A chemical engineer studying the properties of fuels placed2.366 g of a hydrocar

Question

A chemical engineer studying the properties of fuels placed2.366 g of a hydrocarbon in the bomb of a calorimeter and filled itwith O2 gas. The bomb was immersed in 2.723 L of waterand the reaction was initiated. The water temperature rose from13.46 °C to 31.27 °C. If the calorimeter (excluding thewater) had a heat capacity of 420. J/K, what was the heat ofcombustion of the fuel? (Hint: Enter the correctsign.)
Ok so I used this equation -q(sample)=(mcT)H20+(heatcapacity*T)calorimeter but I still keep getting the wrong answer Help? A chemical engineer studying the properties of fuels placed2.366 g of a hydrocarbon in the bomb of a calorimeter and filled itwith O2 gas. The bomb was immersed in 2.723 L of waterand the reaction was initiated. The water temperature rose from13.46 °C to 31.27 °C. If the calorimeter (excluding thewater) had a heat capacity of 420. J/K, what was the heat ofcombustion of the fuel? (Hint: Enter the correctsign.)
Ok so I used this equation -q(sample)=(mcT)H20+(heatcapacity*T)calorimeter but I still keep getting the wrong answer Help?

Explanation / Answer

Heat given out by the combustion of the fuel is absorbed byboth water and the calorimeter.
Mass of water = 2.723 kg (since density ˜ 1 kg/ kg)                        =2723 g q = [ Mass of water * specific heat * T ] +[ Heat capacity of calorimeter * T ]
    = [ 2723 g * 4.18 J/g.K * (31.27 - 13.46) ] + [420 J/ K * (31.27-13.46 ) ]     = [ 202715.91 J] + [ 7480.2 J ]     = 210196.11 J     = 210.19 kJ This is the heat given for 2.366 g of hydrocarbon. So heat given for 1 g of hydrocarbon will be = 210.96 / 2.366= 88.84 kJ/ g So heat of combustion of fuel = - 88.84 kJ/ g Negative sign because heat is liberated.     = 210.19 kJ This is the heat given for 2.366 g of hydrocarbon. So heat given for 1 g of hydrocarbon will be = 210.96 / 2.366= 88.84 kJ/ g So heat of combustion of fuel = - 88.84 kJ/ g Negative sign because heat is liberated.

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