Determining the latent heat of water at vaporization, by investigating the relationship between the mass of the water vaporized and the duration of heating with constant power.

This internal assessment investigates the relationship between the mass of water and the time it takes to vaporize it, in order to determine the latent heat of water at vaporization.

I am a big pasta lover, as it is the easiest thing to cook every few days when I don't feel like standing in the long line in the school canteen. During one of those cooking sessions after physics class that introduced me to energy conversion, I started to wonder where all of the heat goes after raising the temperature to the point of boiling. Clearly, there was heat going into the pot, but I didn't know where would it end up. Would it be used just to keep the temperature constant, or would it go somewhere else? It seemed impossible that water can lose energy quickly enough that it needs more heat instantly, but it also seemed impossible that energy is going anywhere else except into the water. So I have decided to explore what exactly is happening with heat during the process of vaporization.

Determining the latent heat of water at vaporization, by investigating the relationship between the mass of the water vaporized and the duration of heating with constant power.

The latent heat of vaporization is the measure of the energy required when a substance is transformed from its semisolid/liquid state to gaseous vapour in the absence of any significant input or change in temperature.

Here is the equation of latent heat:

\[ Q=L m \]

Where:
Q - heat [J]
L - the specific latent heat of the object [J/g]
m - a mass of the substance [kg]

An electrical kitchen heater will be used, with the constant power of 1000 W. The equation for power is following:

\[ P=\frac{Q}{t} \]

Where:
P - power [W]
Q - heat [J]
t - time [s]

Using (1) and (2), the equation that connects power and latent heat can be obtained:

\[ P=\frac{L m}{t} \]

Latent heat, as well as the boiling point of water, varies with pressure. In the graph below, a variation of boiling point with the pressure is presented:

The evaporated mass of water will be higher as the time increases, and the value of specific latent heat should be within the range of theoretical scope L=2.26×10⁶ J/kg

In the table below, variables have been identified.

1. Electric hotplate, Beko, HIZM 64120 S 1000W
2. Steel cooking vessel
3. Wooden trivet/cutting board
4. Phone stopwatch, ±0.005 s
5. A glass of water
6. Digital Scale, ±0.5 g

1. Turn on the electric heater, and leave it for a few minutes(in order to make sure that it reaches its heating capacity - so that it heats the water up using the same intensity);
2. Measure the weight of the water to 200 g, using the scale.
3. Measure the mass of the pot and the trivet and tare the scale to that mass, so that it measures just the amount of water.
4. Pour the water into the cooking vessel;
5. Put the pot on the heater, and turn on the stopwatch - at the same time!
6. Wait until the water starts boiling. At the moment that the first bubble is visible, turn on the stopwatch.
7. Every 50 seconds, stop the stopwatch, and measure the mass of water using the scale.
8. Return the pot with the water left in it to the heater and resume the stopwatch.
9. Repeat steps 7-8 until the stopwatch reaches 400 seconds.
10. Repeat steps 1 - in 4 trials.

The amount of water should be chosen wisely, in order to be able to conduct the experiment unabstracted, but also to be as efficient as it is possible. Using too much water will not only waste drinkable water, but it also can prolong the experiment in length and therefore make the amount of electricity used bigger.

The person conducting the experiment should be aware of the consequences that occur in contact with hot steam, water, or an electric burner while turned on. Accordingly, they should wear protective gloves and glasses, as well as practice extra efficiency in order to avoid any unpleasant mishaps, especially during the moving of the pot to the scale and returning it as this has to be done quickly.

Also, make sure that the phone used for the stopwatch does not get overheated or wet in order to avoid financial losses and material waste.

Original data does not reflect the mass of water vaporized, but the mass of water left in the pot. Given that, prior to every trial, there was 200 g of water, the amount of vaporized water at each sequence is obtained by subtracting originally collected values from 200 grams. The data obtained is presented in the table below.

The average mass values have been calculated using the following formula:

\[ m_{avg}=\frac{m_1+m_2+m_3+m_4}{4} \]

For example, the calculation for t=100 s was:

\[ m_{avg}=\frac{28+31+30+28}{4}=29.25 \text{ g} \]

The average value had to be rounded to the nearest value that could be recognised by scale (29 grams in this case). This process was done with other average masses as well.

It is noticeable that trial 2 has almost exclusively higher values of mass vaporized than all other trials. This suggests the presence of systemic error during performing it, for example, high human reaction time while turning on the timer for the first time(when the water starts boiling).

