So what kind of intuition do we have about heat and temperature and energy? Discuss the DCI. DCI15.1. Two containers of water are at 20 C initially. One contains 50 mls and the other 100 mls. They are each heated with the same source of heat for the same amount of time. If the final temperature of the 50 mls sample is 50 C what would be the final temperature of the 100 mls sample? A. 50 C B. 80 C C. 25 C D. 100 C E. 35 C 1
DCI15.1. Two containers of water are at 20 C initially. One contains 50 mls and the other 100 mls. They are each heated with the same source of heat for the same amount of time. If the final temperature of the 50 mls sample is 50 C what would be the final temperature of the 100 mls sample? A. 50 C B. 80 C C. 25 C D. 100 C E. 35 C Same amount of heat to both beakers, but different mass. T = 30 for beaker on the lea, so T is half or 15. Mass α 1/ T Two containers each have 50 mls of water at 20 C initially. They are each heated with the same source of heat. One is heated for ten minutes and the other for five minutes. If the container that was heated for five minutes has a final temperature 30 C what would be the final temperature of the other sample? A. 35 C C. 60 C D. 25 C E. 30 C 2
Two containers each have 50 mls of water at 20 C initially. They are each heated with the same source of heat. One is heated for ten minutes and the other for five minutes. If the container that was heated for five minutes has a final temperature 30 C what would be the final temperature of the other sample? A. 35 C C. 60 C D. 25 C E. 30 C Both beakers contain the same amount of water. Twice the heat to one. T is 10 for smaller amount of heat, than T = 20 for larger amount. Q(heat) α T Two containers of water are at 20 C initially. One contains 50 g of water and is heated by a source for a specified time to a final temperature of 30 C. The second container has an unknown amount of water and is heated with the same source to 30 C. However, it takes twice as long to get to this final temperature. How much water is in this container? A. 100 g B. 25 g C. 30 g D. 50 g E. 75 g 3
Two containers of water are at 20 C initially. One contains 50 g of water and is heated by a source for a specified time to a final temperature of 30 C. The second container has an unknown amount of water and is heated with the same source to 30 C. However, it takes twice as long to get to this final temperature. How much water is in this container? A. 100 g B. 25 g C. 30 g D. 50 g E. 75 g Twice the heat is added to one beaker to reach the same final temperature ( T). So the beaker must have twice the mass. Q(heat) α Mass So we have established the following relanonships; Mass α 1/ T q(heat) α T q(heat) α mass So q(heat) α mass T Heat is directly propornonal to the mass Nmes the change in temperature. 4
50 mls of water at 80 C is added to 50 mls of water at 20 C. What would be the final temperature? A. 60 C C. 30 C D. 20 C E. 50 C 50 mls of water at 80 C is added to 50 mls of water at 20 C. What would be the final temperature? A. 60 C C. 30 C D. 20 C E. 50 C 5
50 mls of water at 80 C is added to 50 mls of water at 20 C. What would be the final temperature? A. 60 C C. 30 C D. 20 C E. 50 C q hot water + q cold water = 0 q hot water = q cold water mass hot water T hot water = mass cold water T cold water 50. g T hot water = 50. g T cold water 50. g (T final 80.0 ) = 50. g (T final 20.0 ) 2T final = 100 T final = 50 50 mls of water at 80 C is added to 100 mls of water at 20 C. What would be the final temperature? A. 70 C C. 30 C D. 60 C E. 50 C 6
50 mls of water at 80 C is added to 100 mls of water at 20 C. What would be the final temperature? A. 70 C C. 30 C D. 60 C E. 50 C 50 mls of water at 80 C is added to 100 mls of water at 20 C. What would be the final temperature? A. 70 C C. 30 C D. 60 C E. 50 C q hot water = q cold water mass hot water T hot water = mass cold water T cold water 50. g T hot water = 100. g T cold water 50. g (T final 80.0 ) = 100. g (T final 20.0 ) (T final 80.0 ) = 2 (T final 20.0 ) 3T final = 120 T final = 40 7
