 # Temperature Scales Fahrenheit Celsius Kelvin 32 F freezing

• Slides: 9 Temperature Scales Fahrenheit Celsius Kelvin - 32 F = freezing point of water 212 F = boiling point of water 0 C = freezing point of water 100 C = boiling point of water 273 K = freezing point of water 373 K = boiling point of water Absolute zero The temperature at which it is no longer possible to extract heat from the atoms Absolute zero = 0 K = - 273 C = - 460 F Specific Heat Capacity (c) Specific heat capacity is a measure of the ability of a material to absorb heat energy - specific heat capacity is defined as the amount of heat energy required to change the temperature of 1 gram of a substance by 1 C° Units of Specific Heat Capacity joules / gram·C° or calories / gram·C° Example: The specific heat capacity of water is 1 calorie / gram·C° This means it takes 1 calorie of heat energy to change the temperature of 1 gram of water by 1 C° Molar Heat Capacity (C) - molar heat capacity is defined as the amount of heat energy required to change the temperature of 1 mol of a substance by 1 C° Units of Molar Heat Capacity joules / mol·C° or calories / mol·C° Example: The molar heat capacity of water is 18 calorie / mol·C° This means it takes 18 calorie of heat energy to change the temperature of 1 mol of water by 1 C° Q = mcΔT Q = n. CΔT Where: Q = heat m = mass Joules or calories grams n = mols c = specific heat capacity J/g∙C° or cal/g∙C° C = molar heat capacity J/mol∙C° or cal/mol∙C° ΔT = temperature change C° Latent Heat the heat energy involved in a phase change fusion (hf) – phase change from solid to liquid vaporization (hv) – phase change from liquid to gas Units of Latent Heat joules / gram or calories / gram joules / mol or calories / mol For H 2 O: cwater= 1. 00 cal/gram·Cº cice = 0. 48 cal/gram·Cº csteam = 0. 47 cal/gram·Cº hf = 80 cal/gram hv = 540 cal/gram Cwater= 18. 0 cal/mol·Cº Cice = 8. 65 cal/mol·Cº Csteam = 8. 47 cal/mol·Cº hf = 1440 cal/mol hv = 9720 cal/mol The Heating Curve A → B increase in temperature of the solid (Q=mciceΔT) B → C melting of solid (Q=mhf) C → D increase in temperature of the liquid (Q=mcwaterΔT) D → E vaporization of liquid (Q=mhv) E → F increase in temperature of the vapor (Q=mcsteamΔT) How much heat is required to change a 6 gram ice cube at – 11°C to water at 44 °C? Q = mciceΔT + mhf + mcwaterΔT raise ice melt to 0°C ice raise water (melted ice) to 44°C. Q = (6)(0. 48)(11) + 6(80) + 6(1)(44) = 775. 7 calories THE LAW OF HEAT EXCHANGE Qlost = Qgained When you use the law of heat exchange, the change in temperature (ΔT) must be positive. A 57. 8 gram sample of metal is originally at 200°C. The metal is dropped into 100 grams of cool water at 8°C. The temperature of the water and the metal reaches equilibrium at 14°C. What is the specific heat of the metal? Given: metal: mmetal = 57. 8 g Ti= 200°C Tf= 14°C water: mwater = 100 g Ti= 8°C Tf= 14°C ΔT = 200 – 14 = 186 C° ΔT = 14 – 8 = 6 C° Qlost = Qgained Qmetal = Qwater mcΔT = mcΔT water metal (57. 8)(c)(186) = (100)(1. 00)(6) cmetal = 0. 056 cal/g. C° A 50 gram piece of brass (c= 0. 092 cal/g. C°) at 95°C is dropped into 250 grams of water at 17. 0°C in an 85 gram glass (c= 0. 20 cal/g. C°) beaker. What is the final temperature of everything? Given: metal: water: mmetal = 50. 0 g cmetal= 0. 092 cal/g. C° Ti= 95°C mwater = 250 g cwater = 1. 00 cal/g. C° Ti= 17°C beaker: mbeaker = 85 g cbeaker = 0. 20 cal/g. C° Ti= 17°C 50(0. 092)(95 - Tf) = 250(1. 00)(Tf - 17) + 85(0. 20)(Tf - 17) Qlost = Qgained 4. 6(95 - Tf) = 250(Tf - 17) + 17(Tf - 17) Qmetal = Qwater + Qbeaker DISTRIBUTE mcΔT = mcΔT + mcΔT 437 - 4. 6 Tf = 250 Tf - 4250 + 17 Tf - 289 4976 = 271. 6 Tf Tf = 18. 3°C metal water beaker