Calorimetry
Aim: Calculating the energy absorbed or released during
a chemical reaction.
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Specific heat capacity: the amount of energy required to raise the temperature of 1 g of a substance by 1°C or 1 K. m= mass in grams. Q= heat measured in joules ...
Typology: Exams
2021/2022
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Calorimetry
Aim: Calculating the energy absorbed or released during a chemical reaction.
- Calorimeter – an insulated
device used to measure the
absorption or release of heat
into water.
Unit
- Heat energy is measured in Joules (J) or kilojoules. (kJ) Measuring Heat Energy
Calculating Heat
- To calculate the energy absorbed or released we use the formula:
q=mcΔT
Specific heat capacity: the amount of energy required to raise the temperature of 1 g of a substance by 1°C or 1 K. m= mass in grams. Q= heat measured in joules ΔT= change in temperature in^ ° C or K. The difference between the final and initial temperatures. c = specific heat capacity <-- Found in Table B for water. _The greater the specific heat, the longer it takes a substance to heat up or cool down._
Measuring the Heat Contained in a Cheese Doodle Materials 100 g. H 2 O Initial temperature of water: __________ Aluminum can Final temperature of water: _________ Ring stand Mass of water: _________ Glass stirring rod Specific heat of water, c : ___________ Thermometer (See Reference Table B) Aluminum Foil Q= mcΔT
Practice
- a. 100. grams of water is heated from 10.0 °C to 25.0 °C. Using the calorimetry formula, determine the amount of heat absorbed by the water. Q= x m= 100. g c= 4.18 J/g°C ΔT= 15.0 °C (25.0-10.0) b. Was this change of energy endothermic or exothermic for the water? Explain. Endothermic because the temperature of the water increased due to the absorption of energy. Q= mcΔT Q= (100.)(4.18)(15.0) Q= 6270 J
Practice
- a. An 80.5 g sample of water at 15.0 °C absorbs 1680. J of heat energy. What is the final temperature of the water? (Hint: Solve for ΔT first then adjust the initial temperature to the final temperature based on if heat was lost or gained) b. Explain why the water temperature increased.
Practice
- The temperature of a sample of water changes from 20.0 °C to 5.0 °C when the water releases 420. joules of heat. What is the mass of the sample?
Practice
- The temperature of a sample of water changes from 20.0 °C to 5.0 °C when the water releases 420. joules of heat. What is the mass of the sample? Q= 420. J m= x c= 4.18 J/g°C ΔT= 15.0 °C (20.0-5.0) Q= mcΔT **420. = x(4.18)(15.0)
- = 62.7 x m= 6.7 g**
Practice
- a. 200. g of an unknown substance is warmed from 25.0 °C to 50.0 °C by the addition of 201.6 joules of heat. What is the specific heat capacity of the substance? Q= 201.6 J m= 200. g c= x ΔT= 25.0 °C (50.0-25.0) Q= mcΔT 201.6 = (200.)x(25.0) 201.6 = 5000 x x= 0.0403 J/g°C No because the specific heat capacity of water is 4.18 J/g°C and this substance has a specific heat of 0.0403 J/g°C. b. Is this substance water? Explain.