The main formula you will see in calorimetry is referred to as q equals mCAT.
Where q is heat measured in either kJ or J
m is the mass measured in grams, C or Cs is the specific heat capacity its unit is J per grams celsius.
And our delta T is the change of temperature which is our final temperature
minus our initial temperature which is measured in Celsius.
It is important to check that all of your given values are in these specific units before plugging your values into the formula.
This page along with others I use throughout the video are part of my 50-page guide on Thermochemistry
which has everything you need to know about Thermochemistry and calorimetry
including common test questions with step by step answers,
so if you would like this guide you can find the link is in the description box along with other helpful resources.
Now let’s do 2 different examples, one finding heat and the other finding specific heat capacity.
Example 1: The specific heat of lead is 0.129 J/g C.
Find the amount of heat released when 497 g Pb are cooled from 37.2 C to 22.5 C.
Let’s identify what we're given and what we're finding.
We are given our specific heat capacity, our mass, and our initial and final temperatures.
From and to are our keywords here.
From means, that’s our initial temperature
and to means, that’s our final temperature.
We are finding the heat released.
The fact that this says released means our answer will be negative.
Since everything is in its proper unit we will start with step 1 which is finding our change in temperature.
Our change in temperature is our final temperature minus our initial temperature.
When we do this we get a -14.7 C as our change in temperature.
Step 2: Plug everything into the formula.
Here’s the formula we will use, we have our mass in grams, specific heat capacity and change in temperature.
Multiply all of these together and our units of grams and celsius cancel out
and we are left with our heat in Joules.
Make sure to round to 3 sig figs. Now we could have written our answer as -942 J.
or we can write it without the negative as long as we put 942 J are released.
Both answers are correct here.
Example 2: When a 120 g sample of aluminum absorbs 9612 J of heat energy,
its temperature increases from 25.0 C to 115 C. Find the specific heat capacity of aluminum.
Let’s identify what we're given and what we're finding.
We are given the mass, our heat which is positive since our keyword here is absorbs.
From 25.0 C means that’s our initial temperature,
to 115 C means this is our final temperature and we are finding the specific heat capacity
which has a unit of J divided by grams celsius.
Since everything is in its proper unit we will start with step 1 which is to find our change in temperature.
Final minus initial gives us 90.0 C.
Step 2: Plug everything into the formula.
So we know our heat, our mass, and we are looking for the specific heat capacity
and we just found our change in temperature.
So multiply these two values together so we get this new value with our units of grams Celsius.
Now to get the specific heat capacity by itself we will divide both sides by this value.
And our specific heat capacity is 0.89 J per grams Celsius.
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