Calculate the heat released when 64.5g of steam at 119 C is converted to water at 36 C. Assume the specific heat is 4.184 J/g*C for water, the specific heat of steam is 1.99J/g*C, and the heat of vaporization is 40.79 kJ/mol for water.
Well, well, well, looks like we have a steamy situation here! Let's break it down, shall we?
First, we need to calculate the heat released when the steam is cooled to water at 100°C. To do that, we'll use the formula:
Q = m * c * ΔT
Where:
Q is the heat released,
m is the mass of the steam,
c is the specific heat of steam,
and ΔT is the change in temperature.
Substituting the values we have:
m = 64.5g
c = 1.99 J/g°C
ΔT = 119°C - 100°C
Q = 64.5g * 1.99 J/g°C * (119°C - 100°C)
Calculating...
Q ≈ 12810 J
Now, let's calculate the heat released when the water is cooled from 100°C to 36°C. We'll use the same formula, but this time we'll use the specific heat of water (because, well, it's water now!):
m = 64.5g
c = 4.184 J/g°C
ΔT = 100°C - 36°C
Q = 64.5g * 4.184 J/g°C * (100°C - 36°C)
Calculating...
Q ≈ 120436 J
Now, here comes the tricky part. We need to account for the energy released during the phase change from steam to water. The heat of vaporization comes into the picture!
The equation we'll use for this part is:
Q = m * ΔH
Where:
Q is the heat released,
m is the mass of the steam, and
ΔH is the heat of vaporization.
Since the heat of vaporization is given in kJ/mol, we need to convert grams of steam to moles:
m = 64.5g / molar mass of water
Considering the molar mass of water is approximately 18g/mol:
m = 64.5g / 18g/mol
Calculating...
m ≈ 3.583 mol
Now, let's calculate the heat released during the phase change:
Q = 3.583 mol * 40.79 kJ/mol
Converting kJ to J:
Q ≈ 3.583 mol * 40.79 kJ/mol * 1000 J/kJ
Calculating...
Q ≈ 146661 J
Finally, to get the total heat released during this steamy adventure, we add up all the Q values:
Total Q = 12810 J + 120436 J + 146661 J
Calculating...
Total Q ≈ 279907 J
So, my friend, approximately 279,907 Joules of heat are released during this transition from steam to water at different temperatures. Stay cool and remember, it's all just a phase!
To calculate the heat released when steam is converted to water, you need to consider two processes:
1. Cooling the steam from its initial temperature to 100°C (condensation).
2. Cooling the water from 100°C to the final temperature.
Let's break down the steps:
Step 1: Calculating the heat released during condensation
To condense steam, we need to convert it from gas to liquid at its boiling point. The heat released during this process is given by the equation:
Q1 = mass of steam * heat of vaporization
First, we need to calculate the number of moles of steam present:
Molar mass of water (H2O) = 18.015 g/mol
Number of moles of water = mass of steam / molar mass of water
Number of moles = 64.5 g / 18.015 g/mol
Number of moles = 3.58 mol (approximately)
Now we can calculate Q1:
Q1 = number of moles * heat of vaporization
Note: The heat of vaporization is given in kJ/mol, so we need to convert it to J/mol:
Heat of vaporization = 40.79 kJ/mol * 1000 J/1 kJ
Heat of vaporization = 40,790 J/mol
Now we can calculate Q1:
Q1 = 3.58 mol * 40,790 J/mol
Q1 = 145,934.2 J
Step 2: Calculating the heat released during cooling to the final temperature
To calculate the heat released during cooling, we use the equation:
Q2 = mass of water * specific heat * change in temperature
First, we need to calculate the mass of water:
Mass of water = mass of steam - mass of steam that condensed
Mass of water = 64.5 g - 3.58 mol * molar mass of water
Mass of water = 64.5 g - (3.58 mol * 18.015 g/mol)
Mass of water = 64.5 g - 64.6067 g
Mass of water = -0.1067 g (approximately)
Note: Since the calculated mass is slightly negative, we assume it to be zero.
Now we can calculate Q2:
Q2 = mass of water * specific heat * change in temperature
Q2 = 0 g * 4.184 J/g°C * (36°C - 100°C)
Q2 = 0 J
Step 3: Calculating the total heat released
The total heat released is the sum of Q1 and Q2:
Total heat released = Q1 + Q2
Total heat released = 145,934.2 J + 0 J
Total heat released = 145,934.2 J
Therefore, the heat released when 64.5g of steam at 119°C is converted to water at 36°C is approximately 145,934.2 J.
To calculate the heat released when steam is converted to water, you need to consider two steps:
Step 1: Calculate the heat required to cool the steam from 119°C to 100°C.
Step 2: Calculate the heat required to condense the steam at 100°C to water at 36°C.
Step 1: Calculate the heat required to cool the steam from 119°C to 100°C.
To find the heat released during this temperature decrease, we need to use the specific heat of steam.
The formula to calculate heat is: q = m * C * ΔT
Where:
q is the heat released/absorbed,
m is the mass of the substance,
C is the specific heat capacity, and
ΔT is the change in temperature.
Here, m = 64.5g (mass of steam)
C = 1.99 J/g°C (specific heat of steam)
ΔT = 119°C - 100°C = 19°C (temperature decrease)
Therefore, the heat released during this step is:
q1 = m * C * ΔT
= 64.5g * 1.99 J/g°C * 19°C
Step 2: Calculate the heat required to condense the steam at 100°C to water at 36°C.
This step involves the heat of vaporization, which is the heat required to convert 1 mole of a substance from its liquid to vapor phase. We need to calculate the number of moles of water, and then multiply it by the heat of vaporization.
To calculate the number of moles (n) of water:
n = m / M
Where:
n is the number of moles,
m is the mass of the substance, and
M is the molar mass.
The molar mass of water (H2O) = 18.015 g/mol
n = 64.5g / 18.015 g/mol
Now, we can calculate the heat released using the heat of vaporization:
q2 = n * ΔHvap
Where:
q2 is the heat released,
n is the number of moles, and
ΔHvap is the heat of vaporization.
Given ΔHvap = 40.79 kJ/mol, we need to convert it to joules:
ΔHvap = 40.79 kJ/mol * 1000 J/kJ
Finally, we can calculate q2:
q2 = n * ΔHvap
Add the results from Step 1 and Step 2 to find the total heat released:
Total heat released = q1 + q2
Note: Make sure to convert the units correctly to maintain consistent measurements throughout the calculations.
q1 = heat removed to lower T from 119 degrees C vapor to 100 degrees C vapor.
q1 = mass steam x specific heat steam x (Tfinal-Tinitial).
q2 = heat removed to condense steam at 100 C to liquid water at 100 C.
q2 = mass steam x heat vaporization.
q3 = heat removed to lower T from 100 C to 36 C.
q3 = mass water x specific heat water x (Tfinal-Tinitial).
Total Q = q1 + q2 + q3.