Practice Quiz Solutions Thermal Energy Practice Quiz Solutions

What is thermal energy?

What is thermal energy? Thermal energy is the energy that comes from heat. This heat is generated by the movement of tiny particles within an object. The faster these particles move, the more heat is generated. Stoves and matches are examples of objects that conduct thermal energy.

How is thermal energy related to the concept temperature?

How is thermal energy related to the concept temperature How is thermal energy related to the concept temperature? As thermal energy is added, the particles move quicker and there is an increase in kinetic energy which results in an increase in temperature.

How is the concept of temperature related to kinetic energy?

How is the concept of temperature related to kinetic energy How is the concept of temperature related to kinetic energy? As thermal energy is added, the particles move quicker and there is an increase in kinetic energy.

Explain the three ways heat can be transferred.

Explain the three ways heat can be transferred Explain the three ways heat can be transferred. Conduction occurs when thermal energy is transferred through the interaction of solid particles. This process often occurs when cooking: the boiling of water in a metal pan causes the metal pan to warm as well. Convection usually takes place in gases or liquids (whereas conduction most often takes place in solids) in which the transfer of thermal energy is based on differences in heat. Using the example of the boiling pot of water, convection occurs as the bubbles rise to the surface and, in doing so, transfer heat from the bottom to the top. Radiation is the transfer of thermal energy through space and is responsible for the sunlight that fuels the earth.

5. What happens when heat energy is added or removed from a substance?

5. What happens when heat energy is added or removed from a substance 5. What happens when heat energy is added or removed from a substance? There is a temperature change (positive/negative). A change in temperature “equilibrium”. Phase change

6. Explain the phase diagram in detail.

6. Explain the phase diagram in detail. Temperature increase or decrease on the “slopes”. The temperature stays the same on the plateaus. Phase change Increase in heat from left to right (warmer) Decrease in heat from right to left (colder)

boiling freezing Heat of fusion Heat of vaporization

8a. Write down the equation for thermal energy during a temperature change equation and explain each variable in the equation. (be sure to include units)

8a. Write down the equation for thermal energy during a temperature change equation and explain each variable in the equation. (be sure to include units) Temperature change: Q = m Cp DT heat transfer = mass x (specific heat) x change in temperature kg x J/(kg K) x o C/o K

8b. Write down the equation for thermal energy during a phase change equation and explain each variable in the equation. (be sure to include units)

8b. Write down the equation for thermal energy during a phase change equation and explain each variable in the equation. (be sure to include units) Phase change: Q = m L mass x Latent heat kg x kJ/kg

9. Compute the amount of thermal energy gained during the warming of 3 kg of water from 10o C to 95o C.

9. Compute the amount of thermal energy gained during the warming of 3 kg of water from 10o C to 95o C. Q = m Cp DT Q = (3 kg)(4186 J/(kg K))(95 – 10) = 1067430 J = 1067.43 kJ

10. Compute the amount of thermal energy lost during the cooling of 3 kg of aluminum from 125o C to 15o C.

10. Compute the amount of thermal energy lost during the cooling of 3 kg of aluminum from 125o C to 15o C. Q = m Cp DT Q = (3 kg)(900 J/(kg K))(15 – 125) Q = -297000 J

11. Compute the amount of thermal energy needed to melt 40 kg of gold.

11. Compute the amount of thermal energy needed to melt 40 kg of gold 11. Compute the amount of thermal energy needed to melt 40 kg of gold. Q = mL Q = (40 kg)(64.5 kJ/kg) Q = 2580 kJ

12. Compute the amount of thermal energy needed to vaporize 15 kg of copper.

12. Compute the amount of thermal energy needed to vaporize 15 kg of copper. Q = mL Q = (15 kg)(5065 kJ/kg) Q = 75975 kJ

13. If 91200 J of thermal energy is required to raise the temperature of 1500 g an unknown substance from 80o C to 240o C, determine the substance in question.

13. If 91200 J of thermal energy is required to raise the temperature of 1500 g an unknown substance from 80o C to 240o C, determine the substance in question. Q = m Cp DT 91200 = (1.5 kg)CP (240o C – 80o C) CP = 380 J/(kg K) Brass or Zinc

14. If 20930 kJ is needed to evaporate 2kg of an unknown substance, what is the substance?

14. If 20930 kJ is needed to evaporate 2kg of an unknown substance, what is the substance? Q = mL 20930 KJ = (2 kg)(L) L = 10465 kJ Aluminum

Cooling mercury vapor from 400o C to 357o C. Q = mCP Dt = (0 Cooling mercury vapor from 400o C to 357o C. Q = mCP Dt = (0.5 kg)(140 J/(kg K))(357o C – 400o C) = -3010 J Condensing the mercury at 357o C Q = mL = (0.5 kg)(293000 J/kg) = -146500 J

Cooling liquid mercury from 357o C to -39o C Q = mCP Dt = (0 Cooling liquid mercury from 357o C to -39o C Q = mCP Dt = (0.5 kg)(140 J/(kg K))(-39o C – 357o C) = -27720 J Freezing the mercury Q = mL = (0.5 kg)(11300 J/kg) = -5650 J

Cooling solid mercury from -39o C to -60o C Q = mCP Dt = (0 Cooling solid mercury from -39o C to -60o C Q = mCP Dt = (0.5 kg)(140 J/(kg K))(-60o C – (-39)o C) = -1470 J Total = -184350 J

1. Determine the final temperature when a 25. 0 g piece of iron at 85 1. Determine the final temperature when a 25.0 g piece of iron at 85.0 °C is placed into 75.0 grams of water at 20.0 °C

1. Determine the final temperature when a 25. 0 g piece of iron at 85 1. Determine the final temperature when a 25.0 g piece of iron at 85.0 °C is placed into 75.0 grams of water at 20.0 °C. Qgain = Qlost mCP Dt = mCP Dt (0.075 kg)(4186 J/(kg K)(Tf – 20o C) = (0.025 kg)(450J/(kg K))(85o C – Tf) (313.95 )(Tf – 20o C) = 11.25 (85o C – Tf) 313.95Tf – 6279 = 956.25 – 11.25Tf 325.2Tf – 6279 = 956.25 325.2Tf = 7235.25 Tf = 22.3o C

2. Determine the final temperature when 10. 0 g of aluminum at 130 2. Determine the final temperature when 10.0 g of aluminum at 130.0 °C mixes with 200.0 grams of water at 25.0 °C.

