# Equilibrium between Steam and Ice

Problem Statement:

$$15 \ g$$ of steam at $$373 \ K$$ ($$100^{\circ} \ C$$) is mixed with $$25 \ g$$ of ice at $$273 \ K$$ ($$0^{\circ} \ C$$). What is the final temperature of the mixture.

My Question: The conceptual problem I'm having is how do I break this into separate smaller thermodynamical processes (such as the ice melting up first) in order to find the final temperature.

I'm not quite sure how the ice and ice interact in this mixture. It seems like a very complicated process to me. Help anyone?

• 100° C is not 100 K. – DYZ Feb 4 at 6:03
• Sorry, I just corrected the typo. Thanks for pointing that out! – Apekshik Panigrahi Feb 4 at 6:06
• 273 K is not 100 C. – Pieter Feb 4 at 6:24
• I sound like a broken record now. Sorry! Thanks for pointing that out! – Apekshik Panigrahi Feb 4 at 6:27
• $T_f=\frac{m_1 C_1 T_1+m_2 C_2 T_2} {m_1 C_1+m_2 C_2}$ where $m_i$ are the masses and $C_i$ are heat capacity per unit mass. – Alexander Feb 4 at 11:14

The concept you need to understand this problem is that the amount of internal energy in your system is a state function. That means it depends only on the temperature, pressure etc of your system and not on the path taken to reach that state.

In this case no heat enters or leaves your system so the amount of heat is constant. What actually happens is the melting ice cools the condensing steam and they meet somewhere in the middle. But we could imagine a different path.

Suppose we inject heat into the system to melt the ice, then boil the water to make steam at 100°C. And suppose we have to put in some heat $$H$$ to do this. Now we have 40g of steam at 100°C. Then we let the steam condense, then cool, until we get the same amount of heat $$H$$ out again. Now we reach a state where the net amount of heat hasn't changed, and because internal energy is a state function this must be the same state that we would have reached by letting the steam and ice interact directly.

Or alternatively suppose we let the steam condense, then cool to 0°C, then freeze, and we get out some amount of heat $$H$$ by doing this. This gives us 40g of ice at 0°C. Now we put the same amount of heat $$H$$ back in to melt the ice, then heat the water, and once again we reach the same state. So this will give us the same result as the two processes above.

We often show the process as diagrams like this:

So the left diagram shows us putting in heat $$H$$ to make steam then taking it out again, while the right diagram shows us taking out heat to make ice then putting it back in again. Because in both cases the net change in the heat is zero the two final states must be the same, and both will be the same as if we just let the ice cool the steam directly.

• In the first path where heat $H$ is injected into the system, along with melting ice and boiling the water, won't the steam already present rise up in temperature? – Apekshik Panigrahi Feb 4 at 6:04
• Remember this is a thought experiment, not the way you would actually do it. In the first case I'm imagining keeping the 25g ice separate from the 15g steam while I melt the ice then boil the water. Then the last step is to combine the 25g of steam at 100°C with the original 15g of steam at 100°C to get 40g steam at 100°C - no heat is absorbed or released in this step. – John Rennie Feb 4 at 6:07
• So we use the fact that "heat is a state function" to our advantage. We make u p our own process with the same initial and final conditions. This new process is easier to think of and solve while still keeping the problem same. Cool! That helps a lot! – Apekshik Panigrahi Feb 4 at 6:12
• @ApekshikPanigrahi yes, exactly. – John Rennie Feb 4 at 6:15
• I think that it would be better to replace a few instances of the word "heat" with "thermal energy" in this answer. Most thermo textbooks that I have seen use "heat" to refer to an energy transfer between systems, which is not a state function, and "thermal energy" to refer to the related state function. – user1476176 Feb 4 at 11:34