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Temperature Conversion and Specific Heat Lab Questions

This text discusses temperature conversion and specific heat concepts, with accompanying lab questions and answers.

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Temperature Conversion and Specific Heat Lab Questions

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  1. Ch 16 Heat & Temp Final

  2. 100 TemperatureConvert a temperature of -233.3 C to KelvinsAnswer

  3. Temp 100 K =C + 273.15 K = -233.3 + 273 = 39.7 Answer = 39.7K

  4. 200 TemperatureConvert a temperature of -15.8C to FAnswer

  5. Temp 200 F = 1.8 C + 32 F = 1.8  (-15.8) + 32 = 3.56 Answer = 3.6 F

  6. 300 TemperatureConvert a temperature of 105F to CelsiusAnswer

  7. Temp 300 C = (F – 32) / 1.8 C = (105 – 32) / 1.8 = 40.55555 C Answer = 41 C

  8. 400 TemperatureGive the definition of temperature.Answer

  9. Temperature 400 Answer • Temperature is the average kinetic energy of all particles in a sample.

  10. 500 Temperature Explain what is meant by absolute zero, what is theorized to occur at this temperature, what are the advantages of the Kelvin scale. Answer

  11. Temperature 500 Answer • Absolute zero is the lowest theoretical temperature. • At absolute zero, atomic motion and volume approach zero. • The Kelvin scale is used in scientific calculations because it has no negative numbers. Its T is also equal to T in Celsius.

  12. 100 Specific HeatDefine specific heat and give the SI units for specific heat.Answer

  13. Specific Heat 100 Answer • Specific heat is the amount of energy required to raise the temperature of 1 g of a substance by 1oC. • SI Units are: Joules or Joules g oC g K

  14. Specific Heat 200 Answer • Gold will heat to 200 oC first because it has the lowest Cp and therefore requires less energy to increase its temperature.

  15. 200 Specific Heat The Cp values of Gold, Mercury and Lead are 0.129, 0.14, and 0.160 J/g C respectively. Equal masses of the three metals are placed in an oven at a temperature of 200. C. Which metal will reach 200. C first? Explain. Answer

  16. 300 Specific HeatDetermine the mass of a substance if it has a Cp of 0.385 J/g C and it absorbs 345 calories when heated from 23.5 C to 99.9 C.Answer

  17. Specific Heat 300 • First convert q to Joules • ?J = 345 cal x 4.184J = 1443.5 J cal • Mass = q = 1443.5 J Cp x T (0.385J/goC) (99.9-23.5oC) Answer: Mass = 49.1 g

  18. 400 Specific HeatCalculate the specific heat of a substance if 44.4g of it releases 7850. cal when cooling from 99.0 C to 23.5 C. Report answer in SI unitsAnswer

  19. Specific Heat 400 Answer Cp = q = -7850.cal mass x T (44.4 g) (23.5-99.0oC) = 2.3417 cal x 4.184 J = 9.80 J g C g C cal

  20. 500 Specific HeatCalculate the specific heat of a metal, in SI units, given the following data for a metal that is heated in a hot water bath then poured into a calorimeter containing 35.00 g of water.Mass of metal =12.55 gTemp of water bath = 98.7 CTi of water in calorimeter = 26.5 CTf of water in calorimeter = 32.8 CAnswer

  21. Sp Heat 500 Answer qH2O = 4.184 J/gC x 35.00g x (32.8C -26.5C ) = 922.572J -q H2O = q metal Cp = -922.572 J = 1.115 (12.55g) x (32.8 C -98.7 C ) Answer = 1.1 J/g C

  22. 100 LabApart from type of material, name the 2 criteria that determine the amount of heat energy in a material.Answer

  23. Lab 100 Answer • Mass • Temperature

  24. 200 Lab • Explain why a candle goes out when a jar is placed over it. Answer

  25. Lab 200 • CO2 produced from combustion extinguishes the flame by pushing the oxygen away from the wick.

  26. 300 LabThe Cp of silver is 0.0564 cal/goC. What is this specific heat in J/goC?Answer

  27. Lab 300 Answer • ? J = 0.0564 cal x 4.184 J = goC goC 1 cal Answer = 0.236 J/g oC

  28. 400 LabWhat is visual evidence that a gas is being produced in a chemical change?Describe a lab test that would determine if the gas produced is O2 or CO2.Answer

  29. Lab 400 Answer Fizzing and bubbling indicates production of a gas. A glowing wood splint inserted into the bubbles will relight if the gas is O2 and will extinguish if the gas is CO2.

