Combustion of Ethanol DATA Exact volume of water in soda can 200mL Trial 2 Trial
ID: 635821 • Letter: C
Question
Combustion of Ethanol DATA Exact volume of water in soda can 200mL Trial 2 Trial 3 Trial 1 20.189 20.189 Initial mass of ethanol lamp Final mass of ethanol lamp Initial temperature of calorimeter Final temperature 20.919 Sarme as final mass, Tral Same as final mass. Trial 2 20.189.30 21.0°C 21.5? 32.5 320C 330°C of calorimeter CALCULATIONS Trial 1 Trial 2 Trial 3 Change in mass of ethanol lamp Heat absorbed by calorimeter based on 4Hcomb-26.8 kJ/g for ethanol Change in temperature of calorimeter Heat capacity of calorimeter Ccal, in kJ/C 2 , 99 l1.5° c ll.5°c Average heat capacity (Ceal) of calorimeter Efficiency of calorimeter Calculate the percent efficiency of the calorimeter using the results from the trial for which the calculated heat capacity is closest to the average heat capacity from the three trials Amount of heat generated, based on mass #1 of ethanol burned and AHcomb of ethanol Amount of heat absorbed by water, based equals exactly 1 gram H20) and AT of water #2 | on mass of water in calorimeter (assume that 1 mL H2O #2 ri #3 | Percent efficiency % efficiency #Explanation / Answer
q = m * s * dT
q = heat energy released or absorbed
m = mass of the sample
s = specific heat capacity of the calorie meter
dT = change in temperature
Case(1)
Mass of the chaged = 0.73 g
Change in temperature = 11.5 degrees celecius
enthalpy of Combustion dH com = -26.8 *10^3 J/g = -Q
from the formula q = m * s * dT
26.8 * 10^3 = 0.73 * s * 11.5
s = specific heat capacity = 3.192 kJ/0C
Case(2)
Mass of the chaged = 0.82 g
Change in temperature = 10 degrees celecius
enthalpy of Combustion dH com = -26.8 *10^3 J/g = -Q
from the formula q = m * s * dT
26.8 * 10^3 = 0.82 * s * 10
s = specific heat capacity = 3.268 kJ/0C
Case(3)
Mass of the chaged = 0.99 g
Change in temperature = 11.5 degrees celecius
enthalpy of Combustion dH com = -26.8 *10^3 J/g = -Q
from the formula q = m * s * dT
26.8 * 10^3 = 0.99 * s * 11.5
s = specific heat capacity = 2.354 kJ/0C
Average Heat capacity = (3.192 + 3.268 + 2.354)/3 = 2.938 kJ/0C
Amount of Heat generated q = ((0.73+0.82+0.99)/3) * 2.938 *((11.5+10+11.5)/3)
Q = 27.362 kJ/g
Q of water = 26.8 kJ/mol
Percent efficiency = 26.8/27.362*100 = 97.95 %