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Experiment to investigate the Factors affecting the Energy Transfer Involved in the Cooling of Water

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Introduction

Experiment to investigate the Factors affecting the Energy Transfer Involved in the Cooling of Water By Clare Dutton Introduction Water cools in many different ways due to a variety of different reasons, which depend on the way in which the water is contained. I will be considering how water in a plastic cup cools. If I put hot water in a plastic cup I would expect heat to be lost by radiation from the sides and the top, conduction through the base and evaporation from the top. If however, the sides and base were highly insulated; most heat would be lost by evaporation and radiation from the top. Here are the ways in which heat is lost from cooling water: 1) Radiation Radiation is the movement of heat energy by electro-magnetic waves. These leave the surface of the object and can pass through a gas or vacuum. The hotter the surface of the radiating object the greater the rate of heat loss. The larger the surface area, the greater the heat lost by radiation. Darker surfaces radiate and absorb more than light coloured surfaces. Shiny surfaces reflect radiation. Examples of radiation are the sun radiating heat through space and a central heating radiator radiating heat into a room. 2) Conduction Conduction can take place in solids, liquids and gases. When a material is heated the particles nearest to the heat gain kinetic energy. They then start to vibrate faster due to this energy and as they do they touch other particles and transfer the kinetic energy to them. This process is repeated and the energy is transferred through out the object from hot regions to cooler regions. As the water looses heat from the sides of the cup conduction will cool the mass of the water. Conduction will also occur through the walls of the cup and then radiate and through the base of the cup into the surface on which the cup stands. ...read more.

Middle

Insulation has a 'strength' to prevent heat loss which is measured as its 'U' value. The U value is measured as Watts per m� per �C and is for a given thickness of insulation. Double the thickness and the U value will halve. Rate of heat loss = U value � surface area � temperature difference. So if the insulation doubles in thickness, the U value will halve and the rate heat loss will halve. Diagram Method Room temperature was taken using the thermometer which had been allowed to adjust to the room temperature and this was recorded. The apparatus was arranged so that a plastic cup stood on a layer of foam insulation with a thermometer suspended from a clamp stand so that the bulb of the thermometer would be hung in the middle of the water in the cup. For the first test no insulation was wrapped around the cup. For the later tests, layers of insulation were tightly wrapped around the cup using sellotape so that once the air was trapped between them none could escape. This was also done to the base of the cup. The kettle was boiled and 150ml� of hot water was measured out using the measuring cylinder. This was then poured into the plastic cup. A piece of cling film was then stretched over the top of the cup to form a seal so that no evaporated water could escape. Square pieces of insulation (bigger than the cup) were placed on top of the cup. The number of squares depended on the number of layers of insulation being tested. A hole was pierced through the centre of the lid. A thermometer was then suspended using cotton and the clamp stand and then pushed through the hole so that the end hung in the centre of the cup. The lid was then taped down securely. When the temperature had fallen to 80 �C, the stop clock was started and the temperature of the water was taken every minute for 30 minutes. ...read more.

Conclusion

With this experiment it is not possible to retake or double check readings because the temperature keeps changing. If money were no expense, it would be better to use an automatic measuring system where a computer records the readings at the exact interval rather than allowing for human response. A digital thermometer would be used to record the temperature because it could readings of 0.1 of a degree rather than 0.25. From the results in Fig 1, I can see that the time at which the clock is started at 80�C is very critical because the temperature is only falling at about 1/2 a degree Celsius per minute. This could lead to an error of about 2 minutes. This could possibly explain the discrepancy between 8 and 10 layers. Further investigation I would like to investigate the effect of adding even more insulation to see if the rate of loss of energy increases due to the increase in surface area increasing the amount of radiation. I would do this because as you can see in Fig 2. The curve begins to slope up showing an increase in change of temperature. I would like to investigate this further and see if this trend continues with more layers of insulation. The experiment could also be repeated for different temperature ranges between the water and room temperature. CALCULATIONS Calculation of the surface area of insulation. With no insulation, the outside of the cup is approximately a cylinder of 6.5cm diameter and 9.0cm high. The surface area = 2 � 6.5 � 6.5 � ? / 4 + 6.5 � ? � 9.0 = 66.3 + 183.7 = 250.0 cm� With 10 layers the outside is a cylinder of 10.7cm diameter and 13.1cm high. The surface area = 2 � 10.7 �10.7 � ? / 4 + 10.7 � ? � 13.1 = 179.7 + 440.1 = 619.8 cm�. Calculation of energy loss Volume of water = 150ml Specific heat capacity of water = 4200 Joules/kg/ �C Therefore energy loss per �C = 4200 � 150 /1000 = 630 Joules. ...read more.

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