temperature of tea after 10 mins,,?

Question

CASE1--

make tea,(without adding milk)
then quickly pour milk in it,(normal temperature milk)
the temperature of the solution after 10 mins is "x" degree Celsius,

CASE2--
make tea,(without adding milk)
then pour milk in it after 10 mins,(normal temp. milk)
the temperature of the solution is "y" degree Celsius,

question--
which temperature will be hotter,,
"x" or "y"??

pls give me an approx. example,,

Answer 1

The temperature of mixture in case 2 will be colder. This is what I though. Hmmm?

Okay, so you want some explanation....

We can express the heat equation as
T(t)= T(ambient) – [T(0) – T(ambient)] exp(-kt)
Where
T(t) – Temperature as a function of time
T(ambient) – ambient temperature
T(0) – Initial temperature of tea
k- const
t – time

To answer this question we really need to know what was the initial temperature and mass of tea, how much milk was added and at what temperature, and the heat conduction of the tea container.

Let's do case 2 first
Base on my experience as a tea drinker a cup tea becomes warm (going from 95C to say 60C) in 10 min. This is 60/95 x 100%= 63.2% (So much for number fudging)

I usually add 10% of milk by volume

Cp m(T1- T2) = Cp m ( T2- T3)/ 10 (tea and milk added)
(T1- T2)= ( T2- T3)/ 10
10 T1 – 11T2 + T3=0
T1= 60C
T2= ?
T3= 2C (I like my milk really cold)
T2= (10T1+T3)/11 = 54.7C
That was Case 2

Case1

Quickly pour milk in
T2= (10T1+T3)/11
T1= 95C
T2 -?
T3- 2C

T3= 86.5 C

Now the constant is say the same 10min
T(10 min)= (86.5 C) 63% = 54.5C

Sounds like tug of war LOL where Case 2 is just a bit on a winning side.

But I would say it is a draw....(considering my fudging)

What if we add 20% of milk?
Now T2= (5T1+T3)/6

Case 1
T2=( 475 + 2)/6= 79.5
Cooling for i time constant (10min)
T= T2 x .63= 50.085

Case 2
T1= 95 x .63= 59.9
T2=( 59.9 x 5 + 2)/6 = 50.25

I guess we have a draw again.

Answer 2

I think the temperature should be the same in both cases. here's why:

the rate of cooling is proportional to the difference D between the temperature of the tea and room temperature. this means that the temperature asymptotically approaches room temperature, and the graph is similar to that of exponential decay. in particular there is something like a "half-life": some interval of time where D halves itself. for the sake of simplicity, let's say that the half-life is 10 mins.

also for the sake of simplicity, let's say you're mixing the tea and milk in equal quantities. when you do that, then the resulting temperature is the average of room temperature and the temperature of the tea before you added the milk. in other words, this causes D to half.

since adding the tea and waiting 10 mins have the same effect (halving D), both cases produce the same result, that the difference in temp. is reduced by a factor of 4.

addendum:
I now agree with donald's argument below. the key point is that the coefficient k is smaller for the tea with milk, since it has more volume. his other rationale, that "it lowers the temperature of the liquid so that cooling proceeds at a slower pace from the outset", has no effect on the final temperature, as you can see from his calculations.

Answer 3

The temperature of both will be the same at 10 minutes, the moment you pour the room temperature milk to your second cup of milk the milk will take some of the heat thus making it cooler than the cup that has tea only. Your controlled cup Case 1 in this experiment has no help in cooling or heating after ten minutes so it will remain hotter than your experimantal cup Case 2.

Answer 4

The other guys answer is wrong, they will not be the same. I have taken advanced heat transfer classes and am a combustion engineer.

Case one will be hotter. This is because heat transfer is related to delta T between an object and its surroundings. Since the tea w/o milk is hotter than the tea with milk, it will dump more heat into the room in the same 10 minutes. Then after adding the milk it will be even cooler.

Answer 5

The tea with milk added at the end will be cooler than the tea with milk add initially. The hot tea cools exponentially, more rapidly at first, with the rate of heat loss slowing as it becomes cooler, with horizontal asymptote the ambient temperature. Adding the milk at the start does two things: it lowers the temperature of the liquid so that cooling proceeds at a slower pace from the outset, and it increases the heat capacity of the liquid (more mass), also slowing the cooling. To actually verify this, we need to determine the two exponential curves and compare them; this is just a heuristic explanation.

To analyze this, we express everything in terms of the difference in temperature between the tea and ambient, call this deltaT(t), t in minutes. Then Newton's law of cooling tells us that for tea without milk,

deltaT(t) = deltaT_0 e^(-kt)

where deltaT_0 is the initial difference and k is a positive constant depending on the heat capacity of the fluid. If we add milk, two things happen. The intial value of deltaT is smaller, and the constant k is smaller since we have more fluid to cool off. Let deltaT' be the temp difference for tea with milk; we have

deltaT'(t) = deltaT'_0 e^(-k't)

with deltaT_0 smaller than deltaT_0 and k' smaller than k. In fact, whenever we add milk to the tea, at that instant the temperature difference decreases by a constant factor, call it c<1 , since the new temp difference is the same weighted average of the tea temp and the ambient. So we can write deltaT'_0 = c*deltaT_0, and at t=10 we have for tea with milk at the start:

deltaT'(10) = c*deltaT_0 e^(-10k')

For hot tea at the start we have at t=10:

deltaT(10) = deltaT_0 e^(-10k)

and when we add milk we just multiply the equation by c,

c*deltaT(10) = c*deltaT_0 e^(-10k).

If we now take the ratio of temp differences

(tea with milk later):(tea with milk at start)

we get

c*deltaT_0 e^(-10k)
-------------------------- = e^(-10(k-k'))
c*deltaT_0 e^(-10k')

Since k>k', this number is < 1, telling us that the denominator is larger than the numerator, i.e., the tea with milk first is warmer after 10 minutes, and indeed, for all eternity. This analyis ignores the additional fact that the initially hotter tea cools more by evaporation and convection than does the cooler tea, so that the effect of adding milk is even more pronounced. In other words, we are underestimating the amount by which the milk-first temperature exceeds milk-last.

Answer 6

Case 1, the Tea will be hotter.

Reason, you are allowing the Tea in Case 2 to cool, then you add the cold milk after the Tea has cooled for 10 minutes.

So you are adding cold milk to a cooler tea, and it will be cooled more than when you added milk to the hotter tea.

Answer 7

In Case 2, the hotter water will cool more quickly due to vaporisation and, after 10 minutes, adding the milk will cool it further.
Therefore, tea 'x' will be hotter.