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Heat Happens!

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Odysses John Cantos

Tier 1
For Schools

Lesson Plan

Heat Happens!

Students will be able to calculate the amount of heat lost or gained during a physical change using the formula Q = m Cp (t2-t1) and interpret the results.

Understanding how heat is transferred is fundamental to daily life, from cooking and keeping food warm to understanding weather patterns and industrial processes. This skill helps us predict and control temperature changes around us.

Audience

10th Grade

Time

30 minutes

Approach

Direct instruction, guided practice, and independent problem-solving.

Materials

Smartboard or Projector, Heat Happens! Slide Deck, Heat Happens! Worksheet, Heat Happens! Answer Key, and Calculators

Prep

Teacher Preparation

15 minutes

  • Review the Heat Happens! Slide Deck and practice example problems.
    - Print copies of the Heat Happens! Worksheet (one per student) and the Heat Happens! Answer Key.
    - Ensure calculators are available or students have their own.
    - Set up the projector/smartboard.

Step 1

Warm-Up: What is Heat?

5 minutes

  • Begin by asking students: "What comes to mind when you hear the word 'heat'?"
    - Facilitate a brief discussion, guiding them towards understanding heat as a form of energy transfer.
    - Introduce the lesson's big question: "How do we measure how much heat is transferred?"

Step 2

Introduction to Q = m Cp (t2-t1)

10 minutes

  • Use the Heat Happens! Slide Deck to introduce the concept of heat transfer during physical changes.
    - Explain each variable in the formula Q = m Cp (t2-t1):
    - Q: Heat energy (Joules or calories)
    - m: Mass (grams)
    - Cp: Specific heat capacity (J/g°C or cal/g°C)
    - (t2-t1): Change in temperature (°C or K)
    - Emphasize the units and how to determine if heat is gained (positive Q) or lost (negative Q).

Step 3

Guided Practice: Example Problems

10 minutes

  • Work through 1-2 example problems from the Heat Happens! Slide Deck as a class.
    - Encourage students to participate in identifying variables, setting up the equation, and solving.
    - Address any immediate questions or misconceptions.

Step 4

Independent Practice: Worksheet

5 minutes

  • Distribute the Heat Happens! Worksheet.
    - Students begin working on the problems independently.
    - Circulate around the room to provide individual support and answer questions.
    - If time is short, assign the remaining problems as homework.
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Slide Deck

Heat Happens!

Understanding Heat Transfer in Physical Changes

Warm-Up Question: What comes to mind when you hear the word 'heat'?

Welcome students and introduce the engaging topic of heat transfer. Ask a warm-up question to get them thinking about heat.

What is Heat?

  • Heat is a form of energy that is transferred due to a temperature difference.
  • We'll learn how to calculate the amount of heat lost or gained when a substance changes temperature (a physical change).

Think: When you boil water, where does the heat go?

Briefly review the concept of heat as energy transfer. Explain that today's lesson will focus on quantifying this transfer during physical changes.

The Heat Formula: Q = m Cp ΔT

To calculate the amount of heat energy transferred (lost or gained), we use:

Q = m Cp (t2 - t1)

Where:

  • Q is the heat energy
  • m is the mass
  • Cp is the specific heat capacity
  • (t2 - t1) is the change in temperature (final temp - initial temp)

Introduce the main formula. Explain that this formula allows us to put a number to heat transfer. Mention that delta T (ΔT) is often written as (t2-t1).

Defining Our Variables

Q = Heat Energy

  • Measured in Joules (J) or calories (cal)
  • Positive Q means heat is GAINED (temperature increases)
  • Negative Q means heat is LOST (temperature decreases)

m = Mass

  • Measured in grams (g) or kilograms (kg)

Cp = Specific Heat Capacity

  • The amount of heat required to raise the temperature of 1 gram of a substance by 1°C.
  • Measured in J/g°C or cal/g°C
  • Unique for each substance!

(t2 - t1) = Change in Temperature (ΔT)

  • t2: Final temperature
  • t1: Initial temperature
  • Measured in °C (Celsius) or K (Kelvin)

Go through each variable individually, explaining its meaning and common units. Emphasize that Q will be positive if heat is gained and negative if heat is lost.

What is Specific Heat Capacity (Cp)?

Imagine trying to heat up a swimming pool versus a small cup of water. Which heats up faster?

Specific heat capacity tells us how much energy a substance can store for a given temperature change.

  • Water has a high Cp (4.18 J/g°C) – it takes a lot of energy to change its temperature.
  • Metals often have low Cp – they heat up and cool down quickly.

Further explain specific heat capacity with a simple analogy. Mention that different materials heat up and cool down at different rates.

Example Problem 1: Heating Water

Problem: How much heat energy is needed to raise the temperature of 200 g of water from 20°C to 50°C? (Cp of water = 4.18 J/g°C)

Steps to Solve:

  1. Identify knowns (m, Cp, t1, t2)
  2. Calculate ΔT = (t2 - t1)
  3. Plug values into Q = m Cp ΔT
  4. Calculate Q



Solution:

  1. m = 200 g, Cp = 4.18 J/g°C, t1 = 20°C, t2 = 50°C
  2. ΔT = 50°C - 20°C = 30°C
  3. Q = (200 g) * (4.18 J/g°C) * (30°C)
  4. Q = 25,080 J

Present the first example. Guide students through identifying m, Cp, t1, t2, and then calculating Q. Show step-by-step solution.

Example Problem 2: Cooling Iron

Problem: A 50 g piece of iron cools from 90°C to 25°C. How much heat is lost by the iron? (Cp of iron = 0.45 J/g°C)

Steps to Solve:

  1. Identify knowns (m, Cp, t1, t2)
  2. Calculate ΔT = (t2 - t1)
  3. Plug values into Q = m Cp ΔT
  4. Calculate Q



Solution:

  1. m = 50 g, Cp = 0.45 J/g°C, t1 = 90°C, t2 = 25°C
  2. ΔT = 25°C - 90°C = -65°C
  3. Q = (50 g) * (0.45 J/g°C) * (-65°C)
  4. Q = -1462.5 J

A negative Q indicates heat was LOST.

Present the second example, this time involving heat loss. Guide students to notice the negative change in temperature and negative Q value.

Your Turn! Practice Time!

Now it's your chance to apply the formula!

  • Work through the problems on the Heat Happens! Worksheet.
  • Remember to show your work, include units, and pay attention to whether heat is gained or lost.
  • Ask questions if you get stuck!

Transition to independent practice using the worksheet. Briefly explain that they will apply what they've learned to new problems.

Questions & Review

Any questions about calculating heat transfer?

Key Takeaways:

  • Q = m Cp (t2 - t1) helps us quantify heat transfer.
  • Pay attention to units and signs (+Q = gained, -Q = lost).

Complete the rest of the worksheet for homework if needed!

Conclude the lesson by inviting questions and reiterating the importance of understanding heat transfer. Assign any unfinished worksheet problems as homework.

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Worksheet

Heat Happens! Calculation Worksheet

Instructions: Read each problem carefully and calculate the heat energy (Q) transferred. Show all your work, including the formula, substituted values, and units. Indicate whether heat is gained or lost.

Problem 1

How much heat energy is absorbed by 150 grams of water when its temperature increases from 10°C to 45°C? (Specific heat capacity of water = 4.18 J/g°C)











Problem 2

A 75 gram piece of copper cools from 120°C to 30°C. How much heat energy is lost by the copper? (Specific heat capacity of copper = 0.385 J/g°C)











Problem 3

If 25,000 Joules of heat are added to 500 grams of a substance, and its temperature rises from 20°C to 30°C, what is the specific heat capacity of the substance?











Problem 4

Calculate the final temperature (t2) of a 300 gram aluminum block if it absorbs 15,000 Joules of heat and its initial temperature was 22°C. (Specific heat capacity of aluminum = 0.90 J/g°C)











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Answer Key

Heat Happens! Answer Key

Problem 1

How much heat energy is absorbed by 150 grams of water when its temperature increases from 10°C to 45°C? (Specific heat capacity of water = 4.18 J/g°C)

Thought Process:

  1. Identify the given values:
    • m = 150 g
    • Cp = 4.18 J/g°C
    • t1 = 10°C
    • t2 = 45°C
  2. Calculate the change in temperature (ΔT):
    • ΔT = t2 - t1 = 45°C - 10°C = 35°C
  3. Apply the formula Q = m Cp ΔT:
    • Q = (150 g) * (4.18 J/g°C) * (35°C)
  4. Calculate Q:
    • Q = 21,945 J

Answer: 21,945 J (Heat is gained)



Problem 2

A 75 gram piece of copper cools from 120°C to 30°C. How much heat energy is lost by the copper? (Specific heat capacity of copper = 0.385 J/g°C)

Thought Process:

  1. Identify the given values:
    • m = 75 g
    • Cp = 0.385 J/g°C
    • t1 = 120°C
    • t2 = 30°C
  2. Calculate the change in temperature (ΔT):
    • ΔT = t2 - t1 = 30°C - 120°C = -90°C
  3. Apply the formula Q = m Cp ΔT:
    • Q = (75 g) * (0.385 J/g°C) * (-90°C)
  4. Calculate Q:
    • Q = -2598.75 J

Answer: -2598.75 J (Heat is lost)



Problem 3

If 25,000 Joules of heat are added to 500 grams of a substance, and its temperature rises from 20°C to 30°C, what is the specific heat capacity of the substance?

Thought Process:

  1. Identify the given values:
    • Q = 25,000 J
    • m = 500 g
    • t1 = 20°C
    • t2 = 30°C
  2. Calculate the change in temperature (ΔT):
    • ΔT = t2 - t1 = 30°C - 20°C = 10°C
  3. Rearrange the formula Q = m Cp ΔT to solve for Cp:
    • Cp = Q / (m * ΔT)
  4. Substitute the values and calculate Cp:
    • Cp = 25,000 J / (500 g * 10°C)
    • Cp = 25,000 J / (5000 g°C)
    • Cp = 5 J/g°C

Answer: 5 J/g°C



Problem 4

Calculate the final temperature (t2) of a 300 gram aluminum block if it absorbs 15,000 Joules of heat and its initial temperature was 22°C. (Specific heat capacity of aluminum = 0.90 J/g°C)

Thought Process:

  1. Identify the given values:
    • Q = 15,000 J
    • m = 300 g
    • Cp = 0.90 J/g°C
    • t1 = 22°C
  2. Rearrange the formula Q = m Cp ΔT to solve for ΔT:
    • ΔT = Q / (m * Cp)
  3. Substitute the values and calculate ΔT:
    • ΔT = 15,000 J / (300 g * 0.90 J/g°C)
    • ΔT = 15,000 J / (270 g°C)
    • ΔT ≈ 55.56°C
  4. Use ΔT = t2 - t1 to solve for t2:
    • t2 = ΔT + t1
    • t2 = 55.56°C + 22°C
    • t2 ≈ 77.56°C

Answer: Approximately 77.56°C

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Quiz

Heat Happens! Quiz

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