To calculate the amount of water that can be heated by the combustion of propane, we need to determine the amount of heat released by the combustion reaction.
The molar enthalpy of combustion of propane is given as -2041.9 kJ/mol. This means that when one mole of propane is combusted in an open system, -2041.9 kJ of heat is released.
First, we need to calculate the number of moles of propane in 28.5g. The molar mass of propane (C3H8) is 44.1 g/mol. Therefore, the number of moles of propane is given by:
Number of moles = Mass of propane / Molar mass of propane
Number of moles = 28.5g / 44.1 g/mol
Number of moles = 0.647 moles
Now, the amount of heat released by combustion of 0.647 moles of propane can be calculated using the molar enthalpy of combustion:
Heat released = Number of moles * Molar enthalpy of combustion
Heat released = 0.647 mol * -2041.9 kJ/mol
Heat released = -1320.29 kJ
To calculate how much water can be heated by this amount of heat, we will use the specific heat capacity formula:
Q = m * c * ΔT
Where:
Q = amount of heat absorbed (in Joules)
m = mass of water (in grams)
c = specific heat capacity of water (4.184 J/g°C)
ΔT = change in temperature (in °C)
The change in temperature is given as 60°C and we need to find the mass of water (m). Rearranging the equation, we get:
m = Q / (c * ΔT)
Substituting the values into the equation:
m = -1320.29 kJ / (4.184 J/g°C * 60°C)
m = -1320.29 kJ / 250.44 J/g
m ≈ -5.27 g
Since mass cannot be negative, we will take the absolute value:
m ≈ 5.27 g
Therefore, approximately 5.27 grams of water could be heated by 60°C by the combustion of 28.5 grams of propane.
Given that the molar enthalpy of combustion of propane is -2041.9 in an open system, how much water could be heated by 60 degree Celsius by the combustion of 28.5g of propane.
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1 answer