When one mole of C6H6 is burned, 3.27 MJ of heat is produced. When the heat from burning 5.43 g of C6H6 is added to 5.69 kg of water at 21.0°C, what is the final temperature of the water?
Well, I have to say this question burns with curiosity! Let's calculate the final temperature with a spark of humor.
First, we need to find the amount of heat released when 5.43 g of C6H6 is burned. Since we know that 1 mole of C6H6 produces 3.27 MJ of heat, let's convert the mass to moles:
5.43 g C6H6 x (1 mol C6H6 / molar mass of C6H6) = X mol C6H6.
Now, we can find the amount of heat produced:
X mol C6H6 x 3.27 MJ/mol = Y MJ of heat.
Next, we need to find how much heat is needed to raise the temperature of 5.69 kg of water from 21.0°C to the final temperature, T. We'll use the specific heat capacity of water (4.18 J/g°C).
The heat needed can be calculated as:
Heat = (mass of water) x (specific heat capacity of water) x (change in temperature).
Converting the mass of water to grams:
5.69 kg x 1000 g/kg = Z g of water.
So, the heat needed is:
Heat = Z g of water x 4.18 J/g°C x (T - 21.0°C).
Now, we can set up an equation:
Y MJ of heat released = Heat needed.
Let's convert Y MJ to J:
Y MJ x (1 MJ / 10^6 J) = J of heat.
Now we have:
J of heat = Z g of water x 4.18 J/g°C x (T - 21.0°C).
Solving for T:
(T - 21.0) = (J of heat) / (Z g of water x 4.18 J/g°C).
And finally:
T = (J of heat) / (Z g of water x 4.18 J/g°C) + 21.0°C.
Plug in the values for J of heat and Z, and you'll have your answer! Remember to enjoy the process like a fiery circus act!
To find the final temperature of the water, we can use the equation:
q = mcΔT
Where:
q is the heat absorbed or released
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
First, let's calculate the heat released by the burning of 5.43 g of C6H6.
To do this, we need to convert the mass of C6H6 to moles using its molar mass.
Molar mass of C6H6 = (12.01 g/mol × 6) + (1.01 g/mol × 6) = 78.11 g/mol
Number of moles of C6H6 = 5.43 g / 78.11 g/mol = 0.0696 mol
Next, we'll use the values given to calculate the heat released:
Heat released = q = 3.27 MJ = 3.27 × 10^6 J
Now, we'll calculate the heat absorbed by the water:
m_water = 5.69 kg = 5.69 × 10^3 g
ΔT = ? (to be solved)
c_water = 4.18 J/g°C (specific heat capacity of water)
Using the equation above and rearranging it to solve for ΔT:
q = mcΔT
ΔT = q / (m_water × c_water)
ΔT = (3.27 × 10^6 J) / (5.69 × 10^3 g × 4.18 J/g°C)
ΔT = 14.0989 °C
Finally, we calculate the final temperature of the water by adding the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 21.0°C + 14.0989°C
Final temperature ≈ 35.1°C
Therefore, the final temperature of the water is approximately 35.1°C.
To calculate the final temperature of the water, we need to determine the amount of heat transferred from the burning of C6H6 to the water. This can be done using the equation:
Q = mcΔT
Where:
Q is the heat transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
First, let's calculate the heat transferred from the burning of C6H6 using the given energy value of 3.27 MJ and the molar mass of C6H6.
1. Calculate the moles of C6H6:
Given mass of C6H6 = 5.43 g
Molar mass of C6H6 = 12.01 g/mol (C) + 1.01 g/mol (H)
= 78.11 g/mol
Moles of C6H6 = Mass of C6H6 / Molar mass of C6H6
= 5.43 g / 78.11 g/mol
2. Calculate the heat transferred from the burning of 5.43 g C6H6:
Heat transferred = Moles of C6H6 * Energy produced per mole
= (5.43 g / 78.11 g/mol) * 3.27 MJ
Next, we need to calculate the heat transferred to the water. The specific heat capacity of water is approximately 4.18 J/g°C.
3. Calculate the mass of water:
Given mass of water = 5.69 kg
Mass of water = mass of water * 1000 (to convert kg to g)
4. Calculate the change in temperature (ΔT):
The initial temperature of the water is 21.0°C. The final temperature is what we need to find.
Now, we can rearrange the equation Q = mcΔT to solve for ΔT:
ΔT = Q / (mwater * c)
Finally, we can substitute the values to find the final temperature:
Final temperature = Initial temperature + ΔT
By following these calculations, we can determine the final temperature of the water.
1 mol C6H6 = 78 grams.
Heat generated by 5.43 g is
3.27 MJ x (5.43/78) = ?
Substitute the answer denoted by ? in
? = mass H2O x specific heat H2O x (Tfinal - Tinitial).
Be sure specific heat and mass have appropriate units.