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Learn about specific heat, calorimetry, and how to calculate heat transfer in chemical reactions. Explore examples and practical applications.
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71 Days Until the Final • Catalyst: • What is entropy? • What is the triple point on a phase diagram? • What is the critical point on a phase diagram? End
Today’s Learning Target • 7.6 – Using calorimetry data, I can utilize my knowledge of temperature change and specific heats to calculate the amount of heat transferred for a reaction.
Today’s Focus Question • How do we determine the amount of calories in Flamin’ Hot Cheetos?
Specific Heat • The amount of energy that is transferred to a material is dependent on the nature of the material receiving the energy. • Specific Heat – The amount of energy required to raise the temperature of one gram of substance one degree Celsius • Represent by the symbol Cp
Calorimetry • Calorimetry is the science of measuring the heat of a chemical reaction. • q represents the heat gained/released • Cp is the specific heat • ΔT = Final Temperature – Initial Temp. • If we can measure the mass and temperature, then we can determine the heat gained/lost in a reaction
Class Example • A 4.0 g sample of glass is heated from 274 K to 314 K. Glass has a specific heat of 0.20 J/(g x K). How much heat is gained during this heating?
Table Talk • The specific heat of copper is 0.4 J/oC. How much heat is needed to change the temperature of a 30 g sample of copper from 20.0 oC to 60.0 oC?
Stop and Jot • Determine the specific heat of a material if a 35 g sample absorbed 96 J as it was heated from 293 K to 313 K.
Calories in Flamin’ Cheetos • Q can be measured in calories • Food scientists measure the calories contained within food by measuring the amount of energy it takes to raise 1 g of water 1 oC. • 1000 calories = 1 Calorie • 1 serving of Cheetos has 140 Calories
Lab Intro • Info about lab procedures and equipment.
Work Time • Begin working on your “Homework 7.3” • There will be an exit slip after this activity
Learning Log Assessment Rate yourself 1 – 4 on LTs 7.6
Exit Slip 1. I have a sample of water and I increase the temperature from 10.0 oC to 40 oC. The specific heat of water is 4.18 J/(g x oC). Calculate the heat absorbed for this process.
Learning Log Assessment Using your exit slip score, re-rate yourself on LTs 7.6
Closing Time • Homework 7.3 – Calorimetry Calculations • Rough Draft of Lab Report DUE TOMORROW