find the temperature increase in C expected for 1.00 L of water when it absorbs all of the energy from the combustion of 1.00 g of acetylene, C2H2(g). (Every mole of acetylene releases 1300KJ of energy when combusted)
2C2H2 + 3O2 ==> 2CO2 + 2H2O
dH = 1300 kJ/mol C2H2
How many mols C2H2 do we have? That's mols g/molar mass = 1.0/26 = approx 0.038 but you should redo that since it is an estimate.
The q = dH = mass H2O x specific heat H2O x delta T. Solved for dT.
To find the temperature increase in Celsius (ΔT) for 1.00 L of water when it absorbs the energy from the combustion of 1.00 g of acetylene (C2H2), we can use the concept of specific heat capacity and the equation:
q = m * c * ΔT
Where:
q is the amount of heat transferred
m is the mass of the substance (in this case, water)
c is the specific heat capacity of the substance (for water, it is approximately 4.18 J/g°C)
ΔT is the change in temperature
First, we need to calculate the amount of energy (q) released by combusting 1.00 g of acetylene. Since every mole of acetylene releases 1300 kJ of energy when combusted, we need to convert the mass to moles and then multiply by the energy released per mole.
1.00 g C2H2 * (1 mole C2H2 / 26.04 g C2H2) * (1300 kJ / 1 mole C2H2) = x kJ
Next, convert the energy from kJ to J by multiplying by 1000:
x kJ * (1000 J / 1 kJ) = y J
Now, we can calculate the temperature increase (ΔT) of the water using the specific heat capacity (c). Since the specific heat capacity of water is approximately 4.18 J/g°C and we have 1.00 L (which is equivalent to 1000 g) of water:
q = m * c * ΔT
y J = 1000 g * 4.18 J/g°C * ΔT
Solve the equation for ΔT:
ΔT = y J / (1000 g * 4.18 J/g°C)
Now, calculate ΔT using the calculated value of y:
ΔT = y J / (1000 g * 4.18 J/g°C)
Finally, you will have the temperature increase (in °C) expected for 1.00 L of water when it absorbs all of the energy from the combustion of 1.00 g of acetylene.