Chemistry Unit 3 Lesson 4: Basics of Calorimetry

Chemistry Unit 3 Lesson 4: Basics of Calorimetry

Introduction to Calorimetry

• Calorimetry is the science of measuring heat energy.

• It involves using a calorimeter to measure the heat exchanged during chemical or physical processes.

• This is how we initially discover the heat capacity of new substances as well as the energy contained within the bonds of certain substances.

• Calorimetry is used in various fields, including chemistry, physics, and engineering.

• By understanding calorimetry, we can unlock the secrets of heat energy and its applications.

Introduction to Calorimetry

• Calorimetry is built on the backbone of the law of conservation of mass.

• The law states that energy is never created or destroyed but instead is merely transferred.

• In calorimetry, the idea is that if one item gains heat we know that heat must have come from the other item losing it.

• If we monitor the temperatures of both substances as this takes place, we can determine how that energy affected their temperatures and determine their specific heat capacities.

Open Ended

How do you think the Calorie content of foods is found?

Introduction to Calorimetry

Introduction to Calorimetry

• Another idea that is important for us to understand with calorimetry is one we learned in lesson 1.

• Heat will always flow from higher temperatures to lower temperatures.

• This will continue to occur until the temperature reaches equilibrium.

• This means heat will continue to transfer between both substances until their temperature is the same.

• The tricky part is this is not always in the middle.

Introduction to Calorimetry

• For example, if a red-hot piece of metal is dunked into a bucket of water. The final equilibrium temperature will be much closer to that of the water, not the metal. This is because of the very high heat capacity of the water and the low heat capacity of the metal. \

• The difference in the mass of the two objects can also play a major role in what the final equilibrium temperature will be.

Multiple Choice

If a 12 g 950 ^oC piece of iron with a Cp of 0.789 cal/g^oC is dropped into a 500g 24 ^oC bucket of water with a Cp of 1.0 cal/g^oC, which of the following is a reasonable guess at the final temperature of these items once reaching equilibrium?

• 900 ^oC

• 150 ^oC

• 50 ^oC

• 27 ^oC

Introduction to Calorimetry

• Not only can we make logical assumptions about the final equilibrium temperature, but we can calculate it as long as we know all other variables for both substances.

• Initial temperature

• Cp

• Mass

Introduction to Calorimetry

• We use our earlier formula

  • Q = m x Cp x ΔT

• But since we know the heat lost by one is the same as the heat gained by another we write it as ...

• Q = -Q But we have to substitute the formula itself in for Q. Kind of like writing a system of equations.

Introduction to Calorimetry

• So Q = -Q becomes

• m x C_p x (T_f - T_i) = -m x -C_p x (-T_f + T_i)

• Notice on the heat lost side of the equation I flipped all the signs. This makes it much easier to ensure you calculate the heat lost much easier.

Introduction to Calorimetry

• So Q = -Q becomes

• m x C_p x (T_f - T_i) = -m x -C_p x (-T_f + T_i)

• The item with the larger initial temperature should be on the negative side of the equation. If you accidentally don't have it this way you will calculate the correct answer still it will simply be negative. All you'll have to do is switch the sign.

Open Ended

What do I do if I get a negative answer on a calorimetry problem?

Introduction to Calorimetry

• Let's complete our first practice problem.

• 2 glasses of water are mixed together. The first has a mass of 40 grams and a temperature of 54^oC. The second has a mass of 30 grams and a temperature of 27^oC. Once they have mixed and reached equilibrium what will be the final temperature of the water?

• m x C_p x (T_f - T_i) = -m x -C_p x (-T_f + T_i)

Introduction to Calorimetry

• Let's complete our first practice problem.

• 2 glasses of water are mixed together. The first has a mass of 40 grams and a temperature of 54^oC. The second has a mass of 30 grams and a temperature of 27^oC. Once they have mixed and reached equilibrium what will be the final temperature of the water?

• m x C_p x (T_f - T_i) = -m x -C_p x (-T_f + T_i)

• 30 x 1 x (X - 27) = -40 x -1 x (X + 54)

Introduction to Calorimetry

• Let's complete our first practice problem.

