A freshly brewed cup of coffee has temperature 95°C in a 20°C room. When its temperature is 67°C, it is cooling at a rate of 1°C per minute. When does this occur? Approximate to three decimal places.

- anonymous

- jamiebookeater

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- anonymous

let's say the temperature at time t is C(t).
at t = 0, C(t) = 95
the rate of cooling dC(t)/dt is proportional to the difference in temperature between the coffee and the room.
so, dC(t)/dt is proportional to C(t)-20
at C(t) = 67, dC(t)/dt = 1 = k*67-20or, k = 1/47

- anonymous

why is everyone asking questions in the wrong sections? is it casue no one checks the others?

- anonymous

since it is cooling down, k should be negative, i.e k is -1/47

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## More answers

- anonymous

the final answer the i got was 23.37 but it says its wrong

- anonymous

how did you get your answer?

- anonymous

i started out subtracting 67-20

- anonymous

then i got -1/47 like u did

- anonymous

then?

- dumbcow

another way of looking at it is;
f(0) = 95
f(t) = 67
f'(t) =-1 *the rate of change at time t
using the definition of the derivative
f'(t) =f(t) -f(0)/t
-1 = (67-95)/t
solve for t

- dumbcow

- anonymous

ok my work is all over the place..but i ended up with t=-50ln(47/75)

- anonymous

i got my answer from 67=20+75e^(-t/50)

- anonymous

you need to find the rate of change of the rate of change. I am pretty sure a double derivative is needed here, but my calculus is a bit rusty.

- anonymous

its ok ill just ask my teacher about it

- anonymous

ok

- dumbcow

i think you're making it more complicated than it needs to be.
t=28

- anonymous

how did you get that?

- dumbcow

look at my previous post

- anonymous

it said that answer is wrong

- anonymous

i tried submitting that in my h.w and it didnt work

- dumbcow

oh ok sorry i guess i was wrong

- anonymous

lol its ok

- anonymous

dumbcow your answer is wrong. The rate of cooling is higher initially and it slows down as the temperature of the coffee approaches the room temperature. Thats why I said you need to find the rate of change of the rate of change.

- dumbcow

ah gotcha thank you. i was assuming the cooling was constant. sorry

- anonymous

you are welcome.

- dumbcow

hey if you're still on i think i figured it out.
check my answer t = 21.914 min

- dumbcow

hey if you're still on i think i figured it out.
check my answer t = 21.914 min

- dumbcow

anyway here is my solution:
assume a decay function starting at 95 and going down and approaching 20 over time
f(x) = a*b^x + 20
f(0) = a+20=95 => a = 75
f(x) = 75b^x+20 = 67 => b^x = 47/75
f'(x) = 75ln(b)b^x = -1 => ln(b)(75)(47/75)=-1 =>ln(b) = -1/47
thus b = e^(-1/47)
so b^x = e^(-x/47) = 47/75
take ln both sides
-x/47 = ln(47/75) => x = -47*ln(47/75) = 21.965

- dumbcow

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