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DO NOW: On back of worksheet!! The Calories in a McDonald’s Big Mac© is approximately 540. How many chemistry calories is this? How many kJ is this?

Unit: Thermochemistry Day 2 - Notes Unit: Thermochemistry Calorimetry and q=mcΔT

After today you will be able to… Describe what a calorimeter is and its function Explain what temperature change is dependant on Explain specific heat (c) and use the specific heat equation: q=mcΔT

Calorimetry: Is the accurate and precise measurement of heat change for chemical and physical processes. A calorimeter is a device used to measure the amount of heat absorbed or released during these processes. Example: Styrofoam cups! 

“Coffee Cup”Calorimeter: Constant-Pressure Calorimeter The thermometer records temperature change as the chemicals react in the water. The temperature change is then converted into units of energy. Most helpful for measuring the heat lost/gained in a chemical reaction.

Bomb Calorimeter: Constant-Volume Calorimeter A food sample is lit on fire. It burns until it is completely gone. Heat from the sample is released and heats up the water. The temperature change is then converted into units of energy. Most helpful in measuring the calorie content in food.

Temperature change is dependant on: Amount of heat added Mass of the substances Composition of the substance (specific heat) Example: Boiling water – metal pot gets hotter faster than water.

Specific heat: (c) amount of heat required to raise the temperature of 1 gram of a substance by 1˚C http://ga.water.usgs.gov/edu/heat-capacity.html

Units: Specific heat of H2O: cal J or g˚C g˚C 4.18 J/g˚C or 1.00 cal/g˚C

q=mcΔT q= m= c= ΔT= heat (J or cal) mass (grams) specific heat g˚C cal g˚C J specific heat or change in temperature (˚C) (Tfinal-Tinitial)

q= (525g) (0.21cal/g˚C) (34.8˚C) q= 3800cal Example: How many calories of heat are required to raise the temperature of 525g of aluminum from 13.0˚C to 47.8˚C? (cAl=0.21cal/g˚C) q= ? cal m= 525g c= 0.21cal/g˚C ΔT= 47.8-13.0= 34.8˚C q= (525g) (0.21cal/g˚C) (34.8˚C) q= 3800cal

5100cal= (m) (1.00cal/g˚C) (71.6˚C) m=71g Example: What mass of water would have its temperature raised from 22.5˚C to 94.1˚C with the addition of 5.1kcal of heat? 1000 cal q= 5.1 kcal x = 5100 cal 1 kcal m= ? g c= 1.00 cal/g˚C ΔT= 94.1-22.5= 71.6˚C 5100cal= (m) (1.00cal/g˚C) (71.6˚C) m=71g

Questions? Complete WS!

SILENCE IS GOLD-EN