thermal energy - myp physicsmypphysics.weebly.com/uploads/2/8/5/5/28556033/thermal_energy_… ·...

Thermal EnergyPractice 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?

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?

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.

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?

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

fusionHeat 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.

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

= (0.5 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

= (0.5 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

= (0.5 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 85.0 °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 85.0 °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 130.0 °C mixes with 200.0 grams of water

at 25.0 °C.

2. Determine the final temperature when 10.0 g of

aluminum at 130.0 °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 14.9 °C mixes with 32.2 grams of water at 46.8 °C.

3. Determine the final temperature when 32.2 g of water

at 14.9 °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. 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. 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. 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. 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. 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. 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. 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.