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September 21, 2014

September 21, 2014

Posted by **bob** on Monday, July 23, 2012 at 12:44pm.

a) What type of function best models the cooling of a hot liquid?

Explain your choice.

b) What is the mathematical model for this situation? (i.e. - the equation)

c) If the optimal temperature for drinking a hot liquid is 280C, at what time would Mr. Currie have had to return in order to enjoy his cup of coffee

- math grade 12 -
**Reiny**, Monday, July 23, 2012 at 1:18pmAccording to the Newton's Law of Cooling we have that:

T(t) = Ts + (To - Ts)*e^(-k*t) ;

where:

t is the time in the preferred units (seconds, minutes, hours, etc.)

T(t) is the temperature of the object at time t

Ts is the sorrounding constant temperature

To is the initial temperature of the object

k is a constant to be found

so let's find k

**Before I go any further, there is something very wrong with your data**

since water boils at 100° C, there is no way that the coffee could have a temperature of 350°

nor can you drink coffee that has a temp of 280°, it would be very hot steam.

check your question and your typing.

- math grade 12 -
**ted**, Monday, July 23, 2012 at 1:23pmits 35° and 20° sorry :$

- math grade 12 -
**Reiny**, Monday, July 23, 2012 at 1:44pmLooking at this question again, I think they want this done in a much simpler way.

when t = 35, the T(35) the temperature has decreased by 1.2% each minute,

so temp after 35 minutes = 35(1.015^-35

= 20.7853°

to have a temp of 28°

we need

35(1.015)^-t = 28

1.015^-t = .8

take log of both sides

log (1.015^-t) = log .8

-t = log .8/log 1.015

-t = -1498

t = appr 15

let me know if we have to use Newton's Law of Cooking equation.

- math grade 12 -
**bob**, Monday, July 23, 2012 at 2:00pmwhere did you get 1.015 from in the first equation?

- math grade 12 -
**bob**, Monday, July 23, 2012 at 2:08pmalso what kind of function is this?

- math grade 12 -
**Reiny**, Monday, July 23, 2012 at 2:16pmsorry, that was a typo, should have been 1.012

for 1 + 1.2%

I am getting myself all messed up here, I noticed that I used 1.015 several times instead of 1.012

Almost too deep into the mess to recover, I should have done this:

reduce by 1.2% per minute ---- >1 - .012 = .988

35(.988)^35 = 22.9°

when does 35° become 28° ?

35(.988)^t = 28

.988^t = 28/35 = .8

t log .988 = log .8

t = log .8/log .988 = 18.48 minutes

does that make more sense?

- math grade 12 -
**Reiny**, Monday, July 23, 2012 at 2:19pmYou are dealing with an "exponential function"

- math grade 12 -
**chad**, Monday, July 23, 2012 at 2:29pm1 - .012 = .988 why did you do this?

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