How much energy (in kilojoules) is released when 26.5 of ethanol vapor at 86.0 is cooled to -12.0? Ethanol has = -114.5 , = 78.4 , = 38.56 and = 4.60 . The molar heat capacity is 113 for the liquid and 65.7 for the vapor
How much energy (in kilojoules) is released when 26.5 WHAT of ethanol vapor at 86.0 WHAT is cooled to -12.0 WHAT? Ethanol has = -114.5WHAT , = 78.4 WHAT , = 38.56 WHAT and = 4.60 WHAT . The molar heat capacity is 113 WHAT for the liquid and 65.7 WHAT for the vapor
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To calculate the amount of energy released when cooling ethanol from 86.0°C to -12.0°C, we will use the equation:
q = m * C * ΔT
Where:
q = energy released (in kilojoules)
m = mass of ethanol vapor (in grams)
C = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
First, let's convert the given temperature from Celsius to Kelvin:
Initial temperature (T1) = 86.0°C + 273.15 = 359.15 K
Final temperature (T2) = -12.0°C + 273.15 = 261.15 K
Next, we need to calculate the mass of ethanol vapor (m) using the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in K)
The volume (V) of ethanol vapor is not given, but assuming it is 1 L, we can calculate the number of moles (n) using the ideal gas law equation:
n = PV / RT
Given:
P = 1 atm
V = 1 L
R = 0.0821 L·atm/(mol·K)
T = initial temperature (T1) in K (359.15 K)
Calculating the number of moles:
n = (1 atm * 1 L) / (0.0821 L·atm/(mol·K) * 359.15 K)
n ≈ 0.0320 mol
Now, we can calculate the mass of ethanol vapor using its molar mass:
Molar mass of ethanol (C2H5OH) = (2 * atomic mass of carbon) + (6 * atomic mass of hydrogen) + atomic mass of oxygen
Molar mass of ethanol = (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + 16.00 g/mol
Molar mass of ethanol ≈ 46.07 g/mol
mass of ethanol vapor (m) = n * molar mass of ethanol
mass of ethanol vapor ≈ 0.0320 mol * 46.07 g/mol
mass of ethanol vapor ≈ 1.47 g
Now we can calculate the energy released (q):
q = m * C * ΔT
But before we proceed, we need to convert the specific heat capacity (C) from J/g°C to J/g·K:
C (in J/g·K) = C (in J/g°C) + 273.15
Specific heat capacity of liquid ethanol (C_liquid) = 113 J/g°C
Specific heat capacity of vapor ethanol (C_vapor) = 65.7 J/g°C
C_liquid = 113 J/g°C + 273.15
C_liquid ≈ 386.15 J/g·K
C_vapor = 65.7 J/g°C + 273.15
C_vapor ≈ 338.85 J/g·K
Now we can finally calculate q:
q = m * C_vapor * (T2 - T1)
q = 1.47 g * 338.85 J/g·K * (261.15 K - 359.15 K)
q ≈ -16690 J
Converting from joules to kilojoules:
q (in kilojoules) = -16690 J / 1000
q ≈ -16.69 kJ
Therefore, approximately -16.69 kilojoules of energy is released when 26.5 g of ethanol vapor at 86.0°C is cooled to -12.0°C.