module 2: thermal physics - consumo publishers...equation (22) can be rearranged into the more...

Module 2: Thermal PhysicsAP Physics Sample module

Grades 11 and 12 Consumo Publishers © Author: Tyler Rodrigues

MODULE 2: THERMAL PHYSICS

Heat

The concept of heat is often confused with temperature, especially in everyday life, but in reality they are significantly different concepts. If two objects at different temperatures are brought into contact with each other, energy is transferred from the hotter object (higher temperature) to the colder object (lower temperature) until thermal equilibrium is reached (they get to the same temperature).

It is this transfer of energy that is called heat. Heat is a type of energy transfer, whereas temperature is not an energy at all (it is only an indicator/measure of energy).

Heat: a spontaneous transfer of energy due to a temperature difference. Symbol: 𝑄 Unit: J (joule)

NOTE: substances do not contain heat; they only contain internal energy. Heat is only used when there is a transfer of energy.

Heat interpreted at the molecular level What is observed as a change in temperature of two macroscopic objects in contact, consists of the transfer of kinetic energy from particles with greater kinetic energy (thus higher temperature) to those with lower kinetic energy (thus lower temperature). In this respect, the process can be viewed in terms of collisions. When the two substances are in thermal contact, the particles of both substances can collide with each other. The particle with a higher kinetic energy transfers energy to the particle with a lower kinetic energy. Thus, in the vast majority of the collisions, the particles in the higher temperature substance lose energy, thus lowering its temperature and the particles in the higher temperature substance gain energy, increasing its temperature until they reach thermal equilibrium. NOTE: some energy is also transferred from the lower temperature substance to the higher temperature substance; however, much more energy is transferred in the other direction. When thermal equilibrium is reached, the energy transfer in either direction is on average the same, so there is no longer any change in internal energy, thus the temperature of each substance remains constant.

Interface in thermal contact

High energy before collision

Low energy before collision

Higher temperature

Lower temperature

Heat transfer



The fallibility of our senses Imagine, on a cold winter’s day, you walk outside to find two chairs, one made of metal (e.g. stainless steel) and one made of wood. Which would you rather sit on? Most people would say the wooden one because it is not as cold. Whilst this might appear to be true (this is certainly what we feel!), what is surprising to most people is that both chairs are actually at the exact same temperature. Why, then, does the metal feel colder? When we touch the chair, our body is at a higher temperature than the chair, so heat will transfer from our bodies to the chair; however, the metal is a much better conductor of heat than the wood. This means that heat is transferred out of our body faster when we touch the metal, so the temperature of our body drops faster, and so we feel that it is colder.



Specific Heat Capacity When heat is added to a body, it causes the temperature to increase, due to an increase in the internal energy of the substance. This change in temperature is directly proportional to the amount of heat added, but the ratio of heat added to change in temperature is constant for a particular substance; we measure this ratio as the heat capacity of the substance:

𝐶 =𝑄

Δ𝑇 (21)

Heat capacity obviously depends on the amount of the substance, as more substance means a greater amount of heat is needed to increase the temperature by the same amount; thus, it is called an extensive property (it depends on the amount of substance). It is more practical and useful to use intensive properties, i.e. properties that do not depend on the amount of substance, so that we can apply it to any amount of substance. We thus add mass to equation (21) to get specific heat capacity, which is the amount of heat per unit mass required to raise the temperature of the substance.

𝑐 =𝑄

𝑚Δ𝑇 (22)

Equation (22) can be rearranged into the more common form below:

𝑄 = 𝑚𝑐Δ𝑇 (23) Where, 𝑄 is the amount of heat added (J) 𝑚 is the mass of the substance (kg)

𝑐 is the specific heat capacity (J·kg−1·°C−1 OR J·kg−1·K−1) Δ𝑇 is the change in temperature (°C or K)

Extensive property: a property of a substance that does depend on the amount of substance present (e.g. energy, mass, volume). Intensive property: a property of a substance that does not depend on the amount of substance present (e.g. temperature, melting/boiling point, pressure, concentration, molar mass).

Specific heat capacity: the amount of heat needed to raise the temperature of 1 gram of a substance by 1 °C (or 1 K). NOTE: specific heat capacity is sometimes simply referred to as specific heat.

NOTE: when the temperature decreases, Δ𝑇 will be negative, and so 𝑄 will also be negative, indicating a heat transfer OUT of the substance. When the temperature increases, Δ𝑇 will be positive, and so 𝑄 will also be positive, indicating a heat transfer INTO the substance:

𝑸 < 𝟎 heat transfers out of substance; temperature decreases 𝑸 < 𝟎 heat transfers into substance; temperature increases

NOTE: the SI unit for energy is the joule (J); however, in everyday life, we often use the unit calorie (1 cal = 4,186 J) and the food unit Calorie with a capital C (1 Cal = 1 kcal = 4,186 kJ).

NOTE: this relationship does NOT apply during phase changes, where the heat added is not being used to raise the temperature (see below).



The table below shows specific heat capacities of some common substances at 25 °C.

