Calculate the change in heat when 3.05g of water vapor (steam) at 100.0 degrees celsius condenses to liquid water and then cools to 18.50 degrees celsius.
q1 = heat released by steam at 100 C condensing to water at 100 C..
q1 = mass steam x delta Hvap.
q2 = heat released in cooling from 100 C to 18.5 C.
q2 = mass water x specific heat water x (Tfinal-Tinitial).
Total heat released
qtotal = q1 + q2.
1040kj
To calculate the change in heat when water vapor condenses and cools, we need to consider the different processes involved: condensation and cooling.
First, let's calculate the heat released during the condensation process. The formula for calculating the heat released during condensation is:
Q = m * ΔHv
Where:
Q is the heat released
m is the mass of the substance
ΔHv is the heat of vaporization
The heat of vaporization for water is 40.7 kJ/mol.
To find the moles of water vapor, we need to convert the mass of water vapor to moles using the molar mass of water:
Molar mass of water (H2O) = 18.015 g/mol
moles = mass / molar mass
moles = 3.05 g / 18.015 g/mol
moles ≈ 0.1695 mol
Now, we can calculate the heat released during condensation:
Q_condensation = moles * ΔHv
Q_condensation = 0.1695 mol * 40.7 kJ/mol
Q_condensation ≈ 6.92 kJ
Next, let's calculate the heat released during cooling. The formula for calculating the heat released during cooling is:
Q = m * c * ΔT
Where:
Q is the heat released
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
The specific heat capacity of water is approximately 4.18 J/g°C.
Now, we need to convert the mass of water from grams to kilograms, as the specific heat capacity is in joules per gram-degree Celsius:
mass_water = 3.05 g / 1000
mass_water ≈ 0.00305 kg
Then, we can calculate the heat released during cooling:
Q_cooling = mass_water * c * ΔT
ΔT = final temperature - initial temperature
ΔT = 18.50°C - 100.0°C
ΔT ≈ -81.50°C
Q_cooling = 0.00305 kg * 4.18 J/g°C * -81.50°C
Q_cooling ≈ -10.11 kJ
Finally, we can calculate the total change in heat by summing the heat released during condensation and cooling:
Change in heat = Q_condensation + Q_cooling
Change in heat ≈ 6.92 kJ + -10.11 kJ
Change in heat ≈ -3.19 kJ
Therefore, the change in heat when 3.05 g of water vapor condenses to liquid water and cools to 18.50°C is approximately -3.19 kJ. Note that the negative sign indicates that heat is being released or lost.
To calculate the change in heat when water vapor condenses and cools, we need to consider two separate processes: condensation and cooling.
Step 1: Calculate the heat released during condensation.
The heat released during condensation can be calculated using the formula:
Q = m × ΔHvap
Where:
Q: Heat energy released (in Joules)
m: Mass of the substance (in grams)
ΔHvap: Heat of vaporization (in J/g)
In this case, the mass of water vapor is given as 3.05 grams. The heat of vaporization for water is typically 40.7 J/g.
Q1 = 3.05 g × 40.7 J/g
Step 2: Calculate the heat released during cooling.
The heat released during cooling can be calculated using the formula:
Q = m × C × ΔT
Where:
Q: Heat energy released (in Joules)
m: Mass of the substance (in grams)
C: Specific heat capacity (in J/g°C)
ΔT: Change in temperature (in °C)
In this case, the mass of water is also 3.05 grams. The specific heat capacity of water is approximately 4.18 J/g°C. The change in temperature is from 100.0°C to 18.50°C.
Q2 = 3.05 g × 4.18 J/g°C × (100.0°C - 18.50°C)
Step 3: Calculate the total change in heat.
To find the total change in heat, we add the heat released during condensation (Q1) and the heat released during cooling (Q2).
Total Change in Heat = Q1 + Q2
Now, let's plug in the values and calculate it:
Q1 = 3.05 g × 40.7 J/g = 124.435 J
Q2 = 3.05 g × 4.18 J/g°C × (100.0°C - 18.50°C) = 320.8285 J
Total Change in Heat = 124.435 J + 320.8285 J = 445.2635 J
Therefore, the total change in heat when 3.05g of water vapor condenses and cools is approximately 445.26 Joules.