Bonding, structure and physical properties.
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The aim of this experiment is to calculate the heat energy required to convert 1mole of liquid (water) into steam.
Please see handout for this experiment.
Time (minutes)Temperature of the water (C).
The table shows the temperature recordings done every 30 seconds, and the graph shows an indication of when the temperature had its steepest rising, and when it flattened out. Note that the 30 seconds are written in 0.50 minutes. This was done so that it was possible to make a graph out of the table.
Weight of water before boiling: 97.73g.
Weight of water after boiling: 44.13g.
Starting temperature of water: 23.5A?C.
Boiling time: 4.30 min.
The rising of the waters temperature during the first 4.5min. was:.
95.0A?C-23.5A?C = 71.5A?C.
The rate of rise of temperature the first 4.5 minutes = Change in y/change in x.
71.5A?C/4.5min. = 15.9A?C/min.
If 1 calorie raises the temperature of 1 gram of water by 1A?C, in this experiment 97.73g of water was raised by 71.5A?C. To calculate how much heat was given to the flame each minute you multiply the rate of rise in temperature by the weight of the water: .
15.9 A?C x 97.73g. = 1553.9 cal/min.
The water was boiling for 10 minutes, which means that 1553.9cal/min x 10min. = 15539 cal. during 10 minutes.
During this process some water was converted into steam and was lost to the surrounding atmosphere.
97.3g – 44.13g = 53.17g.
53.17g. of water were converted into steam during the process.
This is equivalent to: 53.17g/18 = 2.9 mole of water.
To figure out how much heat energy was used to convert 1 mole of liquid into steam simply divide 15539 calories by 2.9 moles.