50 g of water at 80 C is added to 50 g of ethyl alcohol at 20 C. What would be the approximate final temperature? A. 60 C C. 30 C D. 20 C E. 50 C 50 g of water at 80 C is added to 50 g of ethyl alcohol at 20 C. What would be the approximate final temperature? A. 60 C C. 30 C D. 20 C E. 50 C TWO DIFFERENT SUBSTANCES! Experimentally the final temperature is determined to be close to 60. 8
q(heat) α mass T How do we make this an equality? We must introduce a constant.in this case the constant is called the specific heat, SH, q(heat) = mass SH T Specific heat is the amount of heat required to raise the temperature of 1 gram of a substance 1 C. Specific Heats of Substances Compound Specific Heat (J C -1 g -1 ) H 2 O(l) 4.184 H 2 O(s) 2.03 Al(s) 0.89 C(s) 0.71 Fe(s) 0.45 Hg(l) 0.14 O 2 (g) 0.917 CH 3 CH 2 OH 2.46 9
q(heat) = mass SH T A 175 g sample of water, initially at 23.45 C absorbs some heat. The final temperature of the sample after absorbing the heat is 26.85 C. Calculate the amount of heat absorbed by the sample of water. (NOTE: The specific heat for water is 4.184 J g -1 C -1.) q = mass water SH water T water q(heat) = mass SH T A 175 g sample of water, initially at 23.45 C absorbs some heat. The final temperature of the sample after absorbing the heat is 26.85 C. Calculate the amount of heat absorbed by the sample of water. (NOTE: The specific heat for water is 4.184 J g -1 C -1.) q = mass water SH water T water q = 175. g 4.184 J g-1 C -1 (26.85 23.45 ) q = 2.49 x 10 3 J 10
q(heat) = mass SH T A piece of iron weighing 80.0 g initially at a temperature of 92.6 C released the same amount of heat to the 175 g sample of water in DCI16.4. Assume the final temperature of the metal is the same as the final temperature of the water in DCI16.4. What is the specific heat for iron? q(heat) = mass SH T A piece of iron weighing 80.0 g initially at a temperature of 92.6 C released the same amount of heat to the 175 g sample of water in DCI16.4. Assume the final temperature of the metal is the same as the final temperature of the water in DCI16.4. What is the specific heat for iron? The metal is absorbing 2.49 x 103 J of heat 11
q(heat) = mass SH T A piece of iron weighing 80.0 g initially at a temperature of 92.6 C released the same amount of heat to the 175 g sample of water in DCI16.4. Assume the final temperature of the metal is the same as the final temperature of the water in DCI16.4. What is the specific heat for iron? The metal is absorbing 2.49 x 103 J of heat q = mass Fe SH Fe T Fe q(heat) = mass SH T A piece of iron weighing 80.0 g initially at a temperature of 92.6 C released the same amount of heat to the 175 g sample of water in DCI16.4. Assume the final temperature of the metal is the same as the final temperature of the water in DCI16.4. What is the specific heat for iron? The metal is absorbing 2.49 x 103 J of heat q = mass Fe SH Fe T fe 2.49 x 10 3 J = 80.0 g SH Fe (26.85 C - 92.6 C) SH Fe = 2.49 x 10 3 J /(80.0 g (-65.75 C)) SH Fe = 0.473 J g -1 C -1 12
q(heat) = mass SH T The four pictures shown below summarize an experiment. A zinc cylinder of mass 57.968 g was placed in boiling water at 100 C then plunged into a beaker containing 169.340 g of water at 24.64 C. The temperature of the water and zinc cylinder finally levels off at 26.91 C. Calculate the specific heat of zinc metal. Check out the movie Check out the movie q(heat) = mass SH T The four pictures shown below summarize an experiment. A zinc cylinder of mass 57.968 g was placed in boiling water at 100 C then plunged into a beaker containing 169.340 g of water at 24.64 C. The temperature of the water and zinc cylinder finally levels off at 26.91 C. Calculate the specific heat of zinc metal. q metal = q water q water = mass water SH water T water q water = 169.340 g 4.814 J g-1 C-1 (26.91 24.64 ) q water = 1608. J q metal = 1608. J mass metal SH metal T metal = 1608. J 57.968. g SH metal (26.91 100.0 ) = 1608. J SH = 1608 J 57.968 g 73.09 C SH metal = 0.380 J g-1 C-1 13
IMPORTANT RELATIONSHIPS q = mass*sh* T (can use q hot, q cold, q metal, q rxn ) q hot = -q cold q metal = -q cold Coffee Cup Calorimeter q (lost by reaction) = -q (gained by water) q (lost by reaction) = -q (H 2 O) = m SH (H 2 O) t (H 2 O) 14