SUBSTANCE SPECIFIC HEAT CAPACITY

(J·kg−1·K−1) Bismuth 123

Gold 129 Lead 127

Tungsten 134 Mercury 142

Silver 235 Sand 290 Brass 390

Copper 387 Zinc 387

Steel/Iron 452 Granite 790

Glass 840 Concrete 840

Aluminium 900 Air (at 50 °C) 1 050

Wood 1 680 Vegetable oil 2 000

Ethanol 2 430 Ice (−10 °C) 2 050

Water (at 15 °C) 4 186 Steam (at 110 °C) 2 100

NOTE: heat capacity is temperature dependent. This means that the heat capacity of a material differs at different temperatures. The values in this table were measured at 25 °C; if another temperature were used, these values would be different. However, we usually assume that heat capacities remain fairly constant over small temperature ranges, and from now on this is what we do.



Example 15 1. 4 190 J of heat are added to 0,500 kg of water with an initial temperature of 12,0 °C. What is the

temperature of the water after it has been heated? 2. 0,300 L of water is placed into a 0,600 kg aluminium pan. The pan is placed on a stove and heated from

25,0 °C to 75,0 °C. Take the density of water to be 1 g·cm−3.

a) How much heat is required to raise the temperature of both the pan and the water? b) What percentage of this heat is used to raise the temperature of the water?

3. Air has a density of about 𝜌 = 1,2 kg·m−3. How much heat, in joules, is needed to raise the temperature

of the air in a 3,4 m × 4,4 m × 6,0 m room by 5 °C? Solutions:



Phase Changes When there is a phase change from solid to liquid or from liquid to gas, particles are being separated and allowed to move more freely. This can only occur if some/all of the intermolecular forces between the particles, i.e. the forces holding them together, are overcome/broken. This breaking of the intermolecular forces requires energy input, so the heat added to the system is used to overcome the intermolecular forces and thus raise the potential energy of the system, and not to raise the temperature, i.e. raise the kinetic energy. So, during a phase change, although heat is continually added to the system, i.e. energy is continuously flowing into the system, this energy input increases the potential energy of the particles (moves them further apart), and the kinetic energy remains constant, thus the temperature remains constant.

Heating and Cooling Curves In Grade 10 Physical Sciences, you would have looked at very common plots of temperature versus heat added/removed (over time). These plots were called heating or cooling curves and an example of the heating curve for water is shown below as a reminder.

Solid Liquid Gas

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Te

mp

era

ture

(°C

)

Energy Added (kJ·kg−1)

0 3 2 1 −20

0

20

40

60

80

100

120

Ice

Water + Steam

Ice + Water

Water

Steam



Boiling versus Evaporation Both evaporation and boiling concern the phase change from a liquid to a gas, that is the vaporisation of a liquid. Evaporation occurs only from the surface of a liquid and can occur at a wide range of temperatures; however, boiling occurs when the entire bulk liquid is changing phase, not just the particles at the surface.

Practical examples of evaporation Water (or other liquids) will evaporated from an open container at any temperature. When it is

hotter, liquids evaporated faster, but they will still evaporate, even below freezing! Have you ever spilled some alcohol/methylated spirits/surgical spirits/acetone (nail polish

remover) on your hand? As the liquid evaporates, our hand starts to become cold. This is because the liquid particles are using kinetic energy from our hand particles to overcome its intermolecular forces, causing the temperature of our hand to decrease. This also happens when we get wet with water. As the water evaporates, it cools our skin; this just happens much slower than alcohol, etc. as water has a higher boiling point and requires much more energy to get into the vapour phase.

NOTE: the temperature of the liquid decreases during evaporation (not constant) as the kinetic energy of the particles is being used to overcome the intermolecular forces between the liquid particles in order to get them into the gas phase.

Evaporation: the vaporisation of liquid particles from the surface of the liquid, usually accompanied by a decrease in temperature. Boiling: the vaporisation of liquid particles from the bulk liquid, occurring only through the addition of heat, but with no change in temperature.



Specific Latent Heat When a substance changes phase, as discussed above, even though heat is constantly added, the temperature of the substance does not change. This is because the heat energy is being used to increase the potential energy of the particles, not the kinetic energy, due to the intermolecular forces being overcome/broken. The number of intermolecular forces that need to be broken as well as their strength determine the amount of heat that needs to be transferred in order to change the phase of the substance. The heat required to change a substance from the solid to the liquid phase is called the latent heat of fusion. This is the same energy that is released when that substance changes phase from liquid to solid. Latent heat is an extensive property, depending on the amount of substance (more substance = more heat required to change its phase), therefore we make it an intensive property by taking the heat per unit mass of that substance, and it thus becomes the specific latent heat of fusion, 𝐿𝑓 .

𝐿𝑓 =𝑄

𝑚 (24)

Rearranging equation (24) gives us the following:

𝑄 = 𝑚𝐿𝑓 (25) Where, 𝑄 is the amount of heat added to melt substance completely (J) 𝑚 is the mass of the substance (kg)

𝐿𝑓 is the specific latent heat of fusion (J·kg−1)

A similar concept is extended to the phase change from liquid