: Calculate the heat evolved in cooling condensed steam (liquid water) from 0°C to -50°C.
Given:
- Specific heat capacity of water (H2O): 4.18 J/g·°C
- Initial temperature (T1): 0°C
- Final temperature (T2): -50°C
We can use the same formula as in Step 2:
Q3 = m × c × ΔT3
Q3 = (18.015 g ÷ 18.015 g/mol) × 4.18 J/g·°C × (-50°C - 0°C)
Step 5: Calculate the total heat evolved.
Now we need to calculate the total heat evolved by summing up the heat from each step:
Total heat evolved (Q total) = Q1 + Q2 + Q3
Q total = Q1 + Q2 + Q3
Please note that when plugging the values into the equations, it's important to keep track of units and convert when necessary (e.g., kJ to J, °C to Kelvin).
I will now perform the calculations for you.
To calculate the heat evolved in converting steam to ice, we need to consider the specific heat capacity and phase changes of water. Here are the steps to find the answer:
1. Determine the specific heat capacity of steam:
The specific heat capacity of steam at constant pressure is approximately 2.03 J/(g*C°).
2. Convert the temperature of steam from Celsius to Kelvin:
Given that the temperature of steam is 145°C, you need to add 273.15 to convert it to Kelvin. So, the temperature becomes 418.15 K.
3. Calculate the heat required to cool the steam to 0°C:
The heat transfer equation is given by:
Heat (q) = mass (m) * specific heat (c) * change in temperature (ΔT)
Since we have 1.00 mol of steam, we need to convert mols to grams by using the molar mass of water.
The molar mass of water (H2O) is approximately 18.015 g/mol.
So, 1.00 mol of water is equal to 18.015 g.
ΔT = 418.15 K - 273.15 K = 145 K
Now, substituting the values into the equation:
q1 = (18.015 g) * (2.03 J/(g°C)) * 145 K
4. Calculate the heat required for the phase change from water to ice at 0°C:
The enthalpy of fusion of water is approximately 6.01 kJ/mol.
To convert this to joules, we multiply by 1000:
Enthalpy of fusion (ΔH) = 6.01 kJ/mol * 1000 = 6010 J/mol
Since we have 1.00 mole of water, we multiply this by ΔH:
q2 = 6010 J/mol
5. Calculate the heat required to cool the ice to -50°C:
The specific heat capacity of ice at constant pressure is approximately 2.09 J/(g°C).
We have already determined that converting 1.00 mole of water to ice requires 18.015 g of water.
ΔT = 273.15 K - (-50°C + 273.15) K = 323.15 K
q3 = (18.015 g) * (2.09 J/(g°C)) * 323.15 K
6. Add up all the values obtained in steps 3, 4, and 5 to find the total heat evolved:
Total heat evolved = q1 + q2 + q3
Make sure to convert all the values to kilojoules (kJ) if necessary.