Remaining+Open+to+Continuous+Learning

__Remaining Open to Continuous Learning:__

__My Understanding of this Habit:__ This habit means that everyone should always take an optimistic approach towards learning, taking pride in the fact that it is impossible to always be correct or all-knowing.

__How I applied this Habit to Chemistry:__ When Ms.Knowles arranged us to meet with a former SAS student, Vincent, in order for him to check our Chemistry lessons, I happily accepted. It turns out that my presentation lacked many details, especially in the areas of clarifying certain concepts. For example, I did not draw a clear distinction between the two commonly confused concepts: heat and temperature. Also, my presentation was about how a calorimeter is used to determine the enthalpy of a reaction, but the key words calorimeter and enthalpy were not defined. Then I realized that I did not fully understand what the term "enthalpy" meant. Vincent helpfully explained how enthalpy is also know as the heat in a reaction or delta H. My chemistry lesson also had flaws in the area of checking whether or not the student had a clear understanding of the topics I reviewed in my presentation. So, he suggested that I have a few practice problems or quiz questions to check for my classmates' understanding of the different concepts. This definitely helped my presentation because I was able to clearly communicate to my classmates the meaning of each key term and point in my presentation. It turns out that one of the practice questions I posted in my presentation was quite difficult, but it was great to be able to teach my classmates something new in this area of energetics.

Here is the practice problem that my classmates were having difficulty solving: A 440 sample of mercury (specific heat = 0.14J/g degrees Celsius) is placed into 134 grams of water (initial temperature 35.00 degrees Celsius). Find the temperature of the system.

See whether you can solve it :D [Solution Below]

A) A heat value is not given, so we can equate two Q=CMT equations to each other since the heat released is the same. B) (0.14J/g degrees Celsius)(440 g)(T-22.00 degrees Celsius) = (4.187 J/g degrees Celsius)(134g)(T-35.00 degrees Celsius) C) The final temperature is equal to 33.71 degrees Celsius after the equation is "foiled" out.