1. Heat Transfer Problems

2. 11-7 – Case #1 Silver Spoon Steel Spoon Boiling Water Baths at 100 o C m = 10.0 g s = 0.233 J/ g o C s = 0.51 J/ g o C Ti = 25.0 o C

3. Graph of Temp v Time Time TempERATURETempERATURE 100 o C Silver silver steel 20 minutes 2) At 20 min, both metals have equilibrated to the temp. of surroundings.

4. 11-7 #3) Which spoon has more stored heat Q? 3) Steel. It takes 0.51 J to raise the temp. of 1 g of Steel by 1 o C. It takes only.23 J of energy to raise the temperature of 1 g of silver by 1 o C.

5. Analogy to Case #1: Is the same energy required to lift a bowling ball and ping ball to the same height? Different masses analogous to different specific heat capacity Height difference is analogous to ∆T Energy required to move each objective is analogous to heat added 10 pounds 1/1760 of a pound Arrows are analogous to force (energy) required to lift

6. 4) Which Spoon’s atom’s have a greater average kinetic energy? Temperature is directly proportional to average kinetic energy. BOTH ARE AT SAME TEMPERATURE THERE KINETIC ENERGIES ARE THE SAME.

7. 5) T i = 100 o C of spoons; placed in equal masses of cold water : T f = 25 o C Q = sm∆T Silver: Q = (0.233 J/g o C)(10.0 g) (25 o C – 100 o C) Q = 174.75 = -170 J (2 SF) Steel: Q = (0.51 J/g o C)(10.0 g) (25 o C – 100 o C) Q = 382.5 = -380 J (2 SF)

8. Case 2 – If two objects absorb the exact same amount of heat energy in joules, will they be at same final temperature? Q = sm∆T ; ∆T = Q/ sm; ∆T = T f – T i ; T f = ∆T + Ti Silver ∆T = 1000. J ( 0.23 J/ g o C)( 10.0 g) =434.7= 430 (2 SF) T f = ∆T + T i ; T f = 430 + 25 = 455 o C Steel: ∆T = 1000. J ( 0.51 J/g oC) (10.0 g) = 196 = 200 (2 SF) T f = ∆T + T i ; T f = 200 + 25 = 225 o C Different s values causes different ∆T

9. Analogy to Case #2: If the same energy is applied to the bowling ball and ping ball will they rise to the same height? Different masses analogous to different specific heat capacity Height difference is analogous to ∆T Energy required to move each objective is analogous to heat added 10 pounds 1/1760 of a pound = Force =

10. Heat Transfer Problem 100.0 g of water at a temperature of 22.4 o C is placed in a calorimeter. A 75.25 g sample of Al is removed from boiling water at 99.3 o C and placed in calorimeter. Final T = 32.9 o C Aluminum cylinder, 75.25 g Boiling water bath at 99.3 o C 100.0 g water at 22.4 o C Final Temp of system = 32.9 o C S of water = 4.18 J/g o C Link to heat transfer video

11. Calorimeter – an insulated container that does not allow heat to enter or leave

12. Data Table Variable/UnitsAluminumWater (in calorimeter) s (J/g o C) m (mass in grams) T f (Final Temp in oC) Ti (Initial Temp in oC) ∆T = (Tf – Ti) x 4.18 75.25 100.0 32.9 99.322.4 10.5 -66.4

13. Heat lost by Metal = Heat Gained by Water in Calorimeter - Q metal = + Q H 2 O - [sm∆T] metal = + [sm∆T] H 2 O - [s Al (75.25)(32.9-99.3)] = + [(4.18)(100.0)(32.9-22.4) - s Al (75.25)( - 66.4) = (418)(10.5) s Al (4996.6 ) = 4389 s Al = 4389/ (4996.6 ) = 0.87787 = s Al = 0.878 J/ g o C (3 SF)