The main formula you will see in calorimetry is referred to as Q equals m count, where Q is heat measured in either kilojoules or joules, m is the mass measured in grams, C or Cs is the specific heat capacity, its unit is joules per gram 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 the specific units before plugging your values into the formula. This page along with others I used 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'd like to get this guide, you can find the link in the description box along with other helpful resources. Now let's do two different examples, one finding heat and the other finding specific heat capacity. Example 1. The specific heat capacity of lead is 0.129 joules per gram Celsius. Find the amount of heat released when 497 grams of lead are cooled from 37.2 degrees Celsius to 22.5 degrees Celsius. 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 key words 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 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 negative 14.7 degrees Celsius 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 in Celsius cancel out, and we are left with heat in joules.
Make sure to round to 3 sig figs. Now, we could have written our answer as negative 942 joules, or we can write it without the negative as long as we put 942 joules are released. Both answers are correct here.
Example 2. When a 120 gram sample of aluminum absorbs 9,612 joules of heat energy, its temperature increases from 25 degrees Celsius to 115 degrees Celsius, 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 absorbed, from 25 degrees Celsius, means this is our initial temperature, right?
115 degrees Celsius means this is our final temperature, and we are finding the specific heat capacity, which has a unit of joules divided by grams Celsius. Since everything is in its proper unit, we will start with step one, which is to find our change in temperature. Final minus initial gives us 90 degrees Celsius.
Step two, 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 joules per gram Celsius. For more help with calorimetry or thermochemistry, make sure to check out my thermochemistry guide which is right over here. I know that can help shorten your study time and get you a better grade. And make sure to check out the description box for all of my other recommended resources like homework help and online tutoring.