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Lecture

CHEM 1310 - Notes 2.docx

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Department
Chemistry
Course
CHEM 1310
Professor
Sarrah Vakili
Semester
Summer

Description
Question:  A drink has 180 calories. Given that an average 150 lb person uses 800 kJ/hour while walking, how long must that person walk to not store any energy? Assume energy only goes into walking Answer  Convert from calories to kJ o 180 Cal x o 5 180000 cal x 10 J 5 2 o OR 7.53 x 10 / 1000 J = x=7.53 x 10 kJ  Now find time o  Note: energy will flow from high (hot) to low (cold), until both objects are at the same temperature. o Tsys < Tsurr - energy flows until Tsys = Tsurr Heat  Heat capacity - C o Amount of heat needed to change the temperature by 1 degree in the heat system  Molar heat capacity - the heat capacity of one mol of substance3  Specific heat capacity o The heat capacity -1 1 g-1f substance o For water = 1 cal g or C = 4.184 J  Heat capacity - proportionality constants between q (quantity of heat) and T (temp) o Are compound specific ie: is a property of the substance (state function)  More ways to store heat; the higher the heat capacity will be for the compound o Note: how 'hot' an object is is NOT related to its heat capacity  Related to its surroundings Law of the Conservation of Energy  Energy is neither created or destroyed  Total energy between the system and the surroundings, or the universe, is constant  Q systemqsurroundings  Thus -qsys= + qsurrR +q sys -qsurr Heat of a Reaction  Amount of heat exchanged or transferred during a chemical reaction in a system, between the system and its surroundings at constant temperature  Heat of combustion = combustion reaction  Measured with isolated systems and thermometers o Gives T, thus it may be used to find the quantity of heat (q), by a technique called calorimetry o Q = quantity of heat transferred o C = heat capactiy o ∆ T = TfinalTinitial Quantity of Heat = q = C∆T  Amount of heat needed to change the temperature of the substance of 1 compound  Depends on the following o Amount of temperature needed (if T = negative than o Amount of substance/compound present - from definition of C o Type of compound or substance  When positive, heat is absorbed or gained by the system  When negative heat is released or lost by the system  Units to q kJ or J Question: A 55 g piece of metal was heated to 99.8ᵒ C and then dropped into an insulated container with H2O. There was 225 mL of H2O (density = 1 g/mL) at 210ᵒ C, before true metal was added. The final temperature of the metal and water was 23.2ᵒ C. What was the specific heat of the metal assuming no heat transfer to the container or atmosphere? Atmosphere:  Metal is our system - we're asked about it o Cooled down, thus qmetal = negative  Surroundings = water - heats up from 21 to 23ᵒ C o Thus qH2O = positive  Universe = container - remains constant o qH2O + qmetal = 0  By law of conservation of energy qH2O = -qmetal o Also know q=mC ap T  Metal m = 55g -1 -1 -1 -1  C specificg or C or Jg K  ∆T = T finalTinitial  Tf= 23.2ᵒC + 273.15 = 296.35 K  Ti= 99.8ᵒC + 273.15 = 273.15 372.95 K  Water - mass = (225 mL (volume)) x (1 g/mL(density)) = 225 g -1 -1  C spconstant = 4.1845 g K  ∆ T o Tf= 296.35 K o T i 21ᵒC + 273.15 = 273.15 294.15 K  Now sub in the numbers o (225g)(4.184 jK g )(296.35 -294.15 K) = -(55g)(C )sp96.35 - 372.95) o Note: addition/subtraction first, then multiply/divide to find this o 2071.08 J = +4213gK/C sp o C sp -1g-1 Question: 3 -1 You have 2 blocks of metal with volume of 10 cm . One block is made of Al with a density of 2.698 gcm , specific heat of .900 jK g and a temperature of 0ᵒ C.
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