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Thermal Physics

Mr. Klapholz Shaker Heights High School. Thermal Physics. Problem Solving. Problem 1. If the thermal capacity of a certain amount of water is 5000 J C -1 , then how much heat is needed to raise its temperature from 20 ˚C to 100 ˚C?. Solution 1. C = Q / D T Q = C • D T

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Thermal Physics

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  1. Mr. Klapholz Shaker Heights High School Thermal Physics Problem Solving

  2. Problem 1 If the thermal capacity of a certain amount of water is 5000 J C-1, then how much heat is needed to raise its temperature from 20 ˚C to 100 ˚C?

  3. Solution 1 C = Q / DT Q = C • DT Q = (5000 J C-1 ) (100˚C – 20˚C) =400,000 J = 400 kJ

  4. Problem 2 How much heat is lost from a block of metal of thermal capacity 800 J ˚C-1 when it cools from 60 ˚C to 20 ˚C?

  5. Solution 2 C = Q / DT Q = C • DT Q = (800 J C-1 ) (20˚C – 60˚C) = -32,000 J = -32 kJ Energy lost by a system is negative.

  6. Problem 3 How much heat do you need to add to 300 g of water to heat it from 10 ˚C to 50 ˚C? The specific heat of water is 4200 J kg-1 ˚C-1.

  7. Solution 3 Q = mc DT Q ={0.300 kg} {4200 J kg-1 ˚C-1} {50 ˚C - 10 ˚C} Q ={300} {4200} {40} Joules = 50,400 J = 50.4 kJ

  8. Problem 4 When the temperature of a brick (1.5 kg) drops from 61 ˚C to 46˚C, the brick loses 20 kJ of heat. What is the specific heat capacity of the brick?

  9. Solution 4 Q = mc DT c = Q / { mDT } c ={20,000 J} / {1.5 kg • 15 ˚C} c =888.9 J kg-1 ˚C-1

  10. Problem 5 The specific latent heat of fusion of water is 3.35 x 105 J kg-1. What is the minimum amount of energy is required to change 500 g of ice into water?

  11. Solution 5 Note: to find the minimum amount of energy to turn ice into liquid, only calculate the energy required to melt the ice, not the energy required to warm the ice up to the melting temperature. Q = LF m Q = {3.35 x 105 J kg-1} {0.500 kg} c =1.675 x 105 J c =167.5 kJ

  12. Problem 6 When 100 g of steam turns to liquid water, 2.27 x 105 J of energy is released. What is the specific latent heat of vaporization of water?

  13. Solution 6 Q = LV m LV = Q / m LV = {2.27 x 105 J } / {0.100 kg} LV =2.27 x 106 J kg-1

  14. Über Problem 7a What is the power being delivered to the water in the first segment (bold) of the graph? The graph shows Temperature vs. Time for 1 kg of water.

  15. T / ˚C 100 60 20 960 480 240 t / s

  16. Solution 7a Power is the rate at which energy changes form. P = Energy ÷ Time So, for the bold segment, how much energy is delivered in how much time? The time is 240 seconds. The energy delivered to the water is Q = mcDT Q = (1 kg) (4200 J kg-1 ˚C-1) (60 ˚C – 20 ˚C) Q = 168000 J P = 168000 ÷ 240 = 700 Watts = 700 J/s

  17. Über Problem 7b Starting from the when the water begins to boil, and continuing to the end of the graph, how much energy went into the water during boiling?

  18. Solution 7b The graph shows that the water boils at 480 s, and continues to 960 s. That’s a durations of 480 s. P = E / t E = P•t Energy = (700 J s-1) (480 s) E = 3.36 x 105 Joules E = 336 kJ

  19. Über Problem 7c During boiling, how much water turned to steam?

  20. Solution 7c The energy required to vaporize a mass is: Q = Lm m = Q ÷ L m = {3.36 x 105 Joules} ÷ {2.27 x 106 J kg-1} m = 0.15 kg

  21. Problem 8 A solar heater is warming some water (2 kg). The water starts out at 10 ˚C. The heater collects energy at the rate of 10 J s-1 for two hours. Set up an equation that you could use to find the final temperature of the water.

  22. Solution 8 2 hour = 7200 seconds Q = (10 J s-1) (7200 s) = 72000 J Also, Q = mcDT 7200 J = (2 kg) (4.18 x 103 J Kg-1 ˚C-1) (T2 – 10 ˚C) From here you could solve for the final temperature.

  23. Problem 10 (The Law of Mixing) This important problem shows how specific heat capacity can be measured. Put a chunk of metal (0.1 kg) in boiling water for a while, and you’ll know that the metal is at 100 ˚C. Then put the chunk into 0.4 kg of water at 10 ˚C and wait for the temperature of the water to stop rising. The equilibrium temperature is 15 ˚C. What is the heat capacity of the metal?

  24. Solution 10 The heat lost by the metal = the heat gained by the water. Heat lost by chunk: -mcDT = -(0.1) c (15 - 100) = +(8.5 kg ˚C)c Heat gained by water: mcDT = (0.4) (4180) (15 - 10) = 8360 J (8.5 kg ˚C)c = 8360 J c = 983 J kg-1 ˚C-1 Note: Use this method anytime a hot thing and a cold thing are put in contact. Any of the variables (m, c, T) could have been the unknown.

  25. Tonight’s HW: Go through the Thermal Physics and Thermodynamics sections in your textbook and scrutinize the “Example Questions” and solutions. Bring in your questions to tomorrow’s class.

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