The purpose of this investigation is to assess the effect of certain factors on heating of water, therefore the variables of my investigation is the relationship between the power supplied to the heater and the temperature increase of a given mass of water being heated at a given time.

EQUIPMENT

For this experiment the following apparatus was used;

1) A thermometer to measure the temperature of the water

2) A plastic cup.

3) Volt meter to measure the voltage

4) A ametre to measure the current

5) A filament to heat the water

6) A power pack

7) A variable resistor to vary the current.

8) A watch.

A plastic cup was weighed empty and the weight was recorded, it was filled with water and weighed again the result was recorded. The difference between the result was the weight of the water.

The apparatus was then set up as shown in the sketch 1. The ambient temperature and the temperature of the water were then recorded. Keeping the mass of the water constant a steady current 3.1 amp was then passed through the heater for 1 minutes with a recorded voltage of 6 volt. The temperature of the water was recorded.

The water was then thrown away and the experiment was the carried out again but this time the water was heated for 2 minutes.

The above sequence was carried out to measure the temperature increase for 3,4 and 5 minutes heating time.

The above experiment was carried out with increase in voltage to 7 with time take up to 30 minutes.

The result is shown in Table 1.

In order to accurately carry out the measurement the same equipment was used, i.e., the same cup, the same voltmeter and ammeter so that the calibration of the equipment remains the same and the same heater.

The water was weighed before each experiment to keep the water constant.

In order for me to keep my results as accurate as possible, I decided to use the same heater as some heaters were burnt thus meaning it could give out less heat through the heaters. I decided to use the same cup, but throughout the investigation I regretted this as the plastic cup is a insulator, thus meaning it would result in temperature change.

Improvement Measures

The features of my experiment which I would change in order to get more accurate result would be to use a digital thermometer to measure the temperature of the water as the will give me exact result at the exact time.

I n order to prevent heat loss through the cup I should have encased the cup with an insulation material such as a blanket type material or foamed material etc. Since the heat escaping from the cup would be proportional to the temperature difference between the ambient and the temperature of the water then my result would not be accurate.

The other changes would be to use a double walled beaker to prevent heat loss/gain due to ambient temperature as the double wall would act as an insulator and prevent heat loss through the beaker.

Analyzing The Results

From the 1st experiment it can be concluded that the longer the heating element is emmersed in water the greater the temperature difference.

The second result concludes that the greater the power the higher the temperature difference.

Therefore it can be concluded that the temperature difference is proportional to the to power supplied to heat the water. It can also be predicted from the equation that the mass of water being heated is inversely proportional to its temperature difference

Equations

The following equation taken from books and class notes have been used through out my investigation to assess the relation ship of power and temperature.

Power = Energy

Time

Also Power = Voltage x Current

Watts = V x A

Energy = Mass x Current x Temperature Change

Q = M x A x ?T

Therefore Energy = Current x Voltage

Time

Energy = Current x Voltage x Time

Therefore Current x Voltage x Time = Mass x Current x ? Temperature

? Temperature = Timex Current xVoltage

Mass x Specific Heat Capacity

Therefore since current x voltage = power then ?Temperature is directly proportional to power.

From the above equation it can also be predicted that ?Temperature is also proportional to time and inversely proportional to its mass

PRODICTIONS

From the above equations and our results we can predict both the power and the mass of the water as follow;

The specific heat capacity of water is 4200 J/kg K

Voltage = 6 V

Current = 3.1 A

Therefor the Power P = 6 x 3.1 = 18.6 Watts

Using the equation ? Temperature = Timex Current xVoltage_____

Mass x Specific Heat Capacity

At 1minute the ? Temperature is

? Temperature = 1x 18.1_____ = 18.6 = 0.086 degree

50 x 4.200 210

At 2minute the ? Temperature is

? Temperature = 2x 18.1_____ = 36.2 = 0.86 degree

50 x 4.200 210

For further results refer to tables

Evaluating Evidence

I believe that my results are accurate as far as possible. I took care when measuring and recording the results

The 1st graph shows that the result obtained at 5 minute period was not in line of the best fit . This could be due to number of reason for example heat gained through the plastic walls of the cup, I could have taken too long to start the stopwatch or time taken from stopping the stopwatch and reading the thermometer. However the trend for each result shows that the change in power is proportional to change in temperature gained in a given time, with a given heater.

Therefore my result is in line with the above prediction.

The investigation could further be extended by considering the length of the heating element at a given current in order to find out the rate at which heat is produced at specific times

Other investigation would be to find the relationship between the mass of water at a given power and the temperature increase at specific times.

Minimum & Maximum Value

I found out the maximum and minimum values I would be using for the two variables from my preliminary work. The minimum value for the variable mass of water was 50g. This is because if the value were any smaller, then the change in temperature would be too large. Therefore when plotting a graph in the Analyzing Evidence and Drawing Conclusions section, the values would make the scales of the axis too large. Also if the minimum value were less than 50g, the heater would not have been submerged totally in water. The heaters which we were using had to be submerged totally in water otherwise they would be have been damaged so it was necessary for the minimum value to be greater than 50g. The maximum value for the mass of water variable is 100g. If the value were any greater, the change in temperature would be too small for the thermometer to measure accurately. For the Power variable, I chose the minimum value to be 18.6W. This is because the apparatus we used did not measure the Power. However I took the measurement of the voltage and current supplied and then used the formula P = VI to find the power. The voltage was then increased by 1V for each time the current recorded.

Conclusions

By plotting the results obtained as indicated in attached table the graph of time against heat gained was plotted. By drawing the best line through the coordinate, it can be seen that the straight lines passes through the origin.

From the graph it can be concluded that if the Power supplied to the heater is increased the temperature will also be increased in proportion.

This increase can be calculated by utilizing formula y = mx+c, where y = is the value of temperature and X, is the value of time and m, is the gradient, and c is the point where the line of best fit intersects the y-axis.

The gradient of the graph = 7-1 = 2

3

The line intersect at 0 on the y-axis, therefore C = 0.

Therefor by substituting the above figures in the equation we can deduce that Y = 2X . From this equation, we can predict the change in temperature change by multiplying the Power supplied by 2.

From the above analysis I have conclude that my prediction in the Planning section of the write up have been proved and that if an increase in Power will increase the temperature proportionally, i.e. the variables are directly proportional.