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You may not know that h is the heat per second flowing from the boiling water to the ice-water mixture. The answer is h=qδt. You may also not know that q and δ are the specific heats of water, ice, and steam respectively, how much heat is being transferred by conduction in this experiment, what period of time it took for these molecules to move across a certain distance in this experiment.

I’m going to show you how much heat per second flows from boiling water to ice-water mixture when a gradually increasing temperature difference between these two liquids exists.
Here’s a diagram showing how the ice and water molecules are moving away from each other as h increases. As time goes on, the temperature difference between ice and water is increasing.

As h increases, you can see that heat flows from the boiling liquid to the much colder liquid. You can also see that heat flows from the hot liquid to the cold liquid as a temperature difference exists between these two liquids.

I’ll show you that more q flows from boiling water to the ice-water mixture as h increases. This means that more heat per second flows from boiling water to the ice-water mixture as h increases. I’ll use the definition of q (heat transferred by conduction) and the definition of h (heat per second flowing from liquid 1 to liquid 2) to prove this statement. I’ll use this information, along with a temperature difference between these two liquids, to solve for q and δ.*
The dissolving of sugar in water is an example of heat transfer by conduction. You can see this in this video: <YouTube>https://www.youtube.com/watch?v=uSV_YLdh4rk</YouTube>

I’ll use the following information to solve for q and δ. This is what I’ll do:

1) How much heat per second flows from liquid 1 to liquid 2 when a gradually increasing temperature difference between these two liquids exists? I’ll call this q.
2) What is the specific heat of liquid 1 (water)? I’ll call this δ.
3) What is the specific heat of liquid 2 (ice and water)? I’ll call this δ’.
Now I’ll use the definition of q to solve for q:
4) How much heat per second flows from liquid 1 to liquid 2 when a gradually increasing temperature difference between these two liquids exists? I’ll call this q.
5) What is the specific heat of liquid 1 (water)? I’ll call this δ.
6) What is the specific heat of liquid 2 (ice and water)? I’ll call this δ’.
7) How much heat per second flows from liquid 1 to liquid 2 when a gradually increasing temperature difference between these two liquids exists? I’ll call this q.
8) What is the specific heat of liquid 1 (water)? I’ll call this δ.
9) What is the specific heat of liquid 2 (ice and water)? I’ll call this δ’.
Now I can use the definition of q to substitute into 7). I can solve for q by rearranging terms: Q = h(deltaT)) + dq/dx, where: h=q(deltaT)). dq/dx = deltaT – Tc \Delta \cdot \Delta t>. Tc is the freezing point of the ice-water mixture.
I’ll use dq/dx to substitute into 9) (specific heat of liquid 2). I’ll solve for δ’. This will give me my values for δ and δ’.
You can see how I used the values given in 6) and 7) to calculate the values for ΔT (the temperature difference between liquid 1 and liquid 2). I also used dq/dx to calculate deltaT.

Just show the information given in the text to each group and they can solve for q and δ.
So the reason that more heat flows from boiling water to ice-water mixture as h increases is because more q flows from liquid 1 (boiling water) to liquid 2 (ice-water mixture). This means that it’s easier for heat to flow from boiling water to ice-water mixture when a temperature difference between these two liquids exists. The definition of q, along with the definition of h, proves this statement: although heat flows from the hot liquid to the cold liquid, more heat per second flows from boiling water to ice-water mixture as h increases.