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anonymous

  • 5 years ago

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.

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  1. anonymous
    • 5 years ago
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    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

  2. anonymous
    • 5 years ago
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    why is everyone asking questions in the wrong sections? is it casue no one checks the others?

  3. anonymous
    • 5 years ago
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    since it is cooling down, k should be negative, i.e k is -1/47

  4. anonymous
    • 5 years ago
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    the final answer the i got was 23.37 but it says its wrong

  5. anonymous
    • 5 years ago
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    how did you get your answer?

  6. anonymous
    • 5 years ago
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    i started out subtracting 67-20

  7. anonymous
    • 5 years ago
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    then i got -1/47 like u did

  8. anonymous
    • 5 years ago
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    then?

  9. dumbcow
    • 5 years ago
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    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

  10. dumbcow
    • 5 years ago
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    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

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

  12. anonymous
    • 5 years ago
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    i got my answer from 67=20+75e^(-t/50)

  13. anonymous
    • 5 years ago
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    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.

  14. anonymous
    • 5 years ago
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    its ok ill just ask my teacher about it

  15. anonymous
    • 5 years ago
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    ok

  16. dumbcow
    • 5 years ago
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    i think you're making it more complicated than it needs to be. t=28

  17. anonymous
    • 5 years ago
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    how did you get that?

  18. dumbcow
    • 5 years ago
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    look at my previous post

  19. anonymous
    • 5 years ago
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    it said that answer is wrong

  20. anonymous
    • 5 years ago
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    i tried submitting that in my h.w and it didnt work

  21. dumbcow
    • 5 years ago
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    oh ok sorry i guess i was wrong

  22. anonymous
    • 5 years ago
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    lol its ok

  23. anonymous
    • 5 years ago
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    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.

  24. dumbcow
    • 5 years ago
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    ah gotcha thank you. i was assuming the cooling was constant. sorry

  25. anonymous
    • 5 years ago
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    you are welcome.

  26. dumbcow
    • 5 years ago
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    hey if you're still on i think i figured it out. check my answer t = 21.914 min

  27. dumbcow
    • 5 years ago
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    hey if you're still on i think i figured it out. check my answer t = 21.914 min

  28. dumbcow
    • 5 years ago
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    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

  29. dumbcow
    • 5 years ago
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    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

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