Calorimetry, Exams of Physics

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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

1. 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

2. 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

3. 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

3. 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)
4. = 62.7 x m= 6.7 g**

Practice

4. 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.