Investigating Specific Latent Heat of Vaporization of Water
Research Question
Investigating specific latent heat of vaporization of water.
Data
Controlled Variables
Power of the kettle: 1000W Intervals of the recording mass: 15s Mass of the water
boiled
Table 1: Mass of water evaporated over time
Mass
of
Water
±1s
evaporated
Trial 1
Trial 2
Trial
Trial 4
Trial 5
±0.1g
15
4.7
4.8
5.2
5.0
5.4
3
30
11.5
10.5
10.7
10.6
10.9
45
16.7
16.6
16.6
16.3
16.7
60
22.4
22.1
22.9
23.0
23.4
75
28.4
28.6
28.5
29.1
29.1
90
34.2
34.3
35.0
34.4
35.1
105
40.5
40.3
40.4
41.1
41.2
120
47.0
46.1
46.3
47.6
47.9
Uncertainty in measuring time was ±0.01s according to the stopwatch but while
measuring mass you have to first look at the time in stopwatch and then the mass in
the electronic balance and because humans cannot react instantly it is estimated to be
±1s.
Uncertainty in measuring mass of the water was ±0.1g because it was measured using a
weighing machine with the ±0.1g uncertainty.
Time
Table 2: Mass of water evaporate over time with uncertainties
Mass of
Water
Time
Evapora
±1s
t ed
Trial 1g) Trial 2
Trial
Trial 4
Trial 5
(±0.1
15
4.7
4.8
5.2
5.0
5.4
3
30
11.5
10.5
10.7
10.6
10.9
45
16.7
16.6
16.6
16.3
16.7
60
22.4
22.1
22.9
23.0
23.4
75
28.4
28.6
28.5
29.1
29.1
90
34.2
34.3
35.0
34.4
35.1
105
40.5
40.3
40.4
41.1
41.2
120
47.0
46.1
46.3
47.6
47.9
(Average of mass of water Evaporated) = (Trial 1 + 2 +3+4+5)/5
= (4.7+4.8+5.2+5.0+5.4)/5
= 4.9 g
Averag
e
Uncertaint
y
4.9
10.8
16.6
22.6
28.7
34.5
40.6
46.8
0.4
0.5
0.2
0.6
0.4
0.4
0.5
0.9
(Uncertainty in mass of water Evaporated)
= ((Max Value)-(Min Value))/2
= (5.4-4.7)/2
= 0.4 g
Variables on the graph
The formula given was
Q = mL
If we divide both sides by ∆t (time)
So,
From the data processed above a graph of mass of water evaporated vs. time can be
plotted whose slope will give
Graph 2: Mass of Water Evaporated vs. Time Graph
Slope of the graph is 0.3985 g/s
Maximum Slope = 0.4110 g/s
Minimum Slope = 0.3870 g/s
So, the slope of the graph is 0.40 g/s ± 0.01 g/s
Calculating the specific latent heat of vaporization
Calculation of Uncertainty in specific latent heat of vaporization
The calculated value for latent heat of vaporization is 2500 J/g ±60 J/g
Conclusion
The graph of mass of water evaporated over time is linear because the best fit line
passes through all error bars. From the calculations the specific latent heat of
vaporization of water is calculated to be 2500 J/g ±60 J/g. The literature value of
specific latent heat of vaporization of water is 2260 J/g, which is quite low. Percentage
error is 100
The total percent error is 10.6% and the total percent uncertainty is 2.5% which is quite
low compared to the percentage error. 2.5% uncertainty means the final result can be
±2.5% off. That means the total error caused by uncertainties is 2.5%, rest is from
systematic errors. One of the biggest systematic errors could be the heat loss from the
water to the atmosphere. A well-insulated plastic kettle was used to boil the water so
there will be minimum heat loss from water to kettle and kettle to surroundings. If the
heat is lost to the surroundings from water, it means that the power supplied by the
kettle is not completely used to boil water as it is lost in the surrounding so the power
supplied is less than 1000W.
While recording the mass of water, the mass of the water in the electronic balance was
not constantly decreasing. Sometimes it increased, sometimes it decreased slowly and
sometimes rapidly and because of this there was a high error in collecting data. An
electronic balance with high mass capturing should have been used for better results.
The electronic balance used did not have a wide base and the kettle used to boil water
was overturning it which also can result in high error. An electronic balance with wide
base should be used for more accurate results.
Works Cited

"Latent Heat." Wikipedia. Wikimedia Foundation, 10 May 2012. Web. 07 Oct.
2012. <http:// en.wikipedia.org/wiki/Latent_heat>.