Heat, Temperature, and Specific Heat

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Presentation Transcript

Heat : 

Heat Total kinetic energy of the particles in a sample q Joules (SI) calories

Measuring Heat : 

Measuring Heat Thermodynamics Heat Temperature Heat Transfer Exothermic Endothermic

Temperature : 

Temperature Average kinetic energy of the particles in a sample. T Degrees Celsius -°C Degrees Fahrenheit - °F Kelvins - K

Which is hotter?(higher temperature) : 

Which is hotter?(higher temperature)

Which contains more heat? : 

Which contains more heat?

Heat Transfer : 

Heat Transfer Heat is always transferred from the hotter substance to the colder. Particles in the hotter substance therefore slow down (decrease in temp.) Particles in the colder substance therefore speed up (increase in temp.)

Exothermic : 

Exothermic A process that releases heat energy

Endothermic : 

Endothermic A process that absorbs energy

Heat Transfer : 

Heat Transfer Every exothermic process is accompanied by an endothermic process. HEAT Exothermic Endothermic

Specific Heat : 

Specific Heat Property of a substance. How much heat it takes to change the temperature of a one gram of it one degree Celsius. Lower the specific heat the easier it is to heat up.

Specific Heat : 

Specific Heat The specific heat of water is higher than concrete. This is why the concrete around a swimming pool is hotter than the water even though it is exposed to equal amounts of sunlight.

Heat Transfer : 

Heat Transfer Temperature change depends on three factors Amount of heat transferred Joules or calories or Calories (kilocalories) Mass of substance grams Specific heat cal/g•°C q = mCpT

Heat Transfer : 

Heat Transfer q = mCpT m is mass measured in grams Cp is specific heat A characteristic property of the substance Cp is given in cal/g•°C T is the change in temperature (°C)

Systems : 

Systems Closed First Law of Thermodynamics q1 = - q2 Put hot metal in water; the amount of heat given off by the metal is the amount of heat absorbed by the water. Open Heat can be transferred to other substances Calculations are not successful

Sample Problem : 

Sample Problem A 5250 gram copper pot is heated from 21.2oC to 99.8oC, How much heat was added to the copper pot? The specific heat of copper is 0.0923 cal/goC.

Sample Problem : 

Sample Problem A 5250 gram copper pot is heated from 21.2oC to 99.8oC, How much heat was added to the copper pot? The specific heat of copper is 0.0923 cal/goC. Identify information given: m = 5250 grams DT = 99.8oC -21.2oC = 78.6oC Cp = 0.0923 cal/goC

Sample Problem : 

Sample Problem A 5250 gram copper pot is heated from 21.2oC to 99.8oC, How much heat was added to the copper pot? The specific heat of copper is 0.0923 cal/goC. Identify information given and plug in numbers. q = mCpDT q = (5250 g) x (0.0923 cal/goC) x (78.6oC)

Sample Problem : 

Sample Problem A 5250 gram copper pot is heated from 21.2oC to 99.8oC, How much heat was added to the copper pot? The specific heat of copper is 0.0923 cal/goC. Cross out units and solve. q = (5250 g) x (0.0923 cal/goC) x (78.6oC) = 38100 cal

Sample Problem : 

Sample Problem A 118 g piece of tin at 85.0 °C is dropped into 100. g of water at 35.0 °C. The specific heat of water is 1.00 cal/g•°C. The final temperature of the system is 38.0 °C. What amount of heat is absorbed by the water?

Sample Problem : 

Sample Problem A 118-g piece of tin at 85.0 °C is dropped into 100. g of water at 35.0 °C. The final temperature of the system is 38.0 °C. The specific heat of water is 1.00 cal/g•°C. What amount of heat is absorbed by the water? q = mCpT q = 100. g x 1.00 cal/g•°C x (Tfinal – Tinitial) q = 100. g x 1.00 cal/g•°C x (38.0 °C – 35.0 °C) q = 100. g x 1.00 cal/g•°C x 3.0 °C q = 300. cal

Sample Problem : 

Sample Problem A 118-g piece of tin at 85.0 °C is dropped into 100. g of water at 35.0 °C. The final temperature of the system is 38.0 °C. The specific heat of water is 4.18 J/g•°C. What amount of heat is released by the tin? qwater = - qtin 300 cal = - 300 cal

Sample Problem : 

Sample Problem A 118-g piece of tin at 85.0 °C is dropped into 100. g of water at 35.0 °C. The final temperature of the system is 38.0 °C. The specific heat of water is 4.18 J/g•°C. What is the specific heat of the tin? q = mCpT Cp = q / m T Cp = - 300. cal / [118 g x (38.0 °C – 85.0 °C)] Cp = - 300. cal / [118 g x (– 47.0 °C)] Cp = 0.0541 cal/g•°C