When one mole of C6H6 is burned, 3.27 MJ of heat is produced. When the heat from burning 7.87 g of C6H6 is added to 5.69 kg of water at 21.0°C, what is the final temperature of the water?
To find the final temperature of the water, we will use the concept of heat transfer.
First, let's find the amount of heat transferred from burning 7.87 g of C6H6. We know that 1 mole of C6H6 produces 3.27 MJ of heat. We need to convert the mass of C6H6 into moles:
1 mole of C6H6 = molar mass of C6H6 in grams
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 of C6H6
= 7.87 g / 78.11 g/mol
= 0.1007 mol
Now, let's calculate the heat transferred:
Heat transferred = moles of C6H6 * heat produced per mole
= 0.1007 mol * 3.27 MJ/mol
= 0.3279 MJ
According to the law of energy conservation, the heat transferred to the water will be equal to the heat absorbed by the water:
Heat transferred to water = mass of water * specific heat capacity of water * change in temperature
We know:
mass of water = 5.69 kg
specific heat capacity of water = 4.18 J/g°C
First, let's convert the mass of water into grams:
mass of water = 5.69 kg * 1000 g/kg
= 5690 g
Now, let's calculate the change in temperature:
Heat transferred to water = 5690 g * 4.18 J/g°C * (final temperature - 21.0°C)
We can rearrange the equation to solve for the final temperature:
(final temperature - 21.0°C) = Heat transferred to water / (5690 g * 4.18 J/g°C)
(final temperature - 21.0°C) = (0.3279 MJ * 10^6 J / 1 MJ) / (5690 g * 4.18 J/g°C)
(final temperature - 21.0°C) = (327900 / 5690 * 4.18)
(final temperature - 21.0°C) = 14.63
final temperature = 21.0°C + 14.63°C
= 35.63°C
Therefore, the final temperature of the water is 35.63°C.
To find the final temperature of the water, we can use the formula:
q = mcΔT
where q is the heat transferred, 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 moles of C6H6 burned:
Molar mass of C6H6 = 6(12.01 g/mol) + 6(1.008 g/mol) = 78.11 g/mol
Moles of C6H6 = mass of C6H6 (in grams) / molar mass of C6H6
= 7.87 g / 78.11 g/mol
≈ 0.101 mol
Now we need to calculate the heat transferred to the water:
Heat transferred = moles of C6H6 burned x heat produced per mole of C6H6
= 0.101 mol x 3.27 MJ/mol = 0.33027 MJ
Note: We converted the moles of C6H6 to megajoules (MJ).
Next, let's calculate the heat absorbed by the water:
Heat absorbed = mass of water x specific heat capacity of water x change in temperature
Given:
Mass of water = 5.69 kg = 5690 g
Specific heat capacity of water (c) = 4.18 J/g°C
Change in temperature (ΔT) = final temperature - initial temperature
Since the initial temperature is given as 21.0°C, we need to find the final temperature.
Substituting the known values into the formula, we can solve for the final temperature:
0.33027 MJ = 5690 g x 4.18 J/g°C x (final temperature - 21.0°C)
Simplifying the equation:
0.33027 MJ = 23754.2 J/°C x (final temperature - 21.0°C)
0.33027 MJ = 23754.2 J/°C x final temperature - 21.0°C x 23754.2 J/°C
0.33027 MJ = 5690 g x 4.18 J/g°C x final temperature - 21.0°C x 23754.2 J/°C
0.33027 MJ = 23754.2 J/°C x final temperature - 21.0°C x 23754.2 J/°C
0.33027 MJ = 5690 x 4.18 x final temperature - 21.0°C x 23754.2 J/°C
0.33027 MJ = 23754.2 J/°C x final temperature - 498244.2 J
Converting the left side of the equation into megajoules:
0.33027 MJ = 0.00033 MJ
0.33027 = 23754.2 x final temperature - 498244.2
Divide both sides of the equation by 23754.2:
0.00001388 = final temperature - 20.93°C
Add 20.93°C to both sides of the equation:
final temperature = 0.00001388 + 20.93°C
final temperature ≈ 20.93001388°C
Therefore, the final temperature of the water is approximately 20.930°C.