Unit 9 Study Guide KEY

Laws of Thermodynamics

1. Can we reach absolute zero? No

2. Finish this statement: Energy can be neither created nor destroyed.

3. Heat energy travels from hot to cold.

4. If I put a hot piece of metal into a glass of water at room temperature, what happens?

Heat energy transfers from the metal to the water until both have the same final temperature.

Heat Quantities

5. A sample of iron was heated from 15°C to 25°C, what was the mass of the sample if it gained 3000 joules of heat?

3000 J = M x 0.449 Jg℃ x (25°C -15°C)

M = 668 grams

6. A sample of copper had a mass of 23 grams. If it lost 850 joules of heat, what was the initial temperature of the sample if it now has a temperature of 16°C?

850 J = 23 grams x 0.384 Jg℃ x ∆T

∆T = 96.2°C 96.2°C + 16°C = 112.2°C

Since the system lost heat, the initial temperature must be higher than the final temperature.

Calorimetry

7. A 207.3 gram sample of metal at 150°C was placed in 400 grams of water. If the water temperature changed from 18°C to 24°C, what was the metal?

Water CopperMass 400 g 207.3 gSH 4.18 0.384

Tinitial 18°C 150°CTfinal 24°C 24°C

Q = 400 grams x 4.18 Jg℃ x (24°C-18°C)

Q = 10032 J

10032 J = 207.3 grams x C x (150°C-24°C)

C = 0.384 COPPER

Heat 10032 J 10032 J8. A 716 gram sample of metal at 100°C was placed in 2000 grams of water. If the

water temperature changed from 19°C to 22°C, what was the metal?

Water IronMass 2000 g 716 gSH 4.18 0.449

Tinitial 19°C 100°CTfinal 22°C 22°CHeat 25080 J 25080 J

Heating Curves

9. If a block of ice has a mass of 20 grams, how much heat will it gain if it is heated from -12°C to 22°C?

Q = 20 g x 2.1 x 12°C = 504 J Q = 20 g x 334 J/g = 6680 J Q = 20 g x 4.18 x 22°C = 1839 J 9023 J

10. If a block of ice has a mass of 20 grams, how much heat will it gain if it is heated from -12°C to 122°C?

Q = 20 g x 2.1 x 12°C = 504 J Q = 20 g x 334 J/g = 6680 J Q = 20 g x 4.18 x 100°C = 8360 J Q = 20 g x 2259 J/g = 45180 J Q = 20 g x 1.7 x 22°C = 748 J

61472 J

Q = 2000 grams x 4.18 Jg℃ x (22°C-19°C)

Q = 25080 J

25080 J = 716 grams x C x (100°C-22°C)

C = 0.449 IRON