2. Determine the final temperature when 10. 0 g of aluminum at 130 2. Determine the final temperature when 10.0 g of aluminum at 130.0 °C mixes with 200.0 grams of water at 25.0 °C. Qgain = Qlost mCP Dt = mCP Dt (0.2 kg)(4186 J/(kg K)(Tf – 25o C) = (0.01 kg)(900 J/(kg K)(130o C – Tf) (837.2 )(Tf – 25o C) = 9 (130o C – Tf) 837.2Tf – 20930 = 1170 – 9Tf 846.2Tf – 20930 = 1170 846.2Tf = 22100 Tf = 26.1o C

3. Determine the final temperature when 32. 2 g of water at 14 3. Determine the final temperature when 32.2 g of water at 14.9 °C mixes with 32.2 grams of water at 46.8 °C.

3. Determine the final temperature when 32. 2 g of water at 14 3. Determine the final temperature when 32.2 g of water at 14.9 °C mixes with 32.2 grams of water at 46.8 °C. Since we have the same materials and the same amount, we can simply find the average of the two temperatures. (14.9 + 46.8)/2 =30.85o C

4. 84.0g of a metal are heated to 112ºC, and then placed in a coffee cup calorimeter containing 60.0g of water at 32ºC. The final temperature in the calorimeter is 42ºC. What is the specific heat of the metal?

4. 84.0g of a metal are heated to 112ºC, and then placed in a coffee cup calorimeter containing 60.0g of water at 32ºC. The final temperature in the calorimeter is 42ºC. What is the specific heat of the metal? Qgain = Qlost mCP Dt = mCP Dt (0.06 kg)(4186 J/(kg K)(42oC – 32oC) = (0.084 kg) CP (112o C – 42oC) 2511.6 = 5.88 CP 427 = CP (best case senario iron/steel if asked)

5. 300 grams of ethanol at 10 °C is heated with 14640 Joules of energy 5. 300 grams of ethanol at 10 °C is heated with 14640 Joules of energy. What is the final temperature of the ethanol? The specific heat of ethanol is 2440 J/kg·°C.

5. 300 grams of ethanol at 10 °C is heated with 14640 Joules of energy 5. 300 grams of ethanol at 10 °C is heated with 14640 Joules of energy. What is the final temperature of the ethanol? The specific heat of ethanol is 2440 J/kg·°C. Q = mCP Dt 14640 J = (0.3 kg)(2440 J/kg °C)(Tf – 10o C) 14640 J = 732 (Tf – 10o C) 20 = Tf – 10o C Tf = 30o C

6. 5 kg of ice is added to an 8 kg bucket of water at 35o C 6. 5 kg of ice is added to an 8 kg bucket of water at 35o C. Determine how much ice is melted by the water and what is the makeup of the final mixture? How much water at 35o C would we need to melt all of the ice?

6. 5 kg of ice is added to an 8 kg bucket of water at 35o C 6. 5 kg of ice is added to an 8 kg bucket of water at 35o C. Determine how much ice is melted by the water and what is the makeup of the final mixture? How much water at 35o C would we need to melt all of the ice? First: Determine how much heat is in the water by cooling it down to 0o C. Q = mCP Dt Q = (8 kg)(4186 J/(kg K)(0 – 35o C) Q = -1172080 J

6. 5 kg of ice is added to an 8 kg bucket of water at 35o C 6. 5 kg of ice is added to an 8 kg bucket of water at 35o C. Determine how much ice is melted by the water and what is the makeup of the final mixture? How much water at 35o C would we need to melt all of the ice? Second: Use the heat from the water to see how much ice we can melt. Q = mL 1172080 J = m (334000 J/kg) m = 3.5 kg leaving 1.5 kg (5 – 3.5) of ice left in the bucket with 11.5 kg (8 + 3.5) of water.

6. 5 kg of ice is added to an 8 kg bucket of water at 35o C 6. 5 kg of ice is added to an 8 kg bucket of water at 35o C. Determine how much ice is melted by the water and what is the makeup of the final mixture? How much water at 35o C would we need to melt all of the ice? To melt all of the ice we would need this much heat. Q = mL Q = (5 kg)(334000 J/kg) Q = 1670000 J And this is how much heat we have to have in the water to melt all of the ice.

6. 5 kg of ice is added to an 8 kg bucket of water at 35o C 6. 5 kg of ice is added to an 8 kg bucket of water at 35o C. Determine how much ice is melted by the water and what is the makeup of the final mixture? How much water at 35o C would we need to melt all of the ice? To melt all of the ice we would need this much heat. Q = mCP Dt 1670000 J = m(4180 J/(kg K))(35o C) m = 11.4 kg So if we put 5kg of ice into a bucket that had 11.4 kg of water at 35o C, we would melt all of the ice and the mixture would be 16.4 kg of water at 0o C.