Lab background: I set up two styrofoam cups attached together by an aluminum tra

Question

Lab background:

I set up two styrofoam cups attached together by an aluminum transfer bar (a conductor). One cup contained 200g of boiling water, while the other cup contained only 100g of ice water. Every two minutes, I took the temperature of both the cups. There were two slits on the top of the enclosed styrofoam cups to allow a thermometer to take the temperature of both of the contents in the two cups. It seems as though, the hot water LOST heat nearly twice as fast as the cold water gained heat.

Question 1: Why did this happen??? Does it have to do with the different volumes/ surface area?

Question 2: I later calculated the heat lost by the hot water and the heat gained by the cold water (which, technically, should be the same consiering the laws of thermodynamics). Heat lost= 27720 J and heat gained= 11340 J. So they are clearly not the same ... what 2 reasons could cause the difference???

Questions 3: How do I estimate at which both containers should stabilize???

thanks

Explanation / Answer

Question 1)

First, you do not have perfect insulatiors. Although the styrofoam is ok, you will have some losses to it. Also, the lids are probably not very efficient, especially with holes in them. The thermometer will take away some of the heat every time you put it in the water.

Mainly, however, the aluminum conductor will take on some of the heat and create the most significant amount of loss. Additionally, the aluminum is probably not insulated during the transfer and will cause losses to the atmosphere.

Question 2)

The difference is related to the above. The main difference, and most significant, are the TWO losses using the aluminum. That is losses of heat into the aluminum itself as it raises its own temperature and losses to the atmosphere from the aluminum as the heat is trying to transfer to the cold water from the hot. Overall, the aluminum is the main culprit. Just dumping the hot water into the cold would have given more accurate results and would lead to the answer to part 3 below.

Question 3)

Apply Q = mc(delta T) = (mc(delta T) + mH

The left side is the hot water, the right is the ice water. Since I don't know the mass of ice specifically and water specifically, I can just show you the main idea. First the ice will melt, then all of the total cold water (assuming icewater at zero Celsius initially - many factors left out of your question here)

Thus (200)(4186)(100 - Tf) = (m(coldwater)(4186)(Tf - 0) + (m(ice)(2.26 X 10^6)

Hard to tell what the estimation will be without the mass of ice and cold water separately, but let me assume a 2 to 1 factor since 200 g of hot and 100 g of cold. Perhaps your teacher is looing for an estimate of 66.6 degrees. That is 2/3 of the way between 100 degrees and 0 degrees and would be a correct estimate if the ice water had no ice at all but was at 0 Celsius to start.

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