  30. 500 LabExplain why the tallest candle goes out first when a jar is placed over different sized burning candles. Include density changes in your explanation.Answer

  31. Lab 500 The CO2 produced by combustion is hot, so it rises (it is less dense than surrounding cooler air) The CO2 hits the top of the jar, cools, becomes more dense and descends. As the CO2 sinks it successively extinguishes candles from top of cylinder to bottom.

  32. 100 Heat What happens to heat in an exothermic change?Describe a chemical change that is exothermic.Answer

  33. Heat 100 Answer • An exothermic change releases heat energy. • Combustion is an exothermic chemical change. Zn + HCl to produce H2 gas, Zn pyrotechnics, KClO3 + sugar are also exothermic chemical changes.

  34. 200 Heat Define the term “Heat”.Answer

  35. Heat 200 Answer • Heat is energy that is transferred between samples due to a difference in temperatures.

  36. 300 HeatDefine “calorie”Answer

  37. Heat 300 Answer • A calorie is the amount of heat required to raise the temperature of 1 g of water by 1 degree Celsius

  38. 400 HeatSix boneless chicken tenders (deep fried) contain 1 448 000 joules of energy. Convert to calories and to Food Calories.Answer

  39. Heat 400 ? cal = 1 448 000 Joules x 1 cal = 1 4.184 Joules Answer = 3.461 x 105 cal ? Food Cal = 3.461 x 105 cal x 1 Food Cal = 346.1 Food Cal 1000 cal

  40. 500 Heat Determine the heat absorbed by 15.00 g of pure iron if it is heated from 25.0 C to 255.5 C. The Cp of Fe is 0.444J/g C.Answer

  41. Heat 500 Q = sp ht x mass x T Q = 0.444 J/g C x 15.00 g x (255.5 C -25.0 C ) Q = 1540 J

  42. 100 PotpourriDefine BTUAnswer

  43. Potpourri 100 BTU: The amount of heat required to raise the temperature of one pound of water by 1oF.

  44. 200 PotpourriName the following phase changes and tell whether they are exothermic or endothermicsolid to gasliquid to solidgas to liquidAnswers

  45. Answers for Potpourri 200 solid to gas– sublimation, endo liquid to solid – freezing, exo gas to liquid – condensing, exo

  46. 300 PotpourriExplain, in terms of heat, phase change and molecular motion, why sweating cools you off.Answer

  47. Potpourri 300 • Sweat evaporates from your skin. • Evaporation is an endothermic process because the molecules must gain energy to speed up and break free of their attraction to each other in order to go from liquid phase to gas phase. • The energy that the molecules absorb is taken from your skin. • When the molecules evaporate to the gas phase and leave your skin surface, they take this absorbed energy away. Your skin has lost energy thus is cooler.

  48. 400 Potpourri10. 00 g of water must absorb 3138 J of energy as it heats from 25oC to 100oC. 10. 00 g of water needs 22,600 J of energy to go from liquid to gas at 100oC. Explain.Answer

  49. Potpourri 400 • A phase change from liquid to gas requires a much greater amount of energy than a simple temperature change. • As temp increases yet phase remains same, molecules are speeding up • As water goes from a liquid to a gas, its molecules must have enough energy to become very spread apart, overcoming the attractions between molecules

  50. 500 PotpourriGive 3 properties of hydrogen.Explain why it was used in the Hindenburg despite obvious hazards.Answer

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