Calorimetry Chapter 5

Formal assessment – analyze student responses to the quiz and problems Objectives Today I will be able to: Apply knowledge of the polyatomic ions to a quiz Distinguish between the terms heat capacity, specific heat capacity and molar heat capacity. Solve calorimetry problems. Informal assessment – monitoring student questions as they complete the practice problems Formal assessment – analyze student responses to the quiz and problems Common Core Connection Make sense of problems and persevere in solving them Build strong content knowledge

Lesson Sequence Evaluate: Polyatomic Ions Quiz Explain: Calorimetry Notes Elaborate: Calorimetry Practice Problems Evalaute: Exit Ticket

Warm Up Take everything off of your desk for your polyatomic ions quiz Place your electronic device in the bin on Ms. Ose’s desk with a sticky note Make sure your device is on silent

objectives Today I will be able to: Apply knowledge of the polyatomic ions to a quiz Distinguish between the terms heat capacity, specific heat capacity and molar heat capacity. Solve calorimetry problems.

Homework Bring lab notebook to school Thursday Reading Read the Calorimetry section (5.5) in the textbook before Monday (Chapter 5) Read through the Handwarmer Lab before Monday Extra Calorimetry Practice Problems Solubility Rules Quiz Thursday, September 18

AGENDA Polyatomic Ions Quiz Calorimetry Notes Calorimetry Practice Problems Exit Ticket

Calorimetry Chapter 5

Calorimetry Calorimetry is the study of the heat released or absorbed during physical and chemical reactions. For a certain object, the amount of heat energy lost or gained is proportional to the temperature change. The initial temperature and the final temperature in the calorimeter are measured and the temperature difference is used to calculate the heat of reaction.

Heat Capacity Symbol = C is the amount of energy required to raise the temperature of an object 1 Kelvin or 1 °C Units = J/K or J/ °C

Heat Capacity Heat capacity is an extensive property, meaning it depends on the amount present a large amount of a substance would require more heat to raise the temperature 1 K than a small amount of the same substance. Heat capacity depends upon the amount of the substance you have.

Molar Heat Capacity = C molar. Symbol = C molar. For pure substances, the heat capacity for one mole of the substance may be specified as the molar heat capacity. Units = C/moles q = molar heat x moles x DT

Specific Heat Capacity = c or s Symbol = c or s The specific heat capacity, c or s, is often used since it is the heat capacity per one gram of the substance Units=J/gK or J/gC. The specific heat capacity of each substance is an intensive property which relates the heat capacity to the mass of the substance.

Remember… An extensive property is a property that changes when the size of the sample changes. Examples are mass, volume, length, and total charge. An intensive property doesn't change when you take away some of the sample. Examples are temperature, color, hardness, melting point, boiling point, pressure, molecular weight, and density.

Specific Heat Capacity “c or s” q = mcDT This is the main equation for calorimetry calculations mass will be in grams Units for “c” are J/g K or J/g °C The specific heat of water is 1 cal/g ºC

Is q positive or negative? If a process results in the sample losing heat energy, the loss in heat is designated as - q. The temperature of the surroundings will increase during this exothermic process If the sample gains heat during the process, then q is positive. The temperature of the surroundings will decrease during an endothermic process

Direct proportionality The amount of heat that an object gains or loses is directly proportional to the change in temperature. remember q = mcΔT q and T are directly proportional

Dt = tfinal – tinitial = 50C – 940C = -890C How much heat is given off when an 869 g iron bar cools from 940C to 50C? s of Fe = 0.444 J/g • 0C Dt = tfinal – tinitial = 50C – 940C = -890C q = msDt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J 6.4

Try this Problem The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.

Answer to Problem q = mcΔT 348 K - 294 K= q = (75 kg ) (1000g/kg) (0.71 J/gºC)(54 ºC) q = 2875500 J or 2875.5 kJ

Extra Problem A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?

Answer to Problem qcu = mcΔT qh2o = mcΔT mcΔT = qcu = qh2o = mcΔT (46.2 g) ccu (73.6 °C) = (75 g) (4.184 J/g°C) (2.2°C) ccu = 0.20 J/g°C

Summary of Definitions: Heat capacity is the amount of energy required to raise the temperature of an object 1 kelvin or 1 °C. Specific heat capacity is the heat capacity of 1 gram of a substance. Molar heat capacity is the heat capacity of 1 mole of a substance.

Heat (q) absorbed or released: The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = msDt q = CDt Dt = tfinal - tinitial 6.4

Practice problems Try the calorimetry problems at the end of Ch 5 # 51,53, 55, 57

Wrap up Calorimetry is an experimental way of determining heat values for substances. Please make sure that you have a bound composition notebook. We will perform a lab on Monday. You will need to wear close toed shoes.