Mixtures of Systems “From here to there” ThermoChemistry Mixtures of Systems “From here to there”

A Most Important Concept Heat will always flow from a region of high “heat concentration” to a region of lower “heat concentration”.

You already knew that… This is why the stove feels “hot”. heat flow Stove has lots of heat energy. heat flow You have less heat energy.

And you also know that… This is why the window feels “cold” during the winter. heat flow Than the outside does. You have more heat…

Consider the Following A quantity of water at a high temperature is added to a quantity of water at a lower temperature. What do you think will happen? Water at high T Water at lower Temperature

You probably guessed… You will end up with a system that has its own temperature somewhere between the two temperatures of the containers. But can we figure out this “equilibrium temperature” of the mixed system? The answer is:

The thought process: Need to start with the Law of Conservation of Energy… Heat lost by the “hot” system is equal to the heat gained by the “cold” system. Using the notations we have previously developed: qlost = qgained

Continuing (and substituting)… Since qlost and qgained are actually heat flow quantities, we can expand both sides like this: mh  Ch  ΔTh = mc  Cc  ΔTc where the notation “h” represents the system that was initially “hot” and the “c” represents the system that was initially “cold”.

A bit confusing here … We need to explore the ΔT notations because they will need to be expanded themselves. First, the hot system will end up at a lower temperature than it starts. To be sure that we have positive numbers in the equation, we will re-define ΔTh as (Th - Tf) where Th is the initial temperature of the hot system and Tf is the final temperature after mixing. (think about that one for a second…

Continuing on this thought: We will also need to re-define the expression for ΔTc. Keep in mind here that the cold system will actually experience a temperature increase, since it is gaining heat energy from the hot system. Therefore, we will replace the ΔTc with the expression (Tf - Tc) where Tf is the final temperature after mixing and Tc is the initial temperature of the “cold” system.

Putting all of this together: When we substitute the expanded expressions for the ΔT terms in the original equation from three slides ago, we get the following: mh  Ch  (Th – Tf) = mc  Cc  (Tf - Tc) Ultimately, even though this looks to be a mess to solve, there will only be one unknown in the entire equation.