• 2 glasses of water are mixed together. The first has a mass of 40 grams and a temperature of 54^oC. The second has a mass of 30 grams and a temperature of 27^oC. Once they have mixed and reached equilibrium what will be the final temperature of the water?

• m x C_p x (T_f - T_i) = -m x -C_p x (-T_f + T_i)

• 30 x 1 x (X - 27) = -40 x -1 x (-X + 54)

​Our Tf is simply X on both sides because we know it will be the same temperature once we reach equilibrium.

Introduction to Calorimetry

• m x C_p x (T_f - T_i) = -m x -C_p x (-T_f + T_i)

• 30 x 1 x (X - 27) = -40 x -1 x (X + 54)

• 30(X - 27) = 40(X + 54)

• 30X - 810 = 40X + 2160

Introduction to Calorimetry

• m x C_p x (T_f - T_i) = -m x -C_p x (-T_f + T_i)

• 30 x 1 x (X - 27) = -40 x -1 x (-X + 54)

• 30(X - 27) = 40(-X + 54)

• 30X - 810 = -40X + 2160

On these problems it is probably easier to distribute through the parenthesis. Now that they are gone we can combine like terms to get all the X values to one side.

Introduction to Calorimetry

• m x C_p x (T_f - T_i) = -m x -C_p x (-T_f + T_i)

• 30 x 1 x (X - 27) = -40 x -1 x (-X + 54)

• 30(X - 27) = 40(-X + 54)

• 30X - 810 = -40X + 2160

  +810 +40X

• 70X = 2970

Introduction to Calorimetry

• m x C_p x (T_f - T_i) = -m x -C_p x (-T_f + T_i)

• 30 x 1 x (X - 27) = -40 x -1 x (-X + 54)

• 30(X - 27) = 40(-X + 54)

• 30X - 810 = -40X + 2160

  +810 +40X

• 70X = 2970

​X = 42.4 ^oC

Open Ended

Our answer fell between our two original temperatures. Should this always happen? Why do you think so?

Introduction to Calorimetry

• m x C_p x (T_f - T_i) = -m x -C_p x (-T_f + T_i)

• 30 x 1 x (X - 27) = -40 x -1 x (-X + 54)

• 30(X - 27) = 40(-X + 54)

• 30X - 810 = -40X + 2160

  +810 +40X

• 70X = 2970

​X = 42.4 ^oC

​The final temperature must fall between my two original temperatures. I would also expect it to be closer to 54 than 27 because the water with the beginning temperature of 54^oC was more massive.

Introduction to Calorimetry

• We don't always have to find final temperature as well.

• Often calorimetry can be used to find the specific heat of an unknown substance since we know the specific heat of water. We simply need to know the mass of both substances as well as the starting temperature of both substances. Once added together the final equilibrium temperature is measured and the Cp is calculated.

• Lets try an example like this.

Introduction to Calorimetry

• If an unknown metal with a mass of 35 grams and a starting temperature of 270^oC is dropped into a cup of 150 grams of water at 28^oC, what must be the Cp of this unknown metal if the final temperature for both substances balances out at 38^oC?

• m x C_p x (T_f - T_i) = -m x -C_p x (-T_f + T_i)

Introduction to Calorimetry

• If an unknown metal with a mass of 35 grams and a starting temperature of 270^oC is dropped into a cup of 150 grams of water at 28^oC, what must be the Cp of this unknown metal if the final temperature for both substances balances out at 38^oC?

• m × C_p × (T_f - T_i) = -m × -C_p × (-T_f + T_i)

• 150 × 1.0 × (38 - 28) = -35 × -X × (-38 + 270)

Introduction to Calorimetry

• m × C_p × (T_f - T_i) = -m × -C_p × (-T_f + T_i)

• 150 × 1.0 × (38 - 28) = -35 × -X × (-38 + 270)

• 1500 = 35X(232)

Introduction to Calorimetry

• m × C_p × (T_f - T_i) = -m × -C_p × (-T_f + T_i)

• 150 × 1.0 × (38 - 28) = -35 × -X × (-38 + 270)

• 1500 = 35X(232)

• /35/232

Introduction to Calorimetry

• m × C_p × (T_f - T_i) = -m × -C_p × (-T_f + T_i)

• 150 × 1.0 × (38 - 28) = -35 × -X × (-38 + 270)

• 1500 = 35X(232)

• /35/232

​X = 0.185 cal/g^oC

Open Ended

Why does it make sense that we calculated a decimal for the Cp of this substance?

