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Maths Assignment 1: Exponential Decay in Reality

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Introduction

Instructions to operate * Kettle to the boil * Pour 500ml of water in cup (plastic) * Place thermometer in water * Take temperature every minute for the 1st 30 mins and then every 5mins for the next 30 min and then every 10 mins until it is back to room temperature. * Experiment conducted 3 x (results averaged) * Used the exact same equipment for each experiment (water type, container, thermometer, kettle) My apparatus is designed to simulate the process of exponential decay. The model was measuring the temperature drop from boiling water. It was considered that the temperature would only ever drop to room temperature, therefore the temperature difference between the water temp and the room temp was measured to get a final reading of zero, which outlines Newton's cooling law. From experimental observations it is known that (up to a ``satisfactory'' approximation) the surface temperature of an object changes at a rate proportional to its relative temperature. That is, the difference between its temperature and the temperature of the surrounding environment. ...read more.

Middle

It was also determined that the shape of the container, determines how fast the temperature decay's. Obtaining a model: Exponential model Exponential Reg. G.C Stat > Enter > Type "Time" into L1, "Temperature Average" into L2 > Stat > Calc > 0: ExpReg This will give you the mathematical model (y = abx ) y= temperature difference average (�C) x= time (minutes) y=80.412 � .964x a= 80.412 b= 0.965 r�=.80 Comparing the exponential model to other models Quad Reg. y=ax� + bx + c y=.0043x� + -1.04x + 58.90 r�=.912 graph Between t = 90 to 152 graph indicates a negative temp, after t=121 graph rises which is unrealistic, therefore this is not an accurate model, even though the r� value is better then the exponential. Cubic Reg. y=ax 3 + bx 2 + cx + d y= -5x 3 + .017x 2 + -1.80x + 66.89 r�=.98 At t = 91 from 136 the graph rises and after t=136 the model drops unrealistically. Therefore this is not an accurate model; even though the r� value is better then the exponential. ...read more.

Conclusion

- 20� 30 10� - 5� 25 26� - 13� 22 52� - 26� 21 56� - 28� Average half-life from table = 22.4 minutes Model Half life y=80.412 � .964x Half life occurs when the initial temp is halved in value and the time is calculated from the model the initial temp is 80.412(let x =0) this value is divided by 2 and substituted for y. 80 / 2 = 40.206 y=80.412 � .964x 40.206=80.412 � .964x .5=.965x Log .5=log .965x = x log .965 Log .5/ log .965 = x x = 19.46 Time equals 19.46 Comparing the half life of the model with that from the table, it does not appear a close approximation Difference =22.4 - 19.46 = 2.94 % error = 2.94/22.4 * 100 = 13.13% error To find a linear relationship between k & b use b= e^-k From G.C Stat > Calc >Press 4(LinReg), Y=> Vars > 5: statistics> EQ 2nd Table or Graph y= ax + b b= -.785k + .987 r�= .995 k b .1 .905 .2 .819 .3 .741 .4 .670 .5 .607 ?? ?? ?? ?? 1 ...read more.

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