Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

how did you get your answer?

Answer 6

i started out subtracting 67-20

Answer 7

then i got -1/47 like u did

Answer 8

then?

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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.

Answer 14

its ok ill just ask my teacher about it

Answer 15

ok

Answer 16

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

Answer 17

how did you get that?

Answer 18

look at my previous post

Answer 19

it said that answer is wrong

Answer 20

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

Answer 21

oh ok sorry i guess i was wrong

Answer 22

lol its ok

Answer 23

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.

Answer 24

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

Answer 25

you are welcome.

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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