Figure 4.6 shows a simple combustion calorimeter. The simple is ignited electric

Question

Figure 4.6 shows a simple combustion calorimeter. The simple is ignited electrically. After a few minutes the temperature of the water and calorimeter is constant at DeltaT higher than the starting temperature. The heat of combustion is defined as DeltaIfcombustion = U final products of cimbustion - U initial fuel + oxygen/m sample Determine the heat of combustion of a sample from the following data: Ignore the heat capacity of the gases in the calorimeter. Make a list of possible sources of error in this experiment.

Explanation / Answer

Calories are units of energy. The water and the calorimeter each gained energy from the combustion. We will find out how much energy. Energy gained by surroundings = MCdelta T for water and for Calorimeter 5000g(1 cal/g C)(5C) + 500g(0.12 cal/g C)(5C) = 25,300 cal 25,300 was gained from the combustion We usually define energy of elements as zero so heat of combustion = 25,300 cal/4g = 6325 cal/g Since we only have 1 sig fig on mass of sample and on change in T, we should report heat of combustion as 6,000 cal/g (this is why I always like to have a double digit increase in temp and have initial temp be approximately as far below room temp as final tmep will be above room temp to average-out heat flow to and from surroundings). Possible sources of error include: Heat leaking out of calorimeter to surroundings (this is why a single digit increase in temp is good) Not waiting for temp to equilibrate (still heating up when final temp was taken_ Stirrer increasing the temperature Precision of mass measurement Precision of temperature measurement

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