1. Worksheet Chapter 3 - 3 Key

2. 1. Which substance must have more energy removed from it to undergo a ten degree temperature drop? One with a low specific heat or one with a high specific heat?

4. 2. A student wants to determine the amount of energy he will need to add to a beaker of water in order to increase the temperature of the water by 15°C. What else must he measure before he can determine the amount of energy required?

5. He will need to measure the mass of the water. Specific heat is needed but can be looked up rather than measured.

6. 3. How much energy, in J and kJ, is required to raise the temperature of 123.4g of aluminum metal from 3.0 º C to 31.6 º C? (C p of Al = 0.899 J/g º C)

7. q = m∆TC p

8. 3. How much energy, in J and kJ, is required to raise the temperature of 123.4g of aluminum metal from 3.0 º C to 31.6 º C? (C p of Al = 0.899 J/g º C) q = m∆TC p = (123.4g)(28.6ºC)(0.899 J/g∙ºC) = 3172.786 J

9. 3. How much energy, in J and kJ, is required to raise the temperature of 123.4g of aluminum metal from 3.0 º C to 31.6 º C? (C p of Al = 0.899 J/g º C) q = m∆TC p = (123.4g)(28.6ºC)(0.899 J/g∙ºC) = 3172.786 J ≈ 3170 J

10. 3. How much energy, in J and kJ, is required to raise the temperature of 123.4g of aluminum metal from 3.0 º C to 31.6 º C? (C p of Al = 0.899 J/g º C) q = m∆TC p = (123.4g)(28.6ºC)(0.899 J/g∙ºC) = 3172.786 J ≈ 3170 J ≈ 3.17 kJ

11. 4. How much energy, in J and kJ, is released when 890.6 g of iron metal is cooled from 456ºC to 22ºC? (C p Fe = 0.444 J/g ºC)

12. q = m∆TC p

13. 4. How much energy, in J and kJ, is released when 890.6 g of iron metal is cooled from 456ºC to 22ºC? (C p Fe = 0.444 J/g ºC) q = m∆TC p = (890.6g)(434ºC)(0.444 J/g∙ºC) = 171,615 J

14. 4. How much energy, in J and kJ, is released when 890.6 g of iron metal is cooled from 456ºC to 22ºC? (C p Fe = 0.444 J/g ºC) q = m∆TC p = (890.6g)(434ºC)(0.444 J/g∙ºC) = 171,615 J ≈ 172,000 J

15. 4. How much energy, in J and kJ, is released when 890.6 g of iron metal is cooled from 456ºC to 22ºC? (C p Fe = 0.444 J/g ºC) q = m∆TC p = (890.6g)(434ºC)(0.444 J/g∙ºC) = 171,615 J ≈ 172,000 J ≈ 172 kJ

16. 5. What increase in temperature will result if 212.0g of copper absorbs 4.08kJ of heat energy? (C p Cu = 0.385 J/g ºC) ∆T = 4080J ÷ [(212.0g) (0.385 J/g∙ºC)] = 49.9877ºC ≈ 50.0 ºC q = m∆TC p ∆T = q ÷ [(m) (C p )] 4.08 kJ = 4080 J

17. 6. An 83.7g sample of metal absorbs 483 J of energy when the temperature increases from 13.8 ºC to 26.8 ºC. What is the specific heat of metal?

18. q = m∆TC p

19. 6. An 83.7g sample of metal absorbs 483 J of energy when the temperature increases from 13.8 ºC to 26.8 ºC. What is the specific heat of metal? q = m∆TC p C p = q ÷ [(m) (∆T)]

20. 6. An 83.7g sample of metal absorbs 483 J of energy when the temperature increases from 13.8 ºC to 26.8 ºC. What is the specific heat of metal? q = m∆TC p C p = q ÷ [(m) (∆T)] C p = q ÷ [(m) (∆T)] = 483 J ÷ [(83.7g) (13.0ºC)] = 0.443893 J/g∙ºC ≈ 0.444 J/g∙ºC

21. 7. A piece of unknown metal with a mass of 23.8g is heated to 100.0°C and dropped into 50.0ml (50.0g) of water at 24.0°C. The temperature of the water rises to 32.5°C. What is the specific heat of the metal? Water Metal q = q m∆TC p = m∆TC p (50.0g)(8.5ºC)(4.184 J/g∙ºC) = (23.8g)(67.5ºC)(C p ) 1778.2 J = (23.8g)(67.5ºC)(C p ) Cp = 1778.2 J ÷ [(23.8g) (67.5ºC)] Cp = 1.106878 J/g∙ºC Cp ≈ 1.11 J/g∙ºC

22. 8. A blacksmith heats an iron bar to 1445°C. The blacksmith then places the iron into 42800ml of water at 22°C. The temperature of the water increases to 45°C. What is the mass of the iron bar? Water Iron q = q m∆TC p = m∆TC p (42800g)(23ºC)(4.184 J/g∙ºC) = m(1400.ºC)(0.444 J/g∙ºC) 4118729 J = m(1400.ºC)(0.444 J/g∙ºC) m = 4118729 J ÷ [(1400.ºC)(0.444J/g∙ºC)] m = 6626g m ≈ 6600g

23. 9. A geologist at a mining company is trying to identify a metal sample. She takes 5.05g of the unknown metal and heats it to 102.3°C. She then drops the hot metal into 13.2ml of water at 22.1°C. The temperature of the water increase to 23.8°C. What is the specific heat of the metal? Water Metal q = q m∆TC p = m∆TC p (13.2g)(1.7ºC)(4.184 J/g∙ºC) = (5.05g)(78.5ºC)(C p ) 93.8896 J = (5.05g)(78.5ºC)(C p ) C p = 93.8896 J ÷ [(5.05g) (78.5ºC)] C p = 0.23684 J/g∙ºC C p ≈ 0.237 J/g∙ºC

24. 10.

25. __________________________ ll