HINT: It was a metal

Introduction to Calorimetry

A 15 g cup of water at 45^oC is added to a 35 g cup of water at 15^oC. What will the new final temperature be? Don't forget the Cp of water is 1.0 cal/g^oC.

Math Response

A 15 g cup of water at 45^oC is added to a 35 g cup of water at 15^oC. What will the new final temperature be? Don't forget the Cp of water is 1.0 cal/g^oC.

Type answer here

Correct Answer: 24

Deg^°

Rad

Introduction to Calorimetry

A 15 g cup of water at 45^oC is added to a 35 g cup of water at 15^oC. What will the new final temperature be? Don't forget the Cp of water is 1.0 cal/g^oC.

m ⋅ C_p ⋅ (T_f - T_i) = -m ⋅ -C_p ⋅ (-T_f + T_i)

Introduction to Calorimetry

A 15 g cup of water at 45^oC is added to a 35 g cup of water at 15^oC. What will the new final temperature be? Don't forget the Cp of water is 1.0 cal/g^oC.

m ⋅ C_p ⋅ (T_f - T_i) = -m ⋅ -C_p ⋅ (-T_f + T_i)
35 ⋅ 1 ⋅ (X - 15) = -15 ⋅ -1 ⋅ (-X + 45)
35X - 525 = -15X + 625
50X = 1200
X = 24^oC

Introduction to Calorimetry

A sample of iron is heated to 150^oC and dropped into a 120 g cup of water at 30^oC. If the final temperature reaches 38^oC and we know iron has a Cp of 0.108 cal/g^oC, What was the mass of this piece of iron?

Math Response

A sample of iron is heated to 150^oC and dropped into a 120 g cup of water at 30^oC. If the final temperature reaches 38^oC and we know iron has a Cp of 0.108 cal/g^oC, What was the mass of this piece of iron?

Round to 3 Sig Figs.

Type answer here

Correct Answer: 79.4

Deg^°

Rad

Introduction to Calorimetry

A sample of iron is heated to 150^oC and dropped into a 120 g cup of water at 30^oC. If the final temperature reaches 38^oC and we know iron has a Cp of 0.108 cal/g^oC, What was the mass of this piece of iron?

m ⋅ C_p ⋅ (T_f - T_i) = -m ⋅ -C_p ⋅ (-T_f + T_i)

Introduction to Calorimetry

A sample of iron is heated to 150^oC and dropped into a 120 g cup of water at 30^oC. If the final temperature reaches 38^oC and we know iron has a Cp of 0.108 cal/g^oC, What was the mass of this piece of iron?

m ⋅ C_p ⋅ (T_f - T_i) = -m ⋅ -C_p ⋅ (-T_f + T_i)
120 ⋅ 1 ⋅ (38 - 30) = -X ⋅ -0.108 ⋅ (-38 + 150)
960 = 0.108X(112)
79.4 g = X

Introduction to Calorimetry

Once again we will be given the below calorimetry formula. We have to apply it no memorize it.

m ⋅ C_p ⋅ (T_f - T_i) = -m ⋅ -C_p ⋅ (-T_f + T_i)

Introduction to Calorimetry

Also if I get a negative answer know that I had the substances flipped and my answer should still be correct if I drop the negative as long as my math is right.

m ⋅ C_p ⋅ (T_f - T_i) = -m ⋅ -C_p ⋅ (-T_f + T_i)

Poll

How would you rank your comfortability with this concept. 1 being I'm totally lost and 5 being I've got this down pat.

• 1

• 2

• 3

• 4

• 5