A 1.55 gram sample of methanol (CH3OH) is combusted in a bomb calorimeter. The molar heat of combustion of methanol is -725KJ/mole, assuming that 2.0L of water initially at 25.0 C absorbs all of the heat of combustion, what is the final temperature of the water?
2CH3OH + 3O2 ==> 2CO2 + 4H2O
How many heat is emitted by 1.55 g CH3OH? That will be
725 kJ x (1.55 g/(2*molar mass CH3OH) kJ. Change that to J, then
J = mass H2O x specific heat H2O x (Tfinal-25). Solve for Tf. I think the answer is close to 27 C.
To find the final temperature of the water, we can use the equation:
q = m * c * ΔT
where:
q is the heat absorbed by the water
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature of the water
First, we need to find the heat absorbed by the water. We know that the heat of combustion of methanol is -725 KJ/mole. However, we have a 1.55 gram sample of methanol, which is not in moles. We need to convert the mass of methanol into moles.
To convert grams to moles, we use the molar mass of methanol (CH3OH):
Molar mass of CH3OH = (1 * 12.01 g/mol) + (4 * 1.01 g/mol) + (16.00 g/mol) = 32.04 g/mol
Now, we can calculate the number of moles of methanol:
moles of CH3OH = mass / molar mass = 1.55 g / 32.04 g/mol
Next, we need to find the heat absorbed by the water using the equation:
q = moles of CH3OH * heat of combustion
Now that we have the heat absorbed by the water, we can solve for the change in temperature (ΔT) of the water using the equation:
ΔT = q / (m * c)
We know that the mass of water is 2.0 L (which is equivalent to 2000 grams, assuming the density of water is 1 g/mL) and the specific heat capacity of water is approximately 4.18 J/g°C.
Substitute the values into the equation to find the change in temperature (ΔT):
ΔT = q / (2000 g * 4.18 J/g°C)
Finally, subtract the change in temperature from the initial temperature of the water (25.0°C) to find the final temperature of the water.