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HEAT AND CALORIMETRY

HEAT AND CALORIMETRY. What is Heat?. Modern definition: energy transferred because of a difference in temperature(energy on the move) 1 calorie raises the temperature of 1 gram of water by 1 Celsius degree. 1 kilocalorie (kcal or Calorie) raises temperature of 1 kg of water by 1 degree Celsius

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HEAT AND CALORIMETRY

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  1. HEAT AND CALORIMETRY

  2. What is Heat? • Modern definition: energy transferred because of a difference in temperature(energy on the move) • 1 calorie raises the temperature of 1 gram of water by 1 Celsius degree. • 1 kilocalorie (kcal or Calorie) raises temperature of 1 kg of water by 1 degree Celsius • 1 British Thermal Unit (BTU) raises temperature of 1 pound of water by one degree Farenheit

  3. Mechanical Equivalent of Heat • Discovered by James Joule • Falling weight makes paddle turn • 4.186 x 103 J = 1 kcal • Interpretation: HEAT IS ENERGY TRANSFER Courtesy W. Bauer http://lecture.lite.msu.edu/~mmp/kap11/cd295.htm

  4. Joule’s Apparatus Link to Joule’s original article

  5. Imagine This • What other situations could be used to demonstrate the mechanical equivalent of heat?

  6. Example • When digested a slice of bread yields 100 kcal. How high a hill would a 60 kg student need to climb to “work off’ this slice of bread? (100 kcal is energy released by burning the bread 100 kcal x 4.186 x 103 J/kcal = 4.2 x 105 J W = mgh h = W/mg = 4.2 x 105 / (60 kg)(9.80 m/s2) = 714m = 7.1 x 102 m If the body is only 20 percent efficient in transforming the bread, how high need they climb?

  7. Bullet in Block • When a 10 g bullet traveling 500 m/s is stopped inside a 1kg wood block nearly all its KE is transformed to heat. How many kcal are released? KE = ½ mv2 = 0.5 x 0.010 kg x (500)2 = 1250 J 1250 J x 1 kcal/4186 J = 0.30 kcal

  8. Internal Energy, Temperature and Heat • The thermal energy or internal energy of an object is the sum total of all the energy in its molecules (symbol: U) • Temperature is measure of average KE per molecule (T) • Heat refers to transfer of energy (symbol: Q)

  9. Internal Energy Does Not Include KE • KE of whole object represents all molecules moving in the same direction • U is random motion only

  10. Internal Energy of an Ideal Gas • U = N (1/2 mv2) • But 1/2 mv2 = 3/2kT • U = 3/2NkT • U = 3/2 nRT N is total number of molecules in sample k = R/NA U = 3/2 nRT Internal Energy of an ideal gas depends only on temperature and number of moles

  11. Specific Heat • How much does temperature rise when heat is put into something? • Depends on the material, the mass and the quantity of heat: Q = m c DT c is specific heat in J/kg Co or Q = n c DT c is specific heat in J/mol Co Specific Heat of water = 4.186 x 103J/kg Co DT = Q/mc Table of Specific Heats

  12. Understanding Specific Heat • For a given amount of heat added, temperature increases more for a smaller specific heat • Water has a very large specific heat and is therefore relatively difficult to heat or cool.

  13. Examples • How much heat is required to heat 1 liter of water from 20 oC to 100 oC? • Q = mcDT = 1Kg x 4186 J/kg oC x 80 oC = 3.35 x 105 J = 80 kcal • How much heat is required to heat 1.0 kg iron from 20 oC to 100 oC • Q = mcDT = 1Kg x 450 J/kg oC x 80 oC = 3.6 x 104 J

  14. On the Stove • In heating 3.0 liters of water to make spaghetti Mike’s stove heats water at a rate of 4500 watts. The initial temperature of the water is 150 C. After five minutes he sticks his hand in the water. Will he get scalded? What is the temperature? DT = Q/mc = (4.5 x 103 x 300s)/(3.0kg x 4186 J/kg oC) = 107 degrees! What actually happened to the water? Started boiling when it reached 100 0C

  15. Oops • A 145 g baseball crashes into a 500g pane of glass while moving at 30 m/s. The specific heat of the glass is 840 J/kg oC. Find the temperature increase of the glass assuming that half the ball’s KE is transformed into heating the glass. KE = ½ mbv2 = 0.5 x 0.145kg x (30)2 = 65.25 J DT = Q/mgc = ½ KE/mgc = (.5)(65.25)/(0.500kg x 840 J/kgoC) = 0.078 degrees

  16. Armageddon • An asteroid crashes into Earth while traveling 60 km/sec. All its KE is transformed into heating a mass of rock (specific heat 860 J/kg 0C) equal to its own mass. What is the temperature increase of the rock? What happens to it? ½ mv2 = mcDT DT = v2/2c DT = (6 x 104 m/s)2 / 2 x 860 J/kg 0C = 2.1 x 1060C VAPORIZED; BECOMES PLASMA

  17. Calorimetry • When objects at different temperature are placed in thermal contact heat flows from the hotter to the cooler • Energy is conserved • Heat lost by one part of system = heat gained by other part or parts • Calorimetry should be done in a container well insulated so that little heat is lost or gained from the outside. Sometimes it is necessary to include heat gained by Calorimeter cup.

  18. Temperature of an Iron Block • 500g of iron is heated on a hot plate. It is placed in 200ml of water at 20 0C. After allowing time for thermal equilibrium to be reached the temperature of the water is found to be 500C. What was the initial temperature of the iron? Heat lost by iron = heat gained by water Let TW be initial temp. of water; TI that of iron; TF final temp of both mIcI(TI-TF) = mW cW (TF – TW)

  19. Temperature of an Iron Block, Continued • mIcITI= mW cW (TF – TW) + mIcITF • TI ={mW cW (TF – TW) + mIcITF}/ mIcI • TI = {0.2 x 4186(50-20) + 0.5 x 450 x 50}/(0.5 x 450) • TI = 161 oC

  20. Method of Mixtures • How can the specific heat of an unknown liquid such as antifreeze be determined? Design an experiment to do this.

  21. Change of phase and Latent Heat Courtesy “Hyperphysics” web site. Georgia State University

  22. Key Facts • Temperature does not change during change of phase • For water 00 C for freezing and melting • For water 1000 C for boiling and condensation • Slope during heating is related to specific heat! • Question: compare the specific heat of ice with that of water. That of ice is less

  23. Q = mL • Q is heat added or released • M is mass • L is latent heat • Heat of fusion; solid to liquid or vica versa • Heat of vaporization: liquid to vapor or vv • Table of Latent Heats • For water LF = 333 kJ/kg = 3.33 x 105 J/kg LV = 22.6 x 105 J/kg

  24. Example • How much heat is required to turn 1 kg of ice at - 200C to steam? • Heat the ice 1 x 2100 x 20 = 4.2 x 104 J • Melt the ice 1 x 3.33 x 105 = 3.33 x 105 J • Heat the water to boiling 1 x 4186 x 100 = 4.19 x 105 • Vaporize the water 1 x 22.6 x 105 =22.6 x 105 J • Total = 3.05 x 106 J • How long would this take on a 4500 w burner? Time = Q/rate = 3.05 x 106 / 4.5 x 103 = 678 sec = 11.3 min

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