CHEM 105N 1st Edition Lecture 9Outline of Last Lecture II.States of Matter-Solids-Liquids-GasesIII.Physical PropertiesIV.Physical ChangeV.Heating Curve & CalculationsVI.Cooling Curve & CalculationsVII.Heat of FusionVIII.Heat of VaporizationIX.Specific HeatOutline of Current Lecture I. Heat EquationII. CalorimetersIII. Energy RequirementsIV. Chemical Properties & ChangeCurrent LectureI. Heat Equation:- Amount of heat lost or gained by a substance- Rearrange the Specific Heat equation to get:Heat= mass xTx specific heatEx: A thermos holds 800g of water at 55°C. If the water cools to 33°C, how many calories of heat could be transferred to someone’s body?1. Given:Mass of water: 800gInitial Temp: 55°CFinal Temp: 33°C2. NEED: Calories of heat transferred3. Calculate Temp Change: T= Tinitial – Tfinal55°C - 33°C = 22°C4. Complete equation and solve:800g x 22°C x (1 cal/ g°C)= 1,6000 calThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.II. Calorimeters:-Indicates the amount of heat lost by food by observing the temp increase in the water- Use label CAPITAL “C”- Equal to 1,000 lowercase “c”’s- Used to calculate the energy value of foodEnergy value of fat= (Heat (kJ))/(mass of sample)Ex.When a 7.2g sample of fat was combusted in a calorimeter, 122 kJ was absorbed. Calculate the energy value of the fat in kJ/g.(122 kJ)/(7.2g)= 16.9 kJ/gIII. Energy Requirements:- The amount of energy needed each day depends on many factors:Age, sex, and physical activity- If food intake exceeds energy use, a person gains weight. - If food intake is less than energy use, a person loses weight.IV. Chemical Properties and Change- Chemical Properties- describe the ability of a substance to change into a new substance- During a Chemical Change a new substance forms that has a new composition and new chemical/physical
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