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Experiment - Why is ice more effective for cooling a drink than cold water?

Extracts from this document...

Introduction

´╗┐Cooling Drinks 1. What is the specific latent heat of fusion of water? Specific latent heat of fusion is the amount of heat absorbed or released by a substance when changing states. This could be boiling or melting. Latent heat of fusion is the amount of energy needed to melt a substance, while latent heat of vaporisation is the amount of energy needed to boil a substance. The specific latent heat of fusion of water is 334 J/g In the graph, you can see that as the temperature increases more energy is needed to heat the water. When the temperature hits the latent heat of fusion, the temperature stops rising as the energy is being used to change the state of water. This also happens when the latent heat of vaporisation 1. What is the specific heat capacity of water? The specific heat capacity is the amount of heat that is needed to raise the temperature of a substance. The specific heat capacity water is 4181 j/kg. This is higher than most metals. Here are some other substances with their specific heat capacity; Substance SHC (J/goC) Air 1.01 Aluminium 0.902 Copper 0.385 Gold 0.129 Iron 0.450 Mercury 0.140 Sodium Chloride 0.864 Ice 2.03 Water 4.18 1. Why energy is needed to melt ice and how this is explained by the structures of ice and water The molecules of H20 behave differently in water than in ice. ...read more.

Middle

can cut yourself when using a thermometer and beakers http://www.fairbornchempage.com/Resources/Lab%20safety%20packet.htm Wear gloves and do not put the glass at the edge of the table Top pan balance Might be dropped onto the floor. Can injure a person. CLEAPSS Laboratory Handbook ? Section 9 ? General Equipment. Keep the top pan balance away from the edge of the table. Results Table 1. Describe any patterns or trends in your results. Comment on any unexpected results. After analysing my result I have found some trends in the data. I also saw some anomalies that came up in the experiment. The most obvious trend is in the line graph and the temperature drop. The more mass of ice I put into the beaker to cool the drink, the faster the temperature dropped. For example, in beaker one, which had 5g of ice, in minute two averaged out at 18.3oC, but beaker 4, which had 35g of ice, averaged out at 14.3oC. Also, in the 3rd minute the temperatures averaged out at 18oC and 12.3oC. One unexpected result was in beaker 4, were the temperature after one minute averaged out at 17.30C, which is higher than beaker 3 as it was 170C. Another anomaly is that beaker 1 and 2 both averaged out at 19oC, which is unexpected as beaker 2 had more ice, so should have had a lower temperature. 1. ...read more.

Conclusion

The less ice we added the less the temperature dropped as less energy was required to melt the ice. 1. Is it possible to use the equations below to predict the temperature drop of the water when a chosen amount of ice is added to it? Will the actual temperature drop, measured in your experiment, be equal to the predicted value? Use relevant scientific explanations in your answer. Use the results of your experiment, appropriate calculations and your research (Part 1) to provide evidence to support your answer. Yes, it is possible to predict the temperature of the water by using these equations. The temperature drop should be equal to the predicted value. This is because the amount of energy used to change the state of the ice is related to the amount of energy lost by the water, thus changing the temperature. So the more ice we add, the more energy is taken from the water to change the state of the ice, therefore reducing the temperature of the water and vice-versa. We used 200cm3 of water which is 200g and was at 21oC. In the first minute in beaker one, which had 5g of ice, the temperature dropped to 19oC, so the energy released by the water was 200(g)x4.181(j/kg, specific heat capacity of water)x2(oC) which is 1671.2j of energy. 5g of ice needs 1670j of energy to change the temperature of ice as 5(g) x334(j/g, specific latent heat of water) is 1670j ...read more.

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