A 100.00 ml sample of 0.200 M CsOH is mixed with 100. ml of 0.200 M HCl in an OSU calorimeter the following reaction occurs CsOH(aq) + HCl(aq) CsCl(aq) + H 2 O(l) The temperature before mixing of both solutions is 24.30 C. After mixing the final temperature is 25.6 C. The heat capacity of the calorimeter is 50. J C -1 and the specific heat of the solution is 4.20 J g -1 C -1. Calculate the heat released in the reaction. A 100.00 ml sample of 0.200 M CsOH is mixed with 100. ml of 0.200 M HCl in an OSU calorimeter the following reaction occurs CsOH(aq) + HCl(aq) CsCl(aq) + H 2 O(l) The temperature before mixing of both solutions is 24.30 C. After mixing the final temperature is 25.6 C. The heat capacity of the calorimeter is 50. J C -1 and the specific heat of the solution is 4.184 J g -1 C -1. Calculate the heat released in the reaction. (Assume the density of the solution is 1.00 g ml -1 ) q rxn = -(q solution + q calorimeter ) 15
A 100.00 ml sample of 0.200 M CsOH is mixed with 100. ml of 0.200 M HCl in an OSU calorimeter the following reaction occurs CsOH(aq) + HCl(aq) CsCl(aq) + H 2 O(l) The temperature before mixing of both solutions is 24.30 C. After mixing the final temperature is 25.6 C. The heat capacity of the calorimeter is 50. J C -1 and the specific heat of the solution is 4.184 J g -1 C -1. Calculate the heat released in the reaction. (Assume the density of the solution is 1.00 g ml -1 ) q rxn = -(q solution + q calorimeter ) q rxn = -(mass soln *SH soln * T solution + HC cal * T calorimeter ) A 100.00 ml sample of 0.200 M CsOH is mixed with 100. ml of 0.200 M HCl in an OSU calorimeter the following reaction occurs CsOH(aq) + HCl(aq) CsCl(aq) + H 2 O(l) The temperature before mixing of both solutions is 24.30 C. After mixing the final temperature is 25.6 C. The heat capacity of the calorimeter is 50. J C -1 and the specific heat of the solution is 4.184 J g -1 C -1. Calculate the heat released in the reaction. (Assume the density of the solution is 1.00 g ml -1 ) q rxn = -(q solution + q calorimeter ) q rxn = -(200.g* 4.184 J g -1 C -1 *(25.6 24.30) + 50. J C-1 * (25.6 24.30) ) q rxn = -(200.g* 4.184 J g -1 C -1 *(1.3 C) + 50. J C-1 * (1.3 C) ) q rxn = -1153 J q rxn = -1153 J/0.02 mol = -57.6 kj/mol 16
A 0.692 g sample of glucose, C 6 H 12 O 6, is burned in a constant volume bomb calorimeter and the following reaction occurs C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(l) The temperature change of the water and the calorimeter is 1.80 C. The calorimeter contains 1.05 kg of water and the dry calorimeter has a heat capacity of 650 J C -1 and the specific heat of the water is 4.184 J g -1 C -1. Calculate the heat released in the reaction. A 0.692 g sample of glucose, C 6 H 12 O 6, is burned in a constant volume bomb calorimeter and the following reaction occurs C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(l) The temperature change of the water and the calorimeter is 1.80 C. The calorimeter contains 1.05 kg of water and the dry calorimeter has a heat capacity of 650 J C -1 and the specific heat of the water is 4.184 J g -1 C -1. Calculate the heat released in the reaction. q rxn = -(q water + q calorimeter ) 17
A 0.692 g sample of glucose, C 6 H 12 O 6, is burned in a constant volume bomb calorimeter and the following reaction occurs C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(l) The temperature change of the water and the calorimeter is 1.80 C. The calorimeter contains 1.05 kg of water and the dry calorimeter has a heat capacity of 650 J C -1 and the specific heat of the water is 4.184 J g -1 C -1. Calculate the heat released in the reaction. q rxn = -(q water + q calorimeter ) q rxn = -(1050.g* 4.184 J g -1 C -1 *(1.8 C) + 650. J C-1 * (1.8 C) ) q rxn = 9078 J q rxn = -9078 J/0.692 g C 6 H 12 O 6 *(180 g C 6 H 12 O 6 /1 mol C 6 H 12 O 6 ) = -2361 kj/ mol IMPORTANT RELATIONSHIPS q = mass*sh* T (can use q hot, q cold, q metal, q rxn ) q hot = -q cold q hot = -(q cold + q calorimeter ) OSU calorimeter q metal = -q cold q metal = -(q cold + q calorimeter ) OSU calorimeter q rxn = -(q solution + q calorimeter ) OSU calorimeter q rxn = -(q water + q calorimeter ) bomb calorimeter 18