Water and Milk
Name: Kelli B.
What is scientific reasoning for why heat loss occurs at
different rates when cold milk is added to boiling water as opposed to
boiling water being added to cold milk?
200ml water boiled in kettle. 50 ml cold milk. When milk is added to cup
(identical china mug used in each instance) first then hot water added the
starting temp is lower than when hot water is added to cup first and then
the cold milk. They seem to cool at similar rates with the second example
retaining its heat for longer period. What is the scientific explanation for
Yours is another example of a whole category of:
mysterious-results" experiments. The difficulty, and the explanation, of
such experiments is that it is very difficult to control the experimental
conditions well enough to make any reliable conclusions. Here is an
incomplete list of conditions that are usually uncontrollable:
1. What is the temperature of the cold and hot fluid at the time of
addition. Liquids in contact with high heat capacity ceramic mugs warm or
cool surprisingly fast, because the mug provides a large "heat sink" that
can distort the final temperature significantly.
2. The heat of vaporization of water is very large, about 550 cal/gm. The
heat capacity of one gram of liquid water is about 1 cal/ C. So a very small
change in the amount of hot water can have a large effect on the final
temperature. In addition, water at 100 C. has a vapor pressure of 1 atm. so
its rate of evaporation is very fast. In an open system, this means that hot
water cools very quickly.
3. Since milk is largely water, its heat capacity is essentially that of
water, so the experimental result does not depend on the identity of the
"other liquid" so long as the "other liquid" is predominantly water.
4. Doing "the experiment" more carefully tends to make the result WORSE not
BETTER, because the errors are systematic, not random. In short the slower,
i.e. more carefully you do it, the worse it gets.
5. Such experiments try to imply that the heat balance demanded by the First
Law of Thermodynamics is somehow flawed, when it is the experimental
conditions that are flawed. The First Law states that the heat given up by a
hot body of mass, Mh; heat capacity, Ch; and initial temperature Th; and
absorbed exclusively by a cold body of mass Mc; heat capacity, Cc; and
initial temperature, Tc must be identical, and the final temperature, Tf is
Mh*Ch*(Tf-Th) = Mc*Cc*(Tc-Tf). Solving for Tf gives:
Tf = (Mc*Cc*Tc + Mh*Ch*Th) / (Mc*Cc + Mh*Th)
This basic equation has been verified millions of times under controlled
I am not sure if I understand the question correctly, however, I attempt a
response and hope it addresses what you are asking.
If you pour hot water into a cup first (a) some water evaporates and (b) the
cup warms up rapidly. Having a large surface area, the cup begins losing heat
to cooler ambient air. Evaporation also takes heat away.
When you add in the cold milk (or water, for that matter), liquids mix, and
evaporation is substantially decreased. Whether at that moment, there is a net
transfer of heat from the mixture to the cup or from the cup to the mixture
depends on the specifics of the problem including the size and weight of the
cup, its material, and the time gap between pouring in the hot and then the
If you pour in the cold liquid first followed by the hot liquid, chances are
that the heat loss is less than the first case. There is less evaporation, and
less heat loss from the cup.
As a result, the temperature of the mixture in the second case will be higher
than the first although the rate of temperature drop should be nearly the same
in both case.
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Update: June 2012