VIEWS: 3 PAGES: 23 POSTED ON: 3/13/2012
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 Molecules move increases 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) 1 kcal These equivalents can be used in dimensional analysis to convert units Heat Capacity Heat Capacity – The amount of energy that can be absorbed before any mass of a substance has increased its temperature by one degree. (This changes for one substance depending on how much) 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 (This stays the same for one substance because it is for a set amount.) 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 Small amount of energy can energy to noticeably noticeably change change temp temperature Heats up slowly Heats up quickly Cools down slowly Cools down quickly Maintains temp better with Quickly readjusts to new small condition changes conditions •A pool takes a long time to warm up and remains fairly warm over night. (high) •The air warms quickly on a sunny day, but cools quickly at night. (low) This is why a desert has extremes of hot and cold due to a lack of water which can maintain the temp. •A cast-iron pan stays hot for a long time after removing from oven. (high) •Aluminum foil can be grabbed by your hand from a hot oven because it cools so quickly (low) 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 H m C p T Energy added Mass of sample Change in temperature or removed Positive & Negative T Change in temperature (T) is always T2 – T1 (final temperature – initial temperature) If temperature increases, T 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, T 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 H Energy must be put in for temperature to increase A “+” T will have a “+” H Energy must be removed for temperature to decrease A “-” T will have a “-” H Let’s Practice #1 Example: How many joules must be removed from 25 g of water at 75.0°C to drop the temperature to 30.0°? Cp water = 4.18 J/g°C Let’s Practice #1 Example: H m C p T How many joules must be removed from 25.0 g of water H = change in energy at 75.0°C to drop the m = mass temperature to 30.0°? Cp Cp = heat capacity water = 4.18 J/g°C T = change in temperature (T2 - T1) H 25 .0 g 4.18 J 30 .0 75 .0 C g C H = - 4702.5 J H = - 4.70 x 103 J 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: H m C p T If 285 J is added to 45 g of water at 25°C, what is the H = change in energy final temperature? Cp water m = mass = 4.18 J/g°C Cp = heat capacity T = change in temperature (T2 - T1) 285 J 45 g 4.18 J T2 25 C 285 J T2 25 C g C 45 g 4.18 J g C 285 J 25 C T2 45 g 4.18 J g C T2 = 26.515°C T2 = 27°C Let’s Practice #2 Example: If the specific heat capacity of aluminum is 0.900 J/g°C, what is the final temperature if 437 J is added to a 30.0 g sample at 15.0°C Let’s Practice #2 Example: H m C p T If the specific heat capacity of aluminum is 0.900 J/g°C, H = change in energy what is the final temperature m = mass if 437 J is added to a 30.0 g Cp = heat capacity sample at 15.0°C T = change in temperature (T2 - T1) 437 J 30 .0 g 0.900 J T2 15 .0 C 437 J T2 15 .0 C g C 30 .0 g 0.900 J g C 437 J 15 .0 C T2 30 .0 g 0.900 J g C 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. Hsurroundings = - Hsystem (m×Cp×T)surroundings = - (m×Cp×T)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 Why is this important? So you know that T2 for the system is the same as T2 for the surroundings! An example of Calorimetry Example: 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 Example: A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24.0°C water. If the final temperature of the water is 32.5°,what is the heat capacity of the metal? Metal: Water: M = 23.8 g M = 50.0 g T1 = 100.0 °C T1 = 24.0 °C H metal H water T2 = 32.5 °C T2 = 32.5 °C Cp = ? Cp = 4.18 J/g°C m C p T metal m C p T water 23 .8 g C p 32 .5 C 100 .0 C 50 .0 g 4.18 J 32 .5 C 24 .0 C gC 50 .0 g 4.18 J 32 .5 C 24 .0 C Cp gC 23 .8 g 32 .5 C 100 .0 C Cp = 1.04 J/g°C Let’s Practice #3 Example: A 10.0 g of aluminum (specific heat capacity is 0.900 J/g°C) at 95.0°C is placed in a container of 100.0 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 0.900 J/g°C) at 95.0°C is placed in a container of 100.0 g of water (specific heat capacity is 4.18 J/g°C) at 25.0°C. What’s the final temperature? H metal H water Metal: m = 10.0 g m Cp T metal m Cp T water T2 95 .0 C 100 .0 g 4.18 J T2 25 .0 C T1 = 95.0°C 10 .0 g 0.900 J T2 = ? g C gC Cp = 0.900 J/g°C 9.0 T2 95 C 418 T2 25 .0 C Water: m = 100.0 g 11305 T1 = 25.0°C 9.0T2 855 418 T2 10450 T2 427.0 T2 = ? Cp = 4.18 J/g°C 427 .0T2 11305 T = 26.5 °C 2