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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.
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let'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*6720or, k = 1/47

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why is everyone asking questions in the wrong sections? is it casue no one checks the others?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since it is cooling down, k should be negative, i.e k is 1/47

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the final answer the i got was 23.37 but it says its wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get your answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i started out subtracting 6720

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then i got 1/47 like u did

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0another 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 = (6795)/t solve for t

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0another 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 = (6795)/t solve for t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok my work is all over the place..but i ended up with t=50ln(47/75)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got my answer from 67=20+75e^(t/50)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0its ok ill just ask my teacher about it

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0i think you're making it more complicated than it needs to be. t=28

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get that?

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0look at my previous post

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it said that answer is wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i tried submitting that in my h.w and it didnt work

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok sorry i guess i was wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dumbcow 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
 5 years ago
Best ResponseYou've already chosen the best response.0ah gotcha thank you. i was assuming the cooling was constant. sorry

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0hey if you're still on i think i figured it out. check my answer t = 21.914 min

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0hey if you're still on i think i figured it out. check my answer t = 21.914 min

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0anyway 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
 5 years ago
Best ResponseYou've already chosen the best response.0anyway 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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