TITLE : THE SPECIFIC LATENT HEAT OF VAPORIZATION OF WATER

OBJECTIVE :

To determine the total power needed to vaporise water and the mass of water vaporized.

THEORY :

The latent heat of vaporization, L, of a substance is the heat per unit mass required to change

the substance from a liquid to a vapour at its boiling temperature.

L=Qm

or Q=mL

With Q is the quantity of heat, if the energy is to be conversed, we say that the heat yielded

by the Bunsen burner must equal the heat gained by the boiling water.

h eat lost=heat gained

Pt=mL

Where P is the power of the Bunsen burner and t is the time in second.

APPARATUS :

Bunsen burner

Stopwatch

Beaker

Thermometer

Weight balance

PROCEDURE :

1. The beaker was weighed, and the mass was recorded.

2. The beaker was filled with tap water, then the mass of beaker and its contain was

measured.

The initial temperature, θi , of the water was also measured.

3. The water was then heated, and simultaneously, the stopwatch was started.

4. As the water started boiling, the boiling process was continued for about ten more

minutes.

5. After ten minutes, the water was removed from the heat and the timing was also

stopped. The water was allowed to cool.

6. The mass of the remaining water was weighed. Assuming the density of water to be

1.0gcm-3

RESULTS :

1. Mass of beaker = 121.97 g

2. Initial mass of beaker + water = 210.88 g

3. Initial mass of water, m1 = (mass of beaker + water) – (mass of beaker)

= 210.88 g – 121.97 g

= 88.91 g

= 0.08891 kg

4. Final mass of the beaker + water = 173.66 g

Final mass of water , m2 = 173.66 g – 121.97 g

= 51.69 g

= 0.05169 kg

5. boiling time=20 min ×60 s

1 min

boiling time=1200 s

6. Initial temperature, θi=26.5° c

Final temperature, θ f=100 ° c

ANALYSIS :

P=Q1+Q2

t

P=mc ∆θ−m Lv

t

P=(0.08891 kg ) ( 4.186 kJ kg−1℃−1 ) (100−26.5℃ )+(0.05169 kg ) (2260 kJ kg−1 )

1200 s

P=27.355 kJ +116.819kJ1200 s

P=0.120 kJ s−1

P=0.120 kW

Mass of water that evaporate = Initial mass of water – Final mass of water

= 88.91 g – 51.67 g

=37.24 g

DISCUSSION :

This experiment was carried out to determine the total power needed to evaporate

water, and to determine the mass of water. First, the mass of empty beaker and the mass

beaker containing water were weighed. This is to find the mass of water in put in the beaker,

that is ;

Mass of water = mass of beaker containing water – mass of empty beaker

After the water was started to boil, the water was continued to be boil for about 10

minutes to let the water evaporate. The temperature was constant from the moment it was

boiled until the next 10 minutes. The temperature had not changed because the heat

consumed was not used to raise the temperature of water; instead, the heat consumed was

used to overcome the intermolecular forces holding the molecules together in a liquid state,

thus forming molecules independent of each other. Where there is a change in the bonding

structure of a material without a change in temperature the energy involved is called latent

heat.

The change of phase of the water to steam :

Q1 Q2

Water, 26.5oC Water, 100oC Steam, 100oC

The Power, P, was then calculated by using;

P=Q1+Q2

t

P=m1 c∆ θ−m2 Lv

t

Where, P is the power

m1 is the initial mass of water

c is the specific heat for water

∆ θ is the change in temperature

m2 is the final mass of water

Lv is the latent heat for waterThe mass of the water that had been evaporate can be known by calculating it using;

Mass of water vaporized = Initial mass of water – Final mass of water

There are some errors that might affect the outcome of this experiment. First, the time taken might me more than that we recorded. The reason for this is that while carrying out the experiment, the time should not be stop, until the boiling process end. However, the stopwatch was stopped for some seconds when the water started to boil, and the thermometer indicated the temperature of the water was 100oC. This can be and should be prevented. While carrying out an experiment, we should be stricter and be more focus. Secondly, the equipments that we had used during the experiment might be not well calibrated. For this, the error can be corrected by using a well-calibrated equipment.

During the experiment, avoid parallax error, especially when reading the measurement, by placing the eye such that the line of view is perpendicular to the scale read. This is one of the precautions in this experiment to avoid any error. While for safety, avoid touching the beaker while it still hot, and other hot apparatus such as the tripod stand. That is why the water was let to be cool before determine the final mass. Besides, weighing the water while it is still hot could cause error.

CONCLUSION :

1. The power needed to vaporize water is P=0.0120 k W2. The total mass of water that vaporized is 37.24 g

REFERENCES :

1. PHY210 LABORATORY MANUAL2. Lim Chok Sang. 2009. New Vision SPM Physics. Marshall Cavendish (M) Sdn. Bhd.

3. https://online.varndean.ac.uk/area/science/physics/absorbphysics/units/050103.html

MARA University of Technology

Sabah Branch

Faculty of Applied Science

Diploma In science

PHY 210

Lab 5

VAPORISATION OF WATER

Prepared by ;

Siti Nurajjar binti Jami

2010673706

Date of experiment :

10th MARCH 2011

Date of submission :

17th MARCH 2011