Mayfield City School District



KEYPractice Thermochemistry Test : 3.3-3.5, 10.530 Points30% of TestThe list of equations for this chapter is attached to the back of the test. Tear them off and use them to help you answer the questions.Math Work Rules Show ALL work to receive full creditCIRCLE final answersInclude units with final answersRound all answers to the 100ths placeLearning Goal: C-Level I can interpret the change in state and energy information contained in a heating/ cooling curve. C-LevelI can explain the relationship between temperature and kinetic energy.Refer to the following Heating Curve for Substance X as you answer Questions 1 thru 4. (11 pts)681076154788120oCWhat is the boiling temperature of this substance? (include units!) ____________ What is the phase of matter for this substance at each of the following temperatures?solidliquid110oC ____________________50oC ____________________freezingWhat phase change is occurring if energy is being removed at 60oC? ____________________Explain why the heating curve remains horizontal at 120oC. Your answer should discuss what is happening to a) temperature, b) kinetic energy, and c) how temperature and kinetic energy are related.Temperature and Ek = both are unchangedAt 120oC Substance X is undergoing a phase change and during a phase change the particles rearrange themselves but they do not gain or lose any EkTemperature is a measure of the average Ek of the particles of a substance, so if the particles don’t change their Ek the temperature won’t change Learning Goal: B-LevelI can calculate the heat of fusion or vaporization for a substance. B-LevelI can use the heat equation to calculate the thermal energy changes of a substance.Calculate the energy required, in kilojoules, for each of the following: Follow ALL math work rules!! (6 pts) The energy change when 20.0 g of steam is cooled from 120. oC to 100oC. q = msteam × SHsteam × ΔTsteamq = 20.0 g × 1.841 Jg-K × (100oC – 120oC) q = 36.82 JK × ─ 20 oC → q = ─ 736.4 J ÷ 1,000 = ─ 0.7364 kJreleased During this process is energy being absorbed or released? ________________________ The energy change when 20.0 g of ice is melted.qvap = mice × ΔHfusH2O qfus = 20.0 g × 334 Jg = 6,680 J ÷ 1,000 = + 6.68 kJabsorbedDuring this process is energy being absorbed or released? ________________________Learning Goal: C-LevelI can explain the relationship between Ek and Heat (thermal) energy transfers between substances. C-LevelI can explain internal energy and heat (thermal) energy of a substance. (a) What is heat (thermal energy) and how is it different from temperature? (2 pts)Heat = energy which is being transferred from one object to another due to a temperature differenceTemperature = a measure of the average EK of an object(b) What is internal energy? (2 pts)the sum of the EK and EP of all of the atoms of a substanceHow are heat and internal energy related? (2 pts)The more internal energy an object has the more “heat” it will have available to transfer Learning Goal: B-LevelI can use the heat equation to calculate thermal energy changes of a substance. C-Level I can define the concept and compare the heat capacity and specific heat of various substances.In the old days, on a cold winter night it was common to bring a hot object to bed with you. Which of the following objects – a 10-kilogram iron brick or a 2-kilogram jug of hot water – would be better to warm up your bed to a nice, cozy temperature of 25.0oC ? Use calculations as part of your answer. Write a few sentences to justify your choice. (7 points)Don't forget to:Use the concept of specific heat to clearly indicate WHICH SUBSTANCE would be the best choice use CALCULATIONS to illustrate WHY this substance would be the best choice write a few sentences in which you COMPARE the results of your calculations and HOW they justify your choice – your answer should include a comparison of the masses and energy of each substance and how this relates to keeping you warmSubstanceMass(kg)Specific Heat J g CoΔT(oC)Water2 4.18425.0Iron100.4525.0 2 kg of water would be the would be the best choice b/ c water has a higher specific heat than ironcalculations to determine the energy which could be released by each substanceqH2O or Fe = mH2O or Fe × SHH2O or Fe × ΔT i) q = 2000 g × 4.184 J g Co × 25.0oC → 209,200 J of energy for waterii) q = 10,000 g × 0.45 J g Co × 25.0oC → 112,500 J of energy for iron 2 kg of water can store more energy than 10 kg of iron (see work above.) The more energy it stores the longer it will be able to keep you warm. Learning Goal: A-LevelI can use the heat equation to calculate energy changes between two substances.BONUS: A 35.2-g sample of a metal heated to 100.0oC is placed in a calorimeter containing 42.5 g of water at an initial temperature of 19.2oC. If the final temperature of the metal and the water is 29.5oC, what is the specific heat of the solid?1st: calculate ΔTwaterΔT = Tf ─ Ti → 29.5oC ? 19.2oC = 10.3oC 2nd : calculate energy gained by water qwater = (42.5 g) (4.184 J g × K ) (10.3oC ) → 1812.29 J3rd : remember that +qwater = ─qmetal ─qmetal = ─ 1812.29 J 4th : calculate cmetal─ 1812.29 J = 35.2 g × SHmetal × (29.5oC – 100oC) = 0.730 J g × KEquation Sheet for Thermodynamics Testcice= 2.03 Jg-KΔHvapH2O = 2260 Jg q = m × SH × ΔTcwater = 4.184 Jg-KΔHfusH2O = 334 Jgqvap = m × ΔHvapcsteam = 1.841 Jg-K qfus = m × ΔHfus ................
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