The heat released by the combustion of 5.50 L of ethanol (molar mass = 46.1 g/mo
ID: 930663 • Letter: T
Question
The heat released by the combustion of 5.50 L of ethanol (molar mass = 46.1 g/mol) is used to heat a 275 kg sample of water that has an initial temperature of 18.0°C. What is the final temperature of the water? The specific heat capacity of H2O is 4.184 J/g°C and the density of ethanol is 0.789 g/mL
2 CH3CH2OH(l) + 7O2(g) 4CO2(g) + 6H2O(l) H°rxn = –1368 kJ
Can someone help me out and explain how to do this? I tried to find the mass of ethanol, then the moles, and then do I use the heat of reaction to find the energy? The answer is supposed to be 74 C
Explanation / Answer
density of ethanol is 0.789 g/mL
mass of ethanol = density *volume= 0.789*5500 = 4339.5g
moles of ethanol= 4339.5/46.1= 94.13 mol
2 moles of ethanol release 1368kJ energy,so 94.13 moles release energy= 1368*94.13/2= 64384.92kJ
To calculate heat you use the formula
Q = mC * ( t2 - t1 ),
wherem is the mass Cand c is the specific heat capacity and t2 - t1 is difference between temperatures.
=275000g * 4.184 J/g°C * (T2-18)
Q=1150600* (T2-18) = 64384.92KJ=64384920
t2-18 = 55.96
t2=73.96C =74C
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