7-4 Exponential Growth and Decay

Newton’s Law of Cooling: T – TS = (T0 – TS)e–kt Ex 7) A temperature probe (thermometer) is removed from a cup of coffee and placed in water that has a temperature of TS = 4.5° C. Temperature readings T, as recorded in the table, are taken after 2 sec, 5 sec, and every 5 sec thereafter. Estimate: a) The coffee’s temperature at the time the temperature probe was removed. b) The time when the temperature probe reading will be 8°C. Time (sec) T (°C) T – TS (°C) 2 64.8 60.3 5 49.0 44.5 10 31.4 26.9 15 22.0 17.5 20 16.5 12.0 25 14.2 9.7 30 7.5 Newton’s Law of Cooling: T – TS = (T0 – TS)e–kt T – 4.5 = (T0 – 4.5)e–kt (need this 3rd column which is 2nd column – 4.5) L1 L2 Use calculator to calculate an exponential regression STAT  CALC  0: ExpReg

y = abx  a = 61.656 b = .928 Now T – 4.5 = (61.656)(.928)t T = 4.5 + (61.656)(.928)t Check it! STAT PLOT 1-ON, 1st type, X: L1, Y: L2 WINDOW: [0, 30] x [0, 65] (Scale = 5) Y1 = 4.5 + (61.656)(.928)t Coffee’s temp when thermometer removed T = 4.5 + (61.656)(.928)0 = 66.156° Time thermometer reads 8° C Graph Y2 = 8 and find intersection t = 38.228  38 sec (need to adjust window)

homework Pg. 361 # 1, 2, 4, 5, 7, 8, 22, 25, 33