Cooling Coffee and a Doubling Culture
A lab technician monitors two processes. Process A: a cup of coffee at 90°C cools in a 20°C room; after 10 minutes it is 65°C. She models temperature with Newton's law form T(t) = 20 + 70e^(−kt). Process B: a bacteria culture doubles every 3 hours, starting at 500 cells. She wants to know when the coffee reaches 30°C and when the bacteria count exceeds 20,000. She also wonders whether the coffee temperature can ever drop below room temperature under her model, and whether the bacteria model, written as an exponential, has a constant doubling time independent of starting size.
1. Find k for the coffee model and determine when T(t)=30°C.
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Using T(10)=65: 65 = 20 + 70e^(−10k) ⇒ 45 = 70e^(−10k) ⇒ e^(−10k) = 45/70 = 9/14. So −10k = ln(9/14) ⇒ k = −(1/10)ln(9/14) = (1/10)ln(14/9) ≈ (1/10)(0.4418) ≈ 0.0442 per minute.
Set T(t)=30: 30 = 20 + 70e^(−kt) ⇒ 10 = 70e^(−kt) ⇒ e^(−kt) = 1/7 ⇒ −kt = ln(1/7) = −ln7 ⇒ t = ln7 / k ≈ 1.9459 / 0.0442 ≈ 44.0 minutes.
2. Address whether T(t) can fall below 20°C, and analyze the bacteria model's doubling time.
Read the worked answer
Coffee: T(t) = 20 + 70e^(−kt). Since 70e^(−kt) > 0 for all real t and → 0 as t→∞, we have T(t) > 20 for all t, approaching 20 asymptotically from above. It never reaches or drops below 20°C - the model has a horizontal asymptote at T=20. Physically this reflects the room being the equilibrium.
Bacteria: N(t) = 500·2^(t/3). Doubling time: N(t+3)/N(t) = 2^(3/3) = 2, independent of t and of the initial 500. Solving 500·2^(t/3) > 20000 ⇒ 2^(t/3) > 40 ⇒ (t/3)ln2 > ln40 ⇒ t > 3ln40/ln2 ≈ 3(3.6889)/0.6931 ≈ 15.97 hours, so after about 16 hours. The doubling time is intrinsic to the base/rate, not the starting count - a key property of exponential growth.