Unit 7 Part 1-Intro to Thermochem

​Thermochemistry

Many chemical reactions occur within your body when you exercise. As these reactions take place, your body gets hot and you sweat. What about these reactions causes such a drastic change in your body?

Lesson Objectives

• Describe the relationship between temperature and a chemical reaction

• Convert temperatures from celsius to kelvin.

• Calculate the specific heat of an object when given known values.

Introduction

• Virtually every chemical reaction is accompanied by a change in energy

• Thermochemistry: the study of the transfers of energy as heat that accompany chemical reactions and physical changes​

  • Measured using a Calorimeter

Collisions in Reactions

• Reactions happen when bonds are broken and atoms of reactants are rearranged

  • Collision Theory: Bonds are broken and/or formed when reactant particles collide with enough energy and the correct orientation​

The Activated Complex

• The Activated Complex: an unstable cluster of atoms that exist during the transition between reactants and products

Systems and Surroundings

• System: the specific part of the universe that a scientist studies

  • Everything outside of the system is the surroundings

  • Energy can transfer between a system and the surroundings

Temperature

• Temperature: the measure of the average kinetic energy of the particles in a sample of matter

  • The greater the kinetic energy, the higher the temperature and the hotter it feels

  • SI Unit for Temperature is Kelvin

    • °C +273

Fill in the Blanks

Convert 25 °C to Kelvin

Fill in the Blanks

Convert 60 °C to Kelvin

Multiple Choice

What does temperature measure?

• Temperature measures the density of a substance.
• Temperature measures the color of an object.
• Temperature measures the volume of gas particles.
• Temperature measures the average kinetic energy of particles.

Drag and Drop

The ​

Placeholder

describes an unstable cluster of atoms that exist during the transition between reactants and products

Drag these tiles and drop them in the correct blank above

activated complec

stable compound

solid state

liquid mixture

Heat

• Heat: The energy transferred from one object to another due to a temperature difference

  • Happens through radiation or the direct collision of particles

  • This will happen until a balance is reached and both objects are at the same temperature

  • Heat always flows from warmer objects to cooler objects

  • Measured in a chemical reaction with a calorimeter

  • SI Unit of heat and other forms of energy is the joule

Heat

Specific Heat

• Materials have differing capacities to absorb and release energy, so a method to compare these capacities is necessary

• Specific Heat: the amount of energy required to raise the temperature of one gram of a substance by 1 °C or 1 kelvin

  • usually measured under constant pressure and shown by the symbol c_p

  • matter with a high specific heat needs more energy to raise the temperature

  • Water has an Extremely high specific heat value

Specific Heat formula

• c_p = the specific heat at a given pressure

• q = the energy lost or gained

• m= mass of the sample

• ΔT= the change in temperature

• Rewritten to find the quantity of energy gained or lost with a change in temperature

  • q=C_p×​m×ΔT

Specific Heat Sample Problem 1 A

• A 4.0 gram sample of glass was heated from 274 K to 314 K and was found to have absorbed 32J of energy as heat

  • What is the specific heat of this type of glass?

• Step 1: Identify what variables we know

  1. c_p

  2. m

  3. Δ​T

  4. q

Specific Heat Sample Problem 1 A

• A 4.0 gram sample of glass was heated from 274 K to 314 K and was found to have absorbed 32J of energy as heat

  • What is the specific heat of this type of glass?

• Step 1: Identify what variables we know

  1. c_p=?

  2. m=4.0 g

  3. Δ​T= 314-274 = 40 K

  4. q=32 J

Specific Heat Sample Problem 1 A

• A 4.0 gram sample of glass was heated from 274 K to 314 K and was found to have absorbed 32J of energy as heat

  • What is the specific heat of this type of glass?

• Step 2: Find the specific heat by plugging into the specific heat equation

Fill in the Blanks

A 4.0-gram sample of glass was heated from 274 K to 314 K and was found to have absorbed 32J of energy as heat. What is the specific heat of this type of glass?

