The document describes two experiments to measure the specific latent heat of fusion and vaporization of water. The first experiment measures the specific latent heat of fusion as ice is added to water, cooling it. Raw temperature and time data is plotted in a graph. The second experiment applies a constant power to heat water and measures the specific latent heat of vaporization. Both experiments measure how thermal energy flows when water changes phase, and the specific latent heats quantify the energy absorbed or released during these phase changes.

1. Lab 4: Measure the specific latent heat of fusion and the s.l.h. of vaporization.
Second experiment: measure the specific latent heat of vaporization of water
Raw data
mcontainer = 0.1619kg ± 0.0005kg
mwaterinit = 0.2957 g ± 0.0005kg
Independent Dependent variable
mwater fin = 0.2372 g ± 0.0005kg
variable
mwatervaporized = 0.0385 g ± 0.0005kg Power(W) Temperature(°C)
We couldn’t write down in a table all the values because they were too many that is why we
sketched the graph above.
We found that the power is 67.718 W from the following procedure:
Q=mcΔΤ=0.2957 ∙4200∙(98.477-23.068)=93653J
Q 93653
P= = = 67.718W
t 1383
Which is the independent variable.
To find the specific latent heat of vaporization of water we will do the following:
Q = P ⋅ t = 67.718 ⋅ (1383 - 920) = 31352J
Where P is the constant power of the heater that we found previously, Q is the thermal
energy needed to boil the water when the temperature is constant and about 100°C.
Q 31352
L= = = 106026kJ ⋅ kg −1
m 0.2957

2. Independent Dependent variable
variable
Power(W) Rate of change of
temperature(°C/t)
In both experiments we had the same variables as we observed how the water changes
temperature when thermal energy flows from an object to another. However the
independent variable (Power) was kept constant in both cases and so one would expect the
rate of change of the temperature to do so as well. The reason why it is not so is the specific
latent heat of fusion and vaporization, so that’s what we are going to measure in the two
experiments.

3. First experiment: Measure the specific latent heat of fusion of water.
Raw Data
We took so many values of temperature per second that I couldn’t write down in a table all
of them and instead drew the graph above.
A small example of our measurement:
t (s) T (°C)
46 38.476
47 38.427 The reading error of the digital stopwatch is ±1s because
48 38.063 its smallest division is the second.
49 37.723
50 37.359 The reading error of the thermometer for the left table is
51 36.873 ±0.001°C as it counts the temperature in Celsius degrees
52 36.144 with three decimal places, whereas for the ice temperature
53 35.610 it is ±0.1°C as the thermometer used in that case
54 35.221 measures the temperature in Celsius degrees with one
55 34.930 decimal place.
56 34.614
The reading error of the weighting scale is ±0.005kg as its
smallest division was the 10grams.
Container mass: 270g
Container + Water mass: 810g
Container + Water + Ice mass: 870g
Ice temperature: -3.5°C

4. Data Collection and Processing
The thermometer and the stopwatch don’t started counting from the time that we put ice
into the water, so we have to get rid of the part of the graph in which water is being heated
up. To do so we will set t=0s by the time value with the highest temperature (which means
before the temperature starts falling due to the ice inserted). Thus, the values after this
process will look like this:
t (s) T (°C)
±1s ±0.001°C
1 38.476
2 38.427
3 38.063
4 37.723
5 37.359
6 36.873
7 36.144 Because the greatest value of
8 35.610 temperature was found in the 46th second
9 35.221 (38.476°C) which was renamed as 1st
10 34.930 second, the 47th was named 2nd, etc.
So the graph will be:
Initial water mass: m w = m c + w - m c = 810 - 270 = 540g
Uncertainties:
m w max = m c + w max - m c min = 820 - 260 = 560g
m w min = m c + w min - m c max = 800 - 280 = 520g
So water mass is 540 g ± 20 g

5. Ice mass: mi = mc + w+ i − mc + w = 870 − 810 = 60 g
m i max = m c + w +i max - m c + w min = 880 - 800 = 80g
m i max = m c + w +i min - m c + w max = 860 - 820 = 40g
So ice mass is 60 g ± 20 g
Q gained = Qlost ⇒ mi ⋅ ci ⋅ ∆Τi + mi ⋅ Li + mi ⋅ c w ⋅ ∆Τw = mw ⋅ c w ⋅ ∆Τw ⇒
0.060 ⋅ 2.1 ⋅ 10 3 ⋅ 0 − 3.5 + 0.060 ⋅ Li + 0.060 ⋅ 4.2 ⋅ 10 3 ⋅ 29.331 − 38.476 = 0.540 ⋅ 4.2 ⋅ 10 3 ⋅ 29.331 − 38.476
441 + 0.060 ⋅ Li + 2304.54 = 20740.86 ⇒
0.060 ⋅ Li + 2745.54 = 20740.86 ⇒
0.060 ⋅ Li = 20740.86 − 2745.54 ⇒
0.060 ⋅ Li = 17995.32 ⇒
17995.32
Li = = 299.922 J ⋅ kg −1 = 300kJ ⋅ kg −1
0.060