Calorimetry CP Unit 9 Chapter 17
CALORIMETRY The enthalpy change associated with a chemical reaction or process can be determined experimentally. Measure the heat gained or lost during a reaction at CONSTANT pressure
Calorimeter Device used to measure the heat absorbed or released during a chemical or physical process Styrofoam cup
Example If you leave your keys and your chemistry book sitting in the sun on a hot summer day, which one is hotter? Why is there a difference in temperature between the two objects?
Because… Different substances have different specific heats (amount of energy needed to raise the temperature of 1 g of a substance by 1 degree Celsius).
What happens in a calorimeter One object will LOSE heat, and the other will ABSORB the heat System loses heat to surroundings = EXO = -q System absorbs heat from surroundings = ENDO = +q When a hot chunk of metal is dropped in a cool glass of water, the metal cools off. Where did the heat from the metal go? Did the metal lose more heat then the water gained? HEAT GAINED = HEAT LOST (ALWAYS!)
To do calorimetry problems… The numbers in these two boxes are always the same, but with different signs (+/-). What heat one lost, the other gained. Make a Chart: Water Object/Reaction Heat Mass Specific Heat 4.184 Final Temp Initial Temp
EXAMPLE: A small pebble is heated and placed in a foam cup calorimeter containing 25.0 g of water at 25.0 C. The water reaches a maximum temperature of 26.4 C. How many joules of heat were released by the pebble? The specific heat of water is 4.184 J/g C. (#13) Water Pebble Heat Mass Specific Heat 4.184 Final Temp Initial Temp The numbers in these two boxes are always the same, but with different signs (+/-). What heat one lost, the other gained. 25.0 g 26.4 oC 25.0 oC
The pebble because the water heated up from 25.0 C to 26.4 C. Pebble loses heat (-q, exothermic) while water gains heat (+q, endothermic) Do you calculation based on water (since the problem gave all the water’s information)
Water Pebble Heat Mass Specific Heat 4.184 Final Temp Initial Temp 25.0 g 26.4 oC 25.0 oC qwater = mwatercwaterTwater qwater = (25.0g)(4.184J/goC)(26.4oC-25.0oC) qwater = 150 J If the water ABSORBED 150 J of heat, then the pebble _______ ______ J of heat. qpebble = - 150 J
Example 2 (LAB type of CALC) Suppose that 100.00 g of water at 22.4 °C is placed in a calorimeter. A 75.25 g sample of Al is removed from boiling water at a temperature of 99.3 °C and quickly placed in a calorimeter. The substances reach a final temperature of 32.9 °C . Determine the SPECIFIC HEAT of the metal.The specific heat of water is 4.184 J/g C. MAKE YOUR CHART
Suppose that 100. 00 g of water at 22. 4 °C is placed in a calorimeter Suppose that 100.00 g of water at 22.4 °C is placed in a calorimeter. A 75.25 g sample of Al is removed from boiling water at a temperature of 99.3 °C and quickly placed in a calorimeter. The substances reach a final temperature of 32.9 °C . Determine the SPECIFIC HEAT of the metal.The specific heat of water is 4.184 J/g C. Make chart Calculate q for water Q for water is the same (but with different sign) as q for metal. Using q metal, calculate c metal Water Pebble Heat Mass Specific Heat 4.184 Final Temp Initial Temp 100.00 g 75.25 g 32.9 oC 32.9 oC 22.4 oC 99.3 oC
0.879 J/g °C
Example 3 A lead mass is heated and placed in a foam cup calorimeter containing 40.0 g of water at 17.0C. The water reaches a temperature of 20.0 C. How many joules of heat were released by the lead? The specific heat of water is 4.184 J/g C.
502 J = 5.0 x 102 J for 2 sig figs