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anonymous
 one year ago
Please help!
You place a cup of 205oF coffee on a table in a room that is 72oF, and 10 minutes later, it is 195oF. Approximately how long will it be before the coffee is 180oF? Use Newton's law of cooling:
T(t)=TA+(To−TA)e−kt
A. 15 minutes
B. 25 minutes
C. 1 hour
D. 45 minutes
anonymous
 one year ago
Please help! You place a cup of 205oF coffee on a table in a room that is 72oF, and 10 minutes later, it is 195oF. Approximately how long will it be before the coffee is 180oF? Use Newton's law of cooling: T(t)=TA+(To−TA)e−kt A. 15 minutes B. 25 minutes C. 1 hour D. 45 minutes

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0plug in your numbers and solve for k. TA is the ambient temp, 72 and T0 is 205 the initial temp \(T(t) = 72+(20572)e^{kt}\) \(T(t) = 72+(178)e^{kt}\) Since we know T = 195 at t = 10, you can solve for k \(195 = 72+(178)e^{10k}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Once you have k, plug it into the equation for T(t) and solve for t when T = 180

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then you can do \[180=72+178e^{0.03696t}\] BTW, the more digits you keep for k, the more accurate your t will be.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah that's what I got. I'm wondering if you're supposed to convert the units to celsius or seconds

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have no idea. it doesn't say
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