A final column represents the uncertainty in mass and is calculated by subtracting the minimum value in the trial from the maximum value, and dividing the result by 2, as per the formula:

\[ \Delta m=\frac{m_{max}-m_{min}}{2} \]

For example, for the value of t=200, the uncertainty was calculated in the following way:

\[ \Delta m=\frac{63-58}{2}=2.5 \]

As the instrument is not precise enough to recognise values less then 1 g, the uncertainty had to be adopted to the nearest value in all cases where the uncertainty was not a full value in grams. If the value is on the half-point between 2 full numbers, it should be changed to the even number. Therefore, in this case, the uncertainty had to be changed to 2.

A graph with the linear fit of mass vs time was plotted, in order to see whether there is a linear relationship between the two variables. Minimum and maximum lines of best fit have been presented as well:

The line of best fit passes through all the error bars, clearly indicating a linear relationship between the mass of water vaporized with the time of vaporization. The largest error bar is at value t=250 s. This suggests a presence of a random error at this point of the experiment. The value of Y intercept is negligible as it is multiple times smaller than the smallest value that could be detected by the scale used in the experiment. This value also discards the presence of most systematic errors that could occur in the experiment as a whole.

Values of the slopes are the following:

\[ s=0.3048 \frac{g}{s}, s_{max}=0.3170 \frac{g}{s}, s_{min}=0.2895 \frac{g}{s} \]

Uncertainty of the slope is obtained from the following equation:

\[ \Delta s=\frac{s_{max}-s_{min}}{2}=\frac{0.3170-0.2895}{2}=0.0137 \frac{g}{s} \]

The final value of the slope is:

\[ s=(0.305 \pm 0.014) \frac{g}{s} \]

Percentage uncertainty of the slope is:

\[ \frac{0.011}{0.305} \times 100=3.6\% \]

Thus, the relationship of mass of vaporized water and the time of boiling can be expressed as follows:

\[ m=0.305t \]

As the relationship between the mass of vaporized water and time of boiling is expressed, the latent heat will be derived:

\[ P=1000 W, m=0.305t \frac{g}{s} \]

From the equation (3), the equation for L can be easily obtained:

\[ L=\frac{P \times t}{m} \]

The values from the slope can be included in this equation:

\[ L=\frac{1000 W \times \frac{s}{0.305 g}}{m}=\frac{1000 Ws}{0.305 g} \]

The latent heat is, therefore, equal to

\[ L=3278.69 \frac{Ws}{g}=3278.69 \frac{J}{g} \]

As the power used in the equation above is the maximum power described in technical specifications of the hotplate, uncertainty is probably very high, as the efficiency of the hotplate is almost never 100%. The efficiency range taken into account will be 75-100%, meaning that the uncertainty of the power variable will be 25%, or 250 W.

Total uncertainty of the specific latent heat of vaporisation can be calculated using the following equation:

\[ L=\frac{P}{s} \Rightarrow \frac{\Delta L}{L}=\frac{\Delta P}{P}+\frac{\Delta s}{s} \]

\[ \frac{\Delta L}{L}=\frac{\Delta P}{P}+\frac{\Delta s}{s}=\frac{250}{1000}+\frac{\frac{1}{0.014}}{\frac{1}{0.305}}=22.04 \]

So the value of specific latent heat is

\[ L=3278.69 \frac{J}{g} \pm 22.04 \frac{J}{g}=3.27869 \times 10^6 \frac{J}{kg} \pm 22040 \frac{J}{kg} \]

The difference between the ideal value from the hypothesis and obtained value:

\[ \frac{L_{ideal}}{L}=\frac{2.26}{3.27869}=0.6893 \Rightarrow e=31\% \]

The research has shown that the mass of water vaporized is directly correlated to the duration of heating with constant power. Their relationship is proved to be linear, increasing with time. As the value of the Y-intercept is small, the possibility of the majority of systemic errors is eliminated, suggesting that no factors are overlooked during the experiment.

The derived value of specific latent heat is relatively close to the value used in the hypothesis. In the graph below, the blue line presents the linear fit of the obtained results, while the green line represents the "ideal" trend which would lead to the value of latent heat from the hypothesis.

The previously calculated error of 31% compared to the value in the hypothesis is evident. "Ideal" masses are higher than obtained values, which is most certainly due to the fact that the power variable used in the calculation is highly unreliable, given that the efficiency of hotplates is rarely above 75%, and that hotplates rarely reach their maximum power provided.

Both values(one from the hypothesis and the one obtained from the research) are in the same order of magnitude, suggesting that the goal of this IA is met.

The table below summarizes errors that have been identified.

This research could be further developed by investigating specific latent heat at other phase changes of water. A really interesting direction in which this research could be taken is investigating the latent heat of water during fusion(melting), which could be later used for predictions of the speed of melting the ice of the Arctic.

Also, an investigation could be done on the effect of different air pressures or room temperatures on the latent heat at vaporization. This could be used to determine how latent heat varies with attitudes.

Finally, knowing the latent heat of water, the research could be slightly modified to measure the power and efficiency of different hotplates instead.