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Name: Kelli B.
Status: student
Age: 30s
Location: N/A
Country: N/A
Date: 1999-2001

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?

The specifics:

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 this?

Yours is another example of a whole category of: "mixing-liquids-at-different-temperatures-that-seem-to-have- 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 given by:

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 conditions.

Vince Calder


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 cold liquids.

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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