water whose temperature is at 100°c is left to cool in a room where the temperature is 60°c. after 3…
water whose temperature is at 100°c is left to cool in a room where the temperature is 60°c. after 3 minutes, the water temperature is 90°. if the water temperature t is a function of time t given by t = 60 + 40e^{kt}, find the time for the water temperature to reach 65°c. round to the nearest hundredth of a minute. minutes question help: ebook
Answer
Explanation:
Step1: Find the value of (k)
Given (T = 60+40e^{kt}), when (t = 3), (T=90). Substitute into the equation: (90=60 + 40e^{3k}). First, simplify: (90-60=40e^{3k}), so (30 = 40e^{3k}). Then (e^{3k}=\frac{30}{40}=\frac{3}{4}). Take the natural logarithm of both sides: (\ln(e^{3k})=\ln(\frac{3}{4})). Using the property (\ln(e^{x})=x), we get (3k=\ln(\frac{3}{4})). So (k=\frac{1}{3}\ln(\frac{3}{4})\approx\frac{1}{3}(- 0.2877)\approx - 0.0959).
Step2: Find the time (t) when (T = 65)
Substitute (T = 65) and (k\approx - 0.0959) into (T = 60+40e^{kt}). We have (65=60 + 40e^{-0.0959t}). Simplify: (65 - 60=40e^{-0.0959t}), so (5 = 40e^{-0.0959t}). Then (e^{-0.0959t}=\frac{5}{40}=\frac{1}{8}). Take the natural logarithm of both sides: (\ln(e^{-0.0959t})=\ln(\frac{1}{8})). Using (\ln(e^{x})=x), we get (-0.0959t=\ln(\frac{1}{8})). Since (\ln(\frac{1}{8})=- \ln(8)\approx - 2.0794), then (t=\frac{-2.0794}{- 0.0959}\approx21.68).
Answer:
(21.68)