Temperature Scales Absolute zero Fahrenheit - Celsius - Kelvin - 32 F = freezing point of water 212  F = boiling point of water Fahrenheit - 0 C = freezing point of water 100 C = boiling point of water Celsius - Kelvin - 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 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 1C° 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 1C°

Q = mcΔT Where: Q = heat m = mass c = specific heat capacity ΔT = temperature change

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 For H2O: cwater= 4.18 J/gram·Cº cice = 2.01 J/gram·Cº csteam = 1.97 J/gram·Cº hf = 334 J/gram hv = 2260 J/gram

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)

Q = mciceΔT + mhf + mcwaterΔT THE LAW OF HEAT EXCHANGE 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 to 0°C melt ice raise water (melted ice) to 44°C Q = (6)(2.01)(11) + 6(334) + 6(4.186)(44) = 3242 J THE LAW OF HEAT EXCHANGE Qlost = Qgained When you use the law of heat exchange, the change in temperature (ΔT) must be positive.

Qlost = Qgained Qmetal = Qwater Cmetal = 0.24 J/gC° Given: 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: water: mmetal = 57.8 g mwater = 100 g Ti= 200°C Ti= 8°C Tf= 14°C Tf= 14°C ΔT = 14 – 8 = 6 C° ΔT = 200 – 14 = 186 C° Qlost = Qgained Qmetal = Qwater Cmetal = 0.24 J/gC° mcΔT = mcΔT water metal (57.8)(c)(186) = (100)(4.186)(6)

Qlost = Qgained Qmetal = Qwater + Qbeaker Given: metal: water: beaker: A 50 gram piece of brass (c= 0.092 cal/gC°) at 95°C is dropped into 250 grams of water at 17.0°C in an 85 gram glass (c= 0.20 cal/gC°) beaker. What is the final temperature of everything? Given: metal: water: beaker: mmetal = 50.0 g mwater = 250 g mbeaker = 85 g cmetal= 0.092 cal/gC° cwater = 1.00 cal/gC° cbeaker = 0.20 cal/gC° Ti= 95°C Ti= 17°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 437 - 4.6Tf = 250Tf - 4250 + 17Tf - 289 mcΔT = mcΔT + mcΔT 4976 = 271.6Tf metal water beaker Tf = 18.3°C