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Heat Review

Understand heat transfer mechanisms through experiments and calculations, explore specific heat concept with interactive examples. Learn about different substances and their specific heat values. Master heat calculations.

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Heat Review

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  1. Heat Review A. When you touch ice, heat is transferred from 1) your hand to the ice 2) the ice to your hand B. When you drink a hot cup of coffee, heat is transferred from 1) your mouth to the coffee 2) the coffee to your mouth

  2. Heat Review A. When you touch ice, heat is transferred from 1) your hand to the ice B. When you drink a hot cup of coffee, heat is transferred from 2) the coffee to your mouth

  3. What is happening?

  4. What do you think? When you heat 200 g of water for 1 minute, the water temperature rises from 10°C to 18°C. 200 g 400 g If you heat 400 g of water at 10°C in the same pan with the same amount of heat for 1 minute, what would you expect the final temperature to be? 1) 10 °C 2) 14°C 3) 18°C

  5. Answer 2) 14°C Heating twice the mass of water using the same amount of heat will raise the temperature only half as much. (4 degrees) 200 g400 g

  6. Specific Heat • Why do some foods stay hot longer than others? • Why is the beach sand hot, but the water is cool on the same hot day?

  7. Specific Heat Some things heat up or cool down faster than others.

  8. Specific Heat This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

  9. Specific Heat Different substances have different capacities for storing energy It may take 20 minutes to heat water to 75°C. However, the same mass of aluminum might require 5 minutes and the same amount of copper may take only 2 minutes to reach the same temperature.

  10. Specific Heat Value Specific heat- is the amount of heat needed to raise the temperature of 1 g of a substance by 1°C cal/g°C J/g°C water 1.00 4.18 aluminum 0.22 0.90 copper 0.093 0.39 silver 0.057 0.24 gold 0.031 0.13

  11. Review A. A substance with a large specific heat 1) heats up quickly 2) heats up slowly B. When ocean water cools, the surrounding air 1) cools 2) warms 3) stays the same C. Sand in the desert is hot in the day, and cool at night. Sand must have a 1) high specific heat 2) low specific heat

  12. Answers A. A substance with a large specific heat 2) heats up slowly B. When ocean water cools, the surrounding air 2) warms C. Sand in the desert is hot in the day, and cool at night. Sand must have a 2) low specific heat

  13. Calculating Heat Heat = mass x temp. chang x specific heat q = grams x T x Sp. Ht.

  14. Heat Calculations How many cal are needed to raise the temperature of 120 g of water from 15°C to 75°C? 1) 1800 cal 2) 7200 cal 3) 9000 cal

  15. Heat Calculations How many cal are needed to raise the temperature of 120 g of water from 15°C to 75°C? 2) 7200 cal 120 g x (75°C - 15°C) x1.00 cal g°C

  16. Heat Calculations A hot-water bottle contains 750 g of water at 65°C. If the water cools to body temperature (37°C), how many calories of heat could be transferred to sore muscles? Heat = g x T x Sp. Ht. (H2O) 750 g x 28°C x 1.00 cal g°C = 21 000 cal

  17. The variation in temperature between night and day on Mars is 150 degrees Fahrenheit, but on Earth it isn’t…….Why?

  18. Guide to Calculations Using Specific Heat

  19. Sample Calculation for Heat A hot-water bottle contains 750 g of water at 65 °C. If the water cools to body temperature (37 °C), how many calories of heat could be transferred to sore muscles? STEP 1Given: 750 g of water cools from 65 °C to 37 °C SHwater = 1.00 cal/g °C Need: calories of heat transferred STEP 2Calculate the temperature change  T: 65 °C – 37 °C = 28 °C

  20. Sample Calculation for Heat STEP 3Write the heat equation: Heat (cal) = mass(g) x  T x SH STEP 4Substitute given values and solve for heat: 750 g x 28 °C x 1.00 cal g °C = 21 000 cal

  21. Learning Check How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 °C to 77.0 °C? 1) 20.2 kJ 2) 84.3 kJ 3) 105 kJ

  22. Solution 2) 84.3 kJ STEP 1Given: 325 g of water warms from 15.0 °C to 77.0 °C SHwater = 4.184 J/g °C 1 kJ = 1000 J Need:kilojoules of heat needed STEP 2 Calculate the temperature change  T: 77.0 °C – 15.0 °C = 62.0 °C STEP 3Write the heat equation: Heat (joules) = mass (g) x  T x SH

  23. Solution (continued) STEP 4Substitute given values and solve for heat: 325 g x 62.0 °C x 4.184 J x 1 kJ g °C 1000 J = 84.3 kJ (2)

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