In this video you will understand
calorimetry. Alright let's do this!
Hello-hello Melissa Maribel here and I
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chemistry together.
Today's topic is calorimetry. Calorimetry
is used to measure amounts of heat
transferred to or from a substance. A
calorimeter measures that heat transfer.
To better understand this concept, let's say that this hot coffee is the
system and your cup is its surroundings.
Your system is whatever you are
investigating or examining and its
surroundings is whatever is coming in
contact with that system. So say if we were to pour our hot coffee into
our cup then the cup becomes warm. Right, that makes sense the cup is now warm.
Because that hot coffee is releasing its
heat to the cup. So your system is
releasing its heat to its surroundings.
So we can say that "q" of our system is
equal to the "-q" of our
surroundings. "q" just refers to heat.
The heat of your system will always be the opposite sign of the heat of its
surroundings. Now let's say if we were to
have this beaker with water at
25 degrees Celsius and a copper penny at 55 degrees Celsius. If you drop that penny
into the water then the penny will
release its heat to the water until they
both have the same final temperature.
Keyword right there, final temperature,
they're both gonna have the same. Alright, so that goes on to another formula where
"q" metal is equal to the "-q" of
your water. Recall that "q" is equal to
MCAT and "q" is your heat which is equal to the mass times the specific heat
capacity times your change in
temperature. And your change in
temperature is the final temperature
minus your initial temperature.
Alright so this formula expanded looks
like this. Okay let's try one.
A 32.5 gram sample of copper at 45.8 degrees Celsius
is placed into 105.3 grams of water at
15.4 degrees Celsius what is the
final temperature? And you're given the
specific heat capacities of copper and
H2O. Let's identify our other givens. Your
first given is the mass of copper which
is 32.5 grams. We're given our
specific heat capacity of copper and
your initial temperature of copper, that
45.8 degrees Celsius. You're
also given your mass of water, specific
heat capacity of water, and your initial
temperature of water. The proper formula
for this specific type of example is our
heat of our metal is equal to the
negative heat of your water. Expanding
this formula because you know that your
heat equals MCAT or the mass of our
metal, the specific heat capacity of our
metal, and our change in temperature.
Remember change in temperature just
means your final temperature minus your
initial temperature. We're going to set
that equal to the negative portion of
our mass of our water, specific heat
capacity of our water, and the change in
temperature of water. Plugging in all the
values that we found before and changing this change in temperature to T final
minus T initial just our initial
temperature for both of these values.
Next we're going to multiply 32.5 and
our 0.385 together as well as these two
values with our negative and this is
what you get. Instead of distributing and
having this big long problem. What
we'll do instead is actually divide by
12.5125 to each side. That'll make this a lot simpler.
We'll cancel these two out and divide
these two values and you get,
you get this. Next I'm gonna distribute
now. Distributing this in just multiplying
we get this we have two T Finals.
Combine like terms just like math. So I'm
going to add over that 35.218 T final to the opposite
side because there's really a 1 in front
of that T final. Combine those two values
and add over the 45.8 to the opposite side and this is what
you get. Next we'll divide both sides by
that 36.2108
dividing both of these, this would then
cancel giving us our T final which is
16.2 degrees Celsius.
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