#AP Chemistry Study Guide: Calorimetry 🌡️

Hey there, future AP Chem master! Let's dive into calorimetry, a topic that's all about measuring heat flow. Think of it as detective work for energy changes! This guide is designed to be your go-to resource, especially when you're doing that last-minute review. Let's make sure you're feeling confident and ready for the exam!

#Introduction to Calorimetry

#What is Calorimetry?

Calorimetry is the science of measuring heat flow. We can't directly measure the absolute enthalpy (H) of a system, but we can measure changes in enthalpy (ΔH). This is done by observing temperature changes when heat is either lost or gained by a system. Basically, we're tracking where the energy goes!

#Types of Calorimeters

There are different types of calorimeters, each with its own setup:

• Bomb Calorimeter: A sealed container where a reaction occurs, and the heat released is measured by the temperature change of the surrounding water. Think of it as a mini explosion in a controlled environment!
• Constant-Pressure Calorimeter: Measures heat changes at constant pressure, where heat transferred equals ΔH. This is super useful for reactions in open beakers.
• Coffee-Cup Calorimeter: This is the simple version we'll focus on. It's like a science experiment in your kitchen, using an insulated cup to measure heat changes.

A typical coffee-cup calorimeter setup.

#Coffee-Cup Calorimetry: The Basics

Our trusty coffee-cup calorimeter has a few key parts:

• Thermometer: 🌡️ To measure temperature changes.
• Reaction Mixture: Where the magic (or chemistry) happens.
• Stirrer: To ensure even temperature distribution.
• Insulated Container: Usually a styrofoam cup to prevent heat loss/gain.
• Heat-Proof Lid: To keep the system closed.

The goal is to insulate the system as much as possible, so that we can accurately measure heat changes during a reaction. The better the insulation, the better the data! 📏

#Quantifying Energy

#The First Law of Thermodynamics

#Measuring Heat Transfer

We use the equation q = mCΔT to quantify heat transfer in a calorimeter:

• q = heat (in Joules)
• m = mass (in grams or kilograms)
• C = specific heat of the substance
• ΔT = change in temperature (in Kelvin or Celsius, since the change is the same)

#Specific Heat

Specific heat (C) is the amount of heat required to raise the temperature of 1 gram of a substance by 1°C. It's like the substance's resistance to temperature change. Think of it as how much "oomph" it takes to heat something up.

#Water vs. Sand 💧🏖️

• Water: Has a high specific heat (4.184 J/g°C). It takes a lot of energy to heat up, which is why it takes forever to boil. Water is like a stubborn mule – it doesn't change temperature easily.
• Sand: Has a low specific heat (around 0.840 J/g°C). It heats up quickly in the sun. Sand is like a race car – it heats up super fast!

#Calorimetry Examples

Let's put this into practice with a couple of examples!

#Example 1

Question: An insulated cup contains 255.0 grams of water, and the temperature changes from 25.2 °C to 90.5 °C. Calculate the amount of heat released by the system. The specific heat capacity of water is 4.184 J/g°C.

Solution:

1. Recognize this is a calorimetry problem, so we use q=mCΔT.
2. Plug in the values: q = (255.0 g)(4.184 J/g°C)(90.5°C - 25.2°C)
3. Solve: q = (255.0 g)(4.184 J/g°C)(65.3°C) q = 69,700 J or 69.7 kJ

#Misconception: q vs. ΔH

#Example 2

Question: (From the Advanced Placement YT Channel) A lab procedure is done in a calorimeter:

(a) What is |ΔT| for the copper? What is |ΔT| for the water?

Solution:

• Copper: |23.6°C - 100.0°C| = 76.4°C
• Water: 23.6°C - 20°C = 3.6°C

(b) A student claims that, since the magnitude of ΔT for the copper is greater than that of the water, it means that the magnitude of heat (q) lost by the copper is greater than the magnitude of heat (q) gained by the water. Do you agree with this claim?

Solution:

No. According to the first law of thermodynamics, the heat lost by the copper must equal the heat gained by the water. The temperature change is different because of the different specific heats of the two substances.

(c) Find the specific heat of copper.

Solution:

1. We know that heat lost by copper = heat gained by water. So, mcΔT (copper) = mcΔT (water).
2. Plug in the values: (50.00 g)(C)(76.4 °C) = (100.00 g)(4.18 J/g°C)(3.6 °C)
3. Solve for C: C = 0.39 J/g°C

#Another Formula

Remember, when dealing with heat loss and gain, you can use:

heat loss = heat gain or mcΔT (1st substance) = mcΔT (2nd substance)

#Final Exam Focus

#High-Priority Topics:

• Calorimetry Basics: Understand the different types of calorimeters and how they work.
• First Law of Thermodynamics: Know that energy is conserved in a closed system.
• q = mCΔT: This equation is your bread and butter for calorimetry problems. Practice it a lot!
• Specific Heat: Understand what it means and how it affects temperature changes.
• Heat Loss = Heat Gain: Use this principle to solve complex calorimetry problems.

#Common Question Types:

• Multiple Choice: Conceptual questions about heat transfer, specific heat, and the first law of thermodynamics.
• Free Response: Calculations involving q=mCΔT, heat loss/gain problems, and experimental design questions.

#Last-Minute Tips:

• Time Management: Don't spend too long on one problem. If you're stuck, move on and come back to it later.
• Units: Always check your units. Make sure they match before you start calculating.
• Practice: The more you practice, the better you'll get. Focus on understanding the concepts, not just memorizing formulas.
• Stay Calm: You've got this! Take a deep breath and remember all the hard work you've put in.

#

Practice Question

Practice Questions

#Multiple Choice Questions

1. A 20.0 g sample of metal at 85.0°C is placed in 50.0 g of water at 22.0°C. The final temperature of the water and metal is 25.6°C. Assuming no heat is lost to the surroundings, what is the specific heat of the metal? (Specific heat of water = 4.184 J/g°C) (A) 0.128 J/g°C (B) 0.450 J/g°C (C) 0.897 J/g°C (D) 1.25 J/g°C

2. Which of the following statements is true regarding a system at thermal equilibrium? (A) The system has reached a state where no heat transfer occurs. (B) The system has reached a state where the temperature is constantly increasing. (C) The system has reached a state where the temperature is constantly decreasing. (D) The system has reached a state where the heat transfer is at its maximum.

3. A 100.0 g sample of a metal is heated to 100.0°C and then placed in 50.0 g of water at 20.0°C. The final temperature of the water and metal is 25.0°C. Which of the following can be concluded about the specific heat of the metal? (A) It is less than the specific heat of water. (B) It is equal to the specific heat of water. (C) It is greater than the specific heat of water. (D) It cannot be determined without additional information.

#Free Response Question

A student performs an experiment to determine the specific heat of an unknown metal. The student heats a 50.0 g sample of the metal to 98.0°C and then places it into 100.0 g of water at 22.0°C in a calorimeter. The final temperature of the water and metal is 25.6°C. Assume no heat is lost to the surroundings.

(a) Calculate the change in temperature of the water. (b) Calculate the amount of heat gained by the water (in Joules). (c) Calculate the change in temperature of the metal. (d) Calculate the specific heat of the metal (in J/g°C). (e) If the student used a calorimeter that was not perfectly insulated, would the calculated specific heat of the metal be higher or lower than the actual value? Explain.

#Free Response Scoring Breakdown

(a) (1 point) ΔT = 25.6°C - 22.0°C = 3.6°C

(b) (2 points) q = mCΔT = (100.0 g)(4.184 J/g°C)(3.6°C) q = 1506 J (or 1.51 kJ)

(c) (1 point) ΔT = 25.6°C - 98.0°C = -72.4°C

(d) (2 points) Heat lost by metal = heat gained by water (50.0 g)(C)(72.4°C) = 1506 J C = 0.417 J/g°C

(e) (2 points) The calculated specific heat of the metal would be higher than the actual value. If the calorimeter is not perfectly insulated, some heat will be lost to the surroundings. Therefore, the calculated heat gained by the water will be lower than the actual value, which will result in a higher calculated specific heat for the metal.