If 5.300 g of C6H6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final temperature of the water?
2C6H6(l) +15O2 (g)----> 12CO2(g)+6H2O(l) +6542
To determine the final temperature of the water, we can use the equation:
q = m * c * ΔT
Where:
- q represents the heat absorbed or released (in joules)
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in J/g°C)
- ΔT is the change in temperature (in °C)
First, we need to calculate the heat produced from the burning of C6H6. The balanced equation tells us that for every 2 moles of C6H6, 6 moles of H2O are produced. Hence, we can convert the mass of C6H6 to moles using its molar mass.
Molar mass of C6H6:
6 carbon atoms * atomic mass of carbon + 6 hydrogen atoms * atomic mass of hydrogen
= (6 * 12.01 g/mol) + (6 * 1.01 g/mol)
= 78.11 g/mol
Moles of C6H6:
mass of C6H6 / molar mass of C6H6
= 5.300 g / 78.11 g/mol
≈ 0.06784 mol
According to the equation, 2 moles of C6H6 release 6542 J of heat. Therefore, we can calculate the heat released from 0.06784 mol of C6H6 as follows:
Heat released = (0.06784 mol * 6542 J) / 2 mol
≈ 225 J
Now we can use the equation q = m * c * ΔT to find the change in temperature for the water. We are given the mass of water (m_water) as 5691 g and the initial temperature (T_initial) as 21 °C. The specific heat capacity of water (c_water) is 4.18 J/g°C.
Substituting the known values:
225 J = (5691 g) * (4.18 J/g°C) * (ΔT)
To solve for ΔT, divide both sides of the equation by (5691 g * 4.18 J/g°C):
225 J / (5691 g * 4.18 J/g°C) = ΔT
≈ 0.0108 °C
Therefore, the final temperature of the water would be approximately 21.0108 °C.
Heat produced by 5.20 g C6H6 is
6542 WHAT. Joules? kJ? I will assume J and you can make th adjustment.
6542 J x (5.30/2*molar mass C6H6 = ? J or kJ. If kJ change to J.
Then q = J = = mass H2O x specific heat H2O x (Tfinial-Tinitial).
Subtitute and solve for Tfinal