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Measuring the Specific Heat Capacity of Water

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Introduction

Measuring the Specific Heat Capacity of Water Plan The experiment being performed is to find out the specific heat capacity of water. The specific heat capacity of a substance is the heat energy needed to raise the temperature of 1kg of the substance by 1 Kelvin, or degree. We already know that the formula for the specific heat capacity is (voltage x current x time)/(mass x temperature rise), so this is the formula we will use. The first part of the formula (VIT), is basically the amount of energy going into the water, and that is divided by the second part of the formula (M?Q), which is the amount of energy that the water receives and makes the temperature rise. When you put them together, you get the formula: VIT M?Q The units of specific heat capacity are Joules/Kg/oC, meaning the amount of energy (Joules), to cause a rise in temperature of 1kg by 1 degree. There are many factors that may affect the accuracy of the experiment. For a start, there will be a lot of heat loss through: 1) the sides of the beaker (via conduction). 2) the surface of the water (via radiation and convection). 3) the actual sides of the heater, because the whole heater is not in the water (via convection and radiation). These factors are very important because it means that not all of the heat is going into the water, so it may change our final result. ...read more.

Middle

I predict that our result for the specific heat capacity of water will be around 4600J/kg/OC. I know already that the specific heat capacity of water is supposed to be 4200J/kg/OC, but I am taking the heat loss into account. I would have predicted a higher value if it was not for 2 reasons: 1) We are not using pure water, but tap water, and when salts are present, the specific heat capacity of water decreases slightly. 2) During the experiment, some of the water will evaporate. We have not studied specific heat capacity in great detail before this experiment, but we were taught the formula. Our previous work on heat transfer helped me, and I also used the text book 'The World of Physics' by John Avison. Results 1st Time: Time(sec) Temp (C) Current Voltage 0 19 4.9 11.72 60 21 4.9 11.72 120 23 4.9 11.72 180 24 4.9 11.72 240 25 4.9 11.72 300 29 4.9 11.72 Mass water= 0.40144kg C= VIT/M x change in temp 11.72 x 4.9 x 300/ 0.40144 x 10= 4292J/kg/OC 2nd Time: Time(sec) Temp (C) Current Voltage 0 27 4.9 11.34 60 29 4.9 11.34 120 31 4.9 11.34 180 32 4.9 11.34 240 34 4.9 11.34 300 36 4.9 11.34 Mass water= 0.40154kg C= VIT/M x change in temp 11.34 x 4.9 x 300/ 0.40154 x 9= 4613J/kg/OC 3rd Time: Time(sec) ...read more.

Conclusion

I think that if I had the opportunity to do the experiment again, I would not. That is because I feel that we performed the experiment to an acceptable level of accuracy. However, I would perform a slightly different experiment that may produce even more accurate results. Here are the things I would change: ? I would wrap the beaker in cotton wool or something, so it insulates the beaker and prevents so much heat loss through the beaker, via conduction. ? I may use a different container for a beaker, made out of a material that does not conduct heat so well, another way of reducing heat loss. ? I would place the heater fully in the water, so no part of it is in the air, to reduce heat loss from the heater, via convection and radiation. ? I would put some form of lid on top of the beaker, to reduce heat loss from the surface of the water via radiation and convection. All these changes would reduce heat loss, and cause the value of specific heat capacity lower, so make it more like what it should be. Overall, I think our result was as good as it could be expected with the method of the experiment that we used. However, I think it would be interesting if we could perform a slightly different experiment, without the flaws in method of this one, and I am sure our result would be very accurate. Faizal Patel 02/05/07 ...read more.

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