Specific Heat Capacity & Calorimetry

Specific Heat Capacity & Calorimetry

Specific Heat Capacity & Calorimetry 1 Temperature and Energy We relate energy and temperature by discussing a substances heat capacity. Heat Capacity = heat required to raise temp. of an object by 1oC more heat is required to raise the temp. of a large sample of a substance by 1 oC than is needed for a smaller sample Specific Heat Capacity a physical property of matter that describes matters resistance to a change in temperature. The symbol for specific heat is Cp.

Not all substances heat up at the same rate. Some get hot quickly and some more slowly. 2 Example If you have ever touched the metal on a car and the fabric on the car seat on a hot day, you have experienced the affect of specific heat. The metal seems much hotter than the fabric seat even if after receiving the same amount of energy from the sun. This is caused by the difference in the

specific heat of each of the materials. The metal has a lower specific heat and gives up its thermal energy at a much higher rate than does the fabric which has a much higher specific heat. 3 High Specific Heat and Water Water has a very high specific heat compared to other matter; therefore ocean water stays about the same temperature throughout day and night despite the differences in temperature between night and day. That also explains why water is used in car radiators to cool the engine.

Low specific heat = less energy required to change the temperature High specific heat = more energy required to change the temperature 4 Practice Which would get hotter if left in the sun? Penny vs. Water Keys vs. soccer ball Plastic recycling bin vs. metal trash can 5 Specific Heat Capacity

Temperature change of a substance depends on three things: Mass, m Amount of energy added, Q Specific Heat, Cp Final temperature Initial temperature Q = m x Cp x (Tf Ti) or

Q = m x Cp x (T) Temperature change 6 Specific Heats of Common Substances 7 Using Q = m x Cp x (Tf Ti) The following problems will show you how to solve for different variables in our equation.

How much energy does it take to raise the temperature of 50 g of aluminum (cp = 0.9025 J/gC0) by 10 0C? Q = (m) (cp) (Tf - Ti) Q = (50g) (0.9025 J/gC0) (100C) Q = 451.25 Joules 8 Using Q = m x Cp x (Tf Ti) If we add 30 J of heat to lead (cp = 0.1276J/gC0) with a mass of 10 g, how much will its temperature increase? Q = (m) (cp) (Tf - Ti) 30J = (10g) (0.1276 J/gC0) (x) 30J = (1.276 J/0C) (x)

23.50C = x = temperature increase 9 Calorimetry Calorimetry is the science of measuring the heat of chemical reactions or physical changes. Calorimetry is also known as a laboratory procedure that measures the amount of heat transferred to the surroundings by a reaction. Calorimetry can be calculated when heat of combustion is given and the mass of the substance is known or, During a calorimetry procedure, the heat released during a chemical or physical change is transferred to another substance, such as water, which undergoes a temperature change.

10 Calorimetry Calculations Example 1: Propane is a commonly used fuel. 1 mol of C3H8 releases 2,220 kJ of heat during combustion. The molar mass of C3H8 is 44.1 g/ mol. How much heat is released if a firework contains 67.8 g of C3H8? 1st convert the grams of C3H8 to moles of C3H8. 67.8 g C3H8 x 1 mol C3H8 = 1.53 mol C3H8 44.1 g C3H8

2nd use the heat of combustion of propane to calculate energy (heat) released 1.53 mol C3H8 x 2,220 kJ = 3413.06 kJ => 3410 kJ released 1 mol 11 Calorimetry Calculations The temperature change, fuel mass, and water volume data from a calorimetry procedure can be used to determine how much heat is transferred during a combustion reaction. The amount of energy transferred from a substance during combustion depends on the identity and mass of the substance.

The equation can be seen as q1 = - q2. One will be losing energy, the other will be gaining energy. 12 Calorimetry Calculations Example 2: 175 grams of hot aluminum (cp = 0.9025 J/gC0 ) ( 100.C) is dropped into an insulated cup that contains 40.0 mL of ice cold water (cp = 4.184 J/gC) (0.0C). by 10 0C? Follow the example above to determine the final temperature, x. 1st set up expressions for energy released and energy absorbed.

Q1 = (40.0 g) (4.184 J/gC) (x -0.0 C) for cold water and Q2 = - (175 g) (0.900 J/gC) (x -100 C) for Al 2nd put expressions together. (40.0 g) (4.184 J/gC) (x -0.0 C) = - [(175 g) (0.900 J/gC) (x -100 C)] 3rd solve for x.

167.4 (x - 0.0) = -157.5 (x 100) 167.4 x = -157.5x + 1575 324.9 x = 1575 => x = 48.5 C 13

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