When one mole of C6H6 is burned, 3.27 MJ of heat is produced. When the heat from burning 7.47 g of C6H6 is added to 5.69 kg of water at 21.0°C, what is the final temperature of the water?
Well, when you're talking about burning mole, I have to ask: did you get a mole with a good sense of humor? Because if you didn't, it might be a blast to burn one. As for your question, let's heat things up!
First, we need to calculate the amount of heat produced when 7.47 g of C6H6 is burned. To do that, we'll use the given value of 3.27 MJ/mol. Now, since we have grams instead of moles, we need to convert that mass to moles.
The molar mass of C6H6 is about 78.11 g/mol, so we can find the number of moles by dividing the mass by the molar mass:
7.47 g / 78.11 g/mol ≈ 0.0956 mol
Now, we can determine the heat produced by multiplying the number of moles by the heat of combustion:
0.0956 mol × 3.27 MJ/mol ≈ 0.312 MJ
Next up, let's calculate the heat absorbed by the water when it increases in temperature. We can use the formula:
q = mcΔT
Where q is the heat absorbed, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The mass of the water is given as 5.69 kg, which can be converted to grams:
5.69 kg × 1000 g/kg = 5690 g
The specific heat capacity of water is approximately 4.18 J/g°C, and the change in temperature is:
Final temperature - Initial temperature
Tf - 21.0°C
Now, assuming that the heat from the burning C6H6 is fully absorbed by the water, we can set up the equation:
0.312 MJ = 5690 g × 4.18 J/g°C × (Tf - 21.0°C)
Solving for Tf:
Tf - 21.0°C ≈ 0.312 MJ / (5690 g × 4.18 J/g°C)
Tf - 21.0°C ≈ 0.0131°C
Therefore, the final temperature of the water would be approximately 21.0131°C. So, don't worry, the water wouldn't start boiling like crazy after this reaction. It's just a small, humorous increase in temperature! Stay cool!
To find the final temperature of the water, we can use the equation:
q = mcΔT
where q is the heat gained or lost by the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
First, let's calculate the heat gained by the water. We know that 3.27 MJ of heat is produced when one mole of C6H6 is burned.
To find the heat produced when 7.47 g of C6H6 is burned, we need to convert the mass of C6H6 to moles and then use the heat of combustion.
First, calculate the moles of C6H6:
molar mass of C6H6 = 6(12.01 g/mol) + 6(1.01 g/mol) = 78.11 g/mol
moles of C6H6 = mass of C6H6 / molar mass = 7.47 g / 78.11 g/mol
Now, calculate the heat produced when burning 7.47 g of C6H6:
heat produced = moles of C6H6 × heat of combustion = moles of C6H6 × 3.27 MJ/mol
Next, calculate the heat gained by the water:
q = heat produced
Now, let's calculate the mass of water:
mass of water = 5.69 kg
The specific heat capacity of water is approximately 4.18 J/g°C.
Now, rearrange the equation q = mcΔT to solve for ΔT:
ΔT = q / (mc)
Substitute the known values:
ΔT = (heat produced) / (mass of water × specific heat capacity of water)
Finally, calculate the final temperature of the water:
final temperature = initial temperature + ΔT
Please provide the initial temperature of the water for a complete solution.
To find the final temperature of the water after adding heat from burning C6H6, we can use the principle of heat transfer. The heat gained by the water will be equal to the heat lost by C6H6.
Let's break down the problem step by step:
Step 1: Calculate the heat gained by the water.
The formula to calculate the heat gained by an object is: Q = mcΔT, where
Q = heat gained (in Joules)
m = mass of the object (in kg)
c = specific heat capacity of the object (in J/g°C or J/kg°C)
ΔT = change in temperature (in °C)
In this case, we want to calculate the heat gained by water, so the values are:
Q = ?
m = 5.69 kg (mass of water)
c = 4.18 J/g°C (specific heat capacity of water)
ΔT = final temperature - initial temperature = final temperature - 21.0°C
Step 2: Calculate the moles of C6H6.
To do this, we need the molar mass of C6H6 (benzene), which is:
C6H6: 6(12.01 g/mol) + 6(1.01 g/mol) = 78.11 g/mol
To convert grams of C6H6 to moles, we will use the molar mass:
moles of C6H6 = mass of C6H6 / molar mass of C6H6
In this case:
mass of C6H6 = 7.47 g
Step 3: Calculate the heat lost by burning C6H6.
The heat lost by burning one mole of C6H6 is given as 3.27 MJ.
To calculate the heat lost by burning 7.47 g (the amount used), we need to convert grams to moles and then multiply by the heat lost per mole.
moles of C6H6 = mass of C6H6 / molar mass of C6H6
heat lost by burning = moles of C6H6 * heat lost per mole
Step 4: Equate the heat gained by water to the heat lost by C6H6:
Since the heat gained by water is equal to the heat lost by C6H6, we can equate the two.
Q_water = heat lost by C6H6
Step 5: Solve for the final temperature of water.
Now we have all the information we need to solve for the final temperature of the water.
Q_water = mcΔT_water
heat lost by C6H6 = Q_water
Rearranging the equation:
mcΔT_water = heat lost by C6H6
Solving for ΔT_water:
ΔT_water = heat lost by C6H6 / (mc)
Finally, we can find the final temperature by adding ΔT_water to the initial temperature:
final temperature = initial temperature + ΔT_water
How much heat do you get from the 7.47 g C6H6 if you get 3.27E6 J from 1 mol C6H6 (78 grams)? That's 3.27E6 J x (7.47/78) = ? = q
Then q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
I would convert kg H2O to g and use specific heat as 4.184 J/g*C and solve for Tfinal after substitution of the numbers.