To find out how much water can be heated by the combustion of ethanol, we need to calculate the amount of heat produced by the combustion reaction and then use this value to determine the change in temperature of the water.
1. Calculate the moles of ethanol used:
Molar mass of ethanol (C2H5OH) = 46.07 g/mol
Moles of ethanol = mass of ethanol / molar mass of ethanol
Moles of ethanol = 48.2 g / 46.07 g/mol
Moles of ethanol = 1.046 mol
2. Calculate the heat produced by the combustion of ethanol:
Heat produced = Moles of ethanol * Molar enthalpy of combustion
Heat produced = 1.046 mol * (-1234.8 kJ/mol)
Heat produced = -1290.08 kJ
3. Calculate the amount of heat transferred to the water:
Q = mcΔT
where Q is the heat transferred, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is approximately 4.18 J/g°C (or 4.18 kJ/kg°C).
Mass of water = molar mass of water * moles of water
molar mass of water = 18.015 g/mol
Mass of water = 18.015 g/mol * Moles of water
Since we don't know the moles of water yet, we have to assume that all the heat produced by the combustion is transferred to the water.
4. Calculate the moles of water using the equation for molar enthalpy of combustion:
Molar enthalpy of combustion of ethanol = heat produced / moles of water
-1290.08 kJ = heat produced / moles of water
moles of water = heat produced / molar enthalpy of combustion of ethanol
moles of water = -1290.08 kJ / -1234.8 kJ/mol
moles of water ≈ 1.045 mol
5. Calculate the mass of water:
Mass of water = moles of water * molar mass of water
Mass of water = 1.045 mol * 18.015 g/mol
Mass of water ≈ 18.8 g
6. Finally, calculate the change in temperature of the water:
Q = mcΔT
Q = -1290.08 kJ
m = 18.8 g
c = 4.18 kJ/kg°C
ΔT = ? (change in temperature)
-1290.08 kJ = (18.8 g * 4.18 kJ/kg°C) * ΔT
ΔT = -1290.08 kJ / (18.8 g * 4.18 kJ/kg°C)
ΔT ≈ -18.6°C
Since the change in temperature is negative, it means that the water will cool down during the combustion of ethanol.
Therefore, the water can be cooled from 10.85°C to approximately -7.75°C by the combustion of 48.2 g of ethanol.
The molar enthalpy of combustion of ethanol is -1234.8kJ/mol. How much water can be heated from 10.85 degrees celsius to 72.6 degrees celsius by the combustion of 48.2 g of ethanol.
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1 answer