coffee drinkers' problem (1 Viewer)

[mojako]

hi..
does anyone have an answer (with the working) to the "coffeee drinkers' problem" in 3U Cambridge mathematics book, page 281 (last question of exercise 7H)?
Well if you don't have the book here's the question.
A couple pour themselves a cup of coffeee each just after the kettle has boled. the woman adds milk from the fridge, stirs it in and then waits for it to col. the man waits for the coffee to cool first, then just be4 drinking adds the milk and stirs. if they both begin drinking at the same time, whose coffee is cooler?
assume that the air temp is colder than the coffee and that th milk is colder still. also assume that after the milk is added and stirred, the temp drops by a fixed percentage [umm i'm not sure what this means]

EDIT: the temp drops by a fixed percentage (from the previous temp) means it experiences exponential decay ^^

Hint: it has do to with exponential growth/decay with a limitting value [which is what section 7H is all about]

Thanks

 

[Rorix]

T1 = A + Be^-kt
T2 = A + Ce^-kt, where C percentage of B.

The coffee must be colder regardless of the values of A, B, C and k, for all t, so taking a=1, b=2, c=1 (thus the effect of the coffee being to halve the temperature), after one unit of time

T1 = 1 + 2e^-1
T2 = 1 + 1e^-1
Then T1 adds milk, so T1 = 1/2 + e^-1 < T2 = 1+e^-1

Hmm, now that I think of it, just having t= large number, t1 approx = t2 approx = 20.
Reducing t1 will mean that it is cooler than t2..

Alternatively you can think of the physics principles, the rate of change in temperature is proportional to the difference in temperature between the two mediums, so the rate of change in T1 is greater than rate of change in T2, so when milk is added to T1 it will reduce it's temperature below T2.

 

[mojako]

I've been trying to understand your answer but I haven't been so successful...
Can you clarify a few things, like what are T1 and T2

>> The coffee must be colder regardless of the values of A, B, C and k, for all t <<
Which coffee? [I'm becoming suspicious that you didn't get the question... sorry if you did]
Bill Pender [the author of the textbook] actually thought that the man's coffee was cooler.

 

[Xayma]

The womans will be warmer, the mans cooler. The rate of change of cooling is proportional to the temperature, hence the mans will cool more quickly. Although the woman's is cooler originally the colder temperature of the milk will be effectively cancelled by the hotter water.

 

[Rorix]

T1 = A + Be^-kt
T2 = A + Ce^-kt, where C percentage of B.

Clearly, t1 is the man's coffee, as it's hotter than the woman's coffee at t=0.

 

[gman03]

At the glance of the problem, I would assume both add significantly little amount of milk.

Now for the woman's coffee, the resultant temperature after adding the cold milk to hot coffee will still be fairly hot (just like in real life?!?) so after t minutes it would still be little bit warmer than the normal temperature ( C > 0)..

NOw for the man's coffee, after t minutes, the coffee would be near the room temperature. So adding cold milk to it will drop coffee temperature slightly to below the room temperature...

So I guess the man's coffee would be cooler.

 

[gman03]

Mathematically

Let initial temperature of the coffee be H, room remperature be R

@ t=0

Man's coffee T_m = H
Woman's coffee T_w = H - delta, where delta represents the drop in temperature by the milk.

So T_w = R + (R - H + delta)e^-kt
T_m = R + (R - H)e^-kt

so after many minutes when the coffees are cooled, e^-kt --> 0

So say T_m = T_w = R

now man add milk to coffee, so T_m < R
so T_m < T_w

 

[mojako]

Thanks all ^^

 

[mojako]

So T_m = R + (R - H + delta)e^-kt
T_w = R + (R - H)e^-kt

Isn't that the other way around [swap Tm and Tw]?

 

[gman03]

Oops.... yes you are right

I will fix it

 

[Sirius Black]

?

where did ur guys get them expression for the temperature?

 

[mojako]

where did ur guys get them expression for the temperature?

hmm firstly, let me introduce myself...
im a male hermione

well, i assume uve read gman's post, #7 [and perhaps #6 as well].
rorix post, #2, is a bit harder to understand.
He said:
Man's coffee Tm = H
Woman's coffee Tw = H - delta, where delta represents the drop in temperature by the milk.

let's pay attention at the woman's coffee.
it's assumed that only a little amount of milk is added and the temp of the milk quickly come down to be equal to the temp of the coffee [which is now a bit cooler than it was before].
we take t = 0 at the instant that the temp of the milk and the temp of the coffee is equal.
of course the temp of the coffee has decreased compared to when milk hadn't been added... that decrease is represented by delta.

so we dont worry about the "decay" that the milk causes to the coffee... and if we do then it will be really complicated... and i think that part of the decay wont be an exponential decay.

im sure this makes it clearer.

 

[Steven12]

is this kind of question ever likely to be in the HSC
if it is..........

 

[mojako]

is this kind of question ever likely to be in the HSC
if it is..........

NOPE
It's suppose to be one of those "interesting" questions in the textbook...
a bit like question 22 of excercise 10C of the Cambridge 3U HSC textbook, on probability... that question is also somewhere in this forum.. not sure if it's the Extension 1 forum...