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Name: Therese C.
Status: student
Grade: 9-12
Country: Australia
Date: Fall 2009


Question:
What effect does the colour of a container have on the retention of heat in the water? I have tried it as an experiment in relation to Newton's law of cooling, but am having trouble getting my head around the scientific understanding. I found that the colour of the tins made very little difference to the results, so am presuming that the temperature outside the tine was what effected the heat loss. Can you assist with any reasons or scientific concepts?


Replies:
Heat transfer from air to containers (or any closed vessel) happens by two paths: radiative and conductive. Radiative heat transfer is energy carried by electromagnetic radiation (like light or infrared). Conductive heat transfer happens when something warmer comes in contact with something colder (this is what Newton's law of cooling describes).

Color can impact radiative heat transfer, but will have basically no effect on conductive heat transfer. If your containers are in normal indoor conditions (no bright lights or sunlight and not glowing-hot), radiative heat transfer will be minimal, and color will matter little. If the containers are in bright sunlight, then radiative color will have more of an effect.

To go in to more detail, I would need to know more about your experimental setup. If you would like more information, feel free to respond, and please describe all the details.

Hope this helps,
Burr Zimmerman


Hi Therese,

Newton's law of cooling applies to heat loss by conduction or convection. It basically says that heat loss from an object to its environment is proportional to the difference in temperature between the object and the environment. This "law" is not very accurate, since in reality heat transfer is FAR more complex than this! Newton's law of cooling is only roughly accurate for small differences in temperature between object and environment.

Color of the object that is cooling has nothing to do with Newton's law of cooling. There are 3 different "mechanisms" of heat transfer:

- Conduction to a cooler object (where the hot and cool object are actually in contact).

- Convection (such as when a hot object heats nearby air causing it to rise, resulting in a rising air current that helps cool the object). This is the major mechanism of heat loss in your experiment, and has nothing at all to do with color.

- Radiation (where a very high temperature object loses heat by infrared radiation).

Regarding radiation cooling, a shiny metal object will radiate less heat than one that is black. BUT (and this is important), the amount of heat lost by radiation can be very large for objects that are red hot, but for objects less than 100 to 200 degrees Celsius, the amount of heat lost by radiation is simply too small to be significant (regardless of what color they are).

It is also important to note that visible "color" itself does not matter at all, because this is the color of visible light. Heat is not lost by radiation of visible light, but rather by radiation of invisible infrared energy. Nearly all colors of paint appear "black" at infrared wavelengths, and so nearly all colors of paint radiate infrared energy (heat) much the same.

But once again, heat loss by radiation is ONLY significant for objects that are at a very high temperature, and is of no significance at the temperatures you are working at.

In your experiment, nearly all the heat is lost by Convection, where the hot object heats the air it comes in contact with, causing the heated air to rise and expose fresh cooler air to the object, further cooling the hot object. The color of your water container has no effect whatsoever on heat loss by convection, and the water is nowhere near hot enough to cause any significant heat loss by radiation.

Regards,
Bob Wilson


There are much larger effects other than container color that influence the heating/cooling of a container. The outside temperature, the heat capacity of water is much larger than the effect you are trying to measure -- the list is long. I do not think you are going to have much luck with simple apparatus that you would have available. The effect is small.

Vince Calder


Therese,

It might be that you are do not have equipment that can measure small variations in cooling rates. You may have to rethink the design of your experiment. Think of all the variables that can affect cooling and remove all those variables except color. This means, you must keep the starting temperature, the amount of water, the material of the container, the size and shape of the container, the external temperature, etc. all the same - except the color of the container. Remember that evaporation is a very effective cooling effect.

When you have achieved this, take temperature readings every 2 minutes or so until the temperature reaches +10deg of the external temperature, and see if color has any effect on how fast the temperature went down.

Greg (Roberto Gregorius)
Canisius College



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