Specific Heat Sample Problem 1 b

• The Same 4.0 gram sample of glass was heated from 274 K to 314 K and was found to have absorbed 32J of energy as heat

  • How much energy will the same glass sample gain when it is heated from 314 K to 344 K

• Step 3: Identify the variables we know

  • c_p=

  • q=

  • m=

  • ΔT=

Specific Heat Sample Problem 1 b

• A 4.0 gram sample of glass was heated from 274 K to 314 K and was found to have absorbed 32J of energy as heat

  • How much energy will the same glass sample gain when it is heated from 314 K to 344 K

• Step 3: Identify the variables we know

  • c_p= 0.20J​

  • q=?

  • m=4.0

  • ΔT=344-314=30

Specific heat Sample Problem 1 b

• A 4.0 gram sample of glass was heated from 274 K to 314 K and was found to have absorbed 32J of energy as heat

  • How much energy will the same glass sample gain when it is heated from 314 K to 344 K

• Step 4: Plug our Variables into our energy equation

Fill in the Blanks

A 4.0-gram sample of glass was heated from 274 K to 314 K and was found to have absorbed 32J of energy as heat. How much energy will the same glass sample gain when it is heated from 314 K to 344 K?

Specific heat Sample Problem 2

• Determine the specific heat of a material if a 35 g sample absorbed 48 J as it was heated from 293 K to 313 K

  • Step 1: Identify what variables we know

    1. c_p

    2. m

    3. Δ​T

    4. q

Specific heat Sample Problem 2

• Determine the specific heat of a material if a 35 g sample absorbed 48 J as it was heated from 293 K to 313 K

  • Step 1: Identify what variables we know

    • c_p= ?

    • m= 35 g

    • Δ​T=313-293= 20

    • q=48 J

  • Step 2: Plug in to our Specific Heat Equation

     

Fill in the Blanks

Determine the specific heat of a material if a 35 g sample absorbed 48 J as it was heated from 293 K to 313 K. Round to 2 decimal places

Specific heat Sample Problem 3a

• A 85 g piece of Copper Alloy is heated from 30 °C to 45°C. In the process it absorbs 523 J of energy as heat. What is the specific heat of this copper alloy?

  • Step 1: Identify what variables we know

    1. c_p

    2. m

    3. Δ​T

    4. q

Specific heat Sample Problem 3a

• A 85 g piece of Copper Alloy is heated from 30 °C to 45°C. In the process it absorbs 523 J of energy as heat. What is the specific heat of this copper alloy?

  • Step 1: Identify what variables we know

    • c_p=?

    • m= 85

    • Δ​T= 45-30=15

    • q=523

  • Step 2: Plug into our Specific heat equation

Fill in the Blanks

A 85 g piece of Copper Alloy is heated from 30 °C to 45°C. In the process it absorbs 523 J of energy as heat. What is the specific heat of this copper alloy? Round to 2 Decimal Places

Specific heat Sample Problem 3b

• An 85 g piece of Copper Alloy is heated from 30 °C to 45°C. In the process it absorbs 523 J of energy as heat. How much energy will the same sample lose if it is cooled from 45°C to 25°C

  • Step 1: Identify what variables we know

    • c_p=

    • m=

    • Δ​T=

    • q=

    ​

Specific heat Sample Problem 3b

• A 85 g piece of Copper Alloy is heated from 30 °C to 45°C. In the process it absorbs 523 J of energy as heat. How much energy will the same sample loose if it is cooled from 45°C to 25°C

  • Step 1: Identify what variables we know

    • c_p=0.41

    • m= 85

    • Δ​T= 25-45=-20

    • q=?

  • Step 2: Plug into our energy equation

Fill in the Blanks

A 85 g piece of Copper Alloy is heated from 30 °C to 45°C. In the process it absorbs 523 J of energy as heat. How much energy will the same sample loose if it is cooled from 45°C to 25°C?

Fill in the Blanks

The temperature of a 74g sample of material increases from 15°C to 45°C when it absorbs 2000.0 J of energy as heat. What is the specific heat of this material? Round to 2 Decimal Places.