To investigate any one of the many factors affecting the rate of heat loss of a liquid. My factor: Surface Area

 Reason of Experiment: To investigate any one of the many factors affecting the rate of heat loss of a liquid. My factor: Surface Area

Hypothesis: I hypothesis that if the base surface area of the container is increased, then the rate of cooling would be higher.

This is because: Higher surface area, results in larger area being exposed to the surrounding. The surrounding being cooler or less energised will get energised or absorb heat away quicker from the liquid. Thus the liquid will get cold faster and so rate of cooling will increase. (Provided other factors are kept constant)

Variables in this experiment:

Independent Variables:  The surface area

Manipulation of variable: this is the independent variable for my investigation, which I would vary to find the result. This I would vary by changing the container I would place the liquid in. each of them would have a different base area.

Dependent Variable: The temperature change per minute

Manipulation or the dependent variable: this is the variable of my investigation and this result I would gather by checking the temperature every minute for each container holding the liquid.

Constant Variable:

     1.       Initial temperature:
     Manipulation of the above constant variable: this is one of the factors which affect the rate of cooling down,( for example, if the initial temperature is higher, then the rate of cooling is faster at the starting few minutes). This variable can be kept constant by using the same heated water together at the same time, so the initial temperature is kept constant, or wait till the water cools down to a certain temperature.

     2.       Surrounding temperature:
      Manipulation of the above constant variable: this is also a variable which can affect.(if the surrounding temperature is high then the rate of cooling is slow, as the air molecules are already energised. This can be kept constant by doing the experiment in the same room at the same time or by using a water bath.

    3.       The liquid used:
     Manipulation of the above constant variable: each liquid has its own heat capacity, so the rate of cooling would differ. Using the same liquid is necessary, generally water. Their density also varies, so the heat capacity will vary along with it.

    4.       Flow rate of air(wind speed)
    Manipulation of the above constant variable: if the wind speed is not constant then the rate of un-energized air molecules occurrence would not be constant. For this, there is no external wind used and the air is at rest.

    5.       The total duration of time:
      this variable is the a constant, so that we can compare the data over a fixed duration of time and equally. So a constant time is taken.

Materials/Apparatus

In my experiment I will be using the following materials and apparatus:

1. 100ml of water of a certain high temperature(less than b. pt. 100^oC)
2. Glass beaker, two different sizes
3. Glass measuring cylinder, two different sizes
4. Alcohol-in-glass thermometer
5. a stop watch
6. notepad and pencil( for noting the data collected)

Specifications:

Thermometer: Least count: 1^oC, Uncertainty value: 0.5 ^oC

                                Highest fixed point: 110^oC

                                Lowest fixed point: -10^oC

Beaker: Material: glass, Uncertainty value: 25ml

                 Least count: 50ml

                Lowest fixed point: 50ml

                Highest fixed point: 250ml

Measuring cylinder: glass, Uncertainty value: 2.5 ml

                Least count: 5ml

                L.F.P. : 50 ml

Highest fixed point: 500ml

Stop watch: Colour: yellow

                        Least count: 0.1seond

Procedure:

1)      Then I decided to first start the experiment with the first container as the measuring cylinder.

2)      A lab assistant provided me with the hot water, which I collected in the beaker for measurement of 100ml of water.

3)      I poured it in the measuring cylinder and placed the alcohol-in-glass thermometer.

4)      The alcohol showed a rise till a point of 80^oC.

5)      Immediately I switched on the stop watch. For 5 minutes I noted the temperature drop every 30 seconds.

6)      After 5 minutes of experiment I stopped the timer and threw the water. It had decreased and I noted all the changes.

7)      Then I carried the same process for the other measuring cylinder, and the other beakers

8)      For each container, I would repeat the steps 2-6.

Diagram

DATA PROCESSING

	OD= 41 53 = Height		55 mm = OD 360 mm = Height		88mm = OD 122mm = Height		AVERAGE
	Try1	Try2	Try1	Try2	Try1	Try2	1320.25 mm2	2375.83 mm2	6082.12 mm2
00:00	80	80	80	80	80	80	80	80	80
00:30	79	79	78	78	78	78	79	78	78
01:00	77.5	78	75.5	76	75	75	77.75	75.75	75
01:30	76	77	73	75	72	72	76.5	74	72
02:00	75	75.5	72	73.5	70	71	75.25	72.75	70.5
02:30	74	74	71	72	69.5	70	74	71.5	69.75
03:00	73	73.9	70	71	68	68.5	73.45	70.5	68.25
03:30	71.5	72	69	70	67	67	71.75	69.5	67
04:00	71	72	68	69	65.5	65.5	71.5	68.5	65.5
04:30	70	71	67	68	64.5	64	70.5	67.5	64.25
05:00	69	70	66	67	64	63	69.5	66.5	63.5
	Surface Area (mm2):	1320.25	Surface Area (mm2):	2375.83	Surface Area (mm2):	6082.12

From the above table and averages, we can construct a graph which looks like:

As per my graph, the lines with higher surface area have a greater fall in temperature, than the one with less surface area. Their decreasing almost linearly, however at one point the blue and green lines curves slightly, which shows some errors, occurred. However, their lines have ended approximately 3 degree difference.

Rate of cooling: final - initial temperature/ total time taken

       1.       For 6082.12 mm² the rate of cooling: (80 - 69.5)/5 min = 2.1 ^oC fall per minute

       2.       For 2375.83 mm² the rate of cooling: (80 - 66.5)/5 min = 2.7 ^oC fall per minute

       3.       For 1320.25 mm² the rate of cooling: (80 - 63.5)/5 min = 3.3 ^oC fall per minute

Conclusion

From the above observations and by recalling my hypothesis, it is clearly proven that my hypothesis is proven correct. My hypothesis states, if the base surface area of the container is increased, then the rate of cooling would be higher. As per my data, the container with 6082.12 mm² has the rate of cooling of 3.3 ^oC fall per minute whereas for surface area of 1320.25 mm² had the rate of cooling is 2.1 ^oC fall per minute. This proves that there is higher rate of cooling for a larger base than a lower surface are of the base.

Evaluation

The practical investigation, resulted with expected results, and they are quite accurate, however, at one point, there might have been some mistake, due to the fact that each of the average values have ended up to be different by 3 degrees approx. Due to uncertainty, some temperatures, might be approximate and this affected the results slightly. One drawback of the materials, were that one measuring cylinder was used and two beakers. Assuming beakers and the cylinder having same density, still have some varying results, because the base type of the measuring cylinder had some extra glass matter, (for a good base) but the beakers, did not have. This might have affected the rate of cooling. The initial temperatures of the water were not accurately 80 degrees, because the thermometer had a least count of 1 degree, not more. A more drawback is the human responding system, which is not accurate to milliseconds, affecting the results, nominally.  Some of the errors, are uncontrollable, however some of them can be improved if I get to carry the investigation once again.

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