Section 7.2—Calorimetry & Heat Capacity

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Section 7.2—Calorimetry & Heat Capacity
Why do some things get hot more quickly than others?

Temperature Temperature – proportional to the average kinetic energy of the molecules Energy due to motion (Related to how fast the molecules are moving) As temperature increases Molecules move faster

Heat & Enthalpy Heat (q)– The flow of energy from higher temperature particles to lower temperature particles Enthalpy (H)– Takes into account the internal energy of the sample along with pressure and volume Under constant pressure (lab-top conditions), heat and enthalpy are the same…we’ll use the term “enthalpy”

Energy Units The most common energy units are Joules (J) and calories (cal) Energy Equivalents = 4.18 J 1.00 cal 1000 J = 1 kJ = 1000 cal 1 Cal (food calorie) These equivalents can be used in dimensional analysis to convert units

Heat Capacity Specific Heat Capacity (Cp) – The amount of energy that can be absorbed before 1 g of a substance’s temperature has increased by 1°C Cp for liquid water = 1.00 cal/g°C or 4.18 J/g°C

Heat Capacity High Heat Capacity Low Heat Capacity
Takes a large amount of energy to noticeably change temp Small amount of energy can noticeably change temperature Heats up slowly Heats up quickly Cools down slowly Cools down quickly Maintains temp better with small condition changes Quickly readjusts to new conditions A pool takes a long time to warm up and remains fairly warm over night. The air warms quickly on a sunny day, but cools quickly at night A cast-iron pan stays hot for a long time after removing from oven. Aluminum foil can be grabbed by your hand from a hot oven because it cools so quickly

What things affect temperature change?
Heat Capacity of substance The higher the heat capacity, the slower the temperature change Mass of sample The larger the mass, the more molecules there are to absorb energy, so the slower the temperature change Specific heat capacity of substance Energy added or removed Change in temperature Mass of sample

Positive & Negative DT Change in temperature (DT) is always T2 – T1 (final temperature – initial temperature) If temperature increases, DT will be positive A substance goes from 15°C to 25°C. 25°C - 15°C = 10°C This is an increase of 10°C If temperature decreases, DT will be negative A substance goes from 50°C to 35°C 35°C – 50°C = -15°C This is a decrease of 15°C

Positive & Negative DH Energy must be put in for temperature to increase A “+” DT will have a “+” DH Energy must be removed for temperature to decrease A “-” DT will have a “-” DH

Example Example: If 285 J is added to 45 g of water at 25°C, what is the final temperature? Cp water = 4.18 J/g°C

Example Example: If 285 J is added to 45 g of water at 25°C, what is the final temperature? Cp water = 4.18 J/g°C DH = change in energy m = mass Cp = heat capacity DT = change in temperature (T2 - T1) T2 = 27°C

Let’s Practice #1 Example:
How many joules must be removed from 25 g of water at 75°C to drop the temperature to 30°? Cp water = 4.18 J/g°C

Let’s Practice #1 Example:
How many joules must be removed from 25.0 g of water at 75.0°C to drop the temperature to 30.0°? Cp water = 4.18 J/g°C DH = change in energy m = mass Cp = heat capacity DT = change in temperature (T2 - T1) DH = J

Let’s Practice #2 Example:
If the specific heat capacity of aluminum is J/g°C, what is the final temperature if 437 J is added to a 30.0 g sample at 15°C

Let’s Practice #2 Example:
If the specific heat capacity of aluminum is J/g°C, what is the final temperature if 437 J is added to a 30.0 g sample at 15.0°C DH = change in energy m = mass Cp = heat capacity DT = change in temperature (T2 - T1) T2 = 31.2°C

Calorimetry

Conservation of Energy
1st Law of Thermodynamics – Energy cannot be created nor destroyed in physical or chemical changes This is also referred to as the Law of Conservation of Energy If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa

Calorimetry Calorimetry – Uses the energy change measured in the surroundings to find energy change of the system Because of the Law of Conservation of Energy, The energy lost/gained by the surroundings is equal to but opposite of the energy lost/gained by the system. DHsurroundings = - DHsystem (m×Cp×DT)surroundings = - (m×Cp×DT)system Don’t forget the “-” sign on one side Make sure to keep all information about surroundings together and all information about system together—you can’t mix and match!

Two objects at different temperatures
Thermal Equilibrium – Two objects at different temperatures placed together will come to the same temperature So you know that T2 for the system is the same as T2 for the surroundings!

An example of Calorimetry
A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of the metal?

An example of Calorimetry
A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of the metal? Metal: m = 23.8 g T1 = 100.0°C T2 = 32.5°C Cp = ? Water: m = 50.0 g T1 = 24°C Cp = 4.18 J/g°C Cp = 1.04 J/g°C

Let’s Practice #3 Example:
A 10.0 g of aluminum (specific heat capacity is J/g°C) at 95.0°C is placed in a container of g of water (specific heat capacity is 4.18 J/g°C) at 25.0°. What’s the final temperature?

Let’s Practice #3 Example:
A 10.0 g of aluminum (specific heat capacity is J/g°C) at 95.0°C is placed in a container of g of water (specific heat capacity is 4.18 J/g°C) at 25.0°C. What’s the final temperature? Metal: m = 10.0 g T1 = 95.0°C T2 = ? Cp = J/g°C Water: m = g T1 = 25.0°C Cp = 4.18 J/g°C T2 = 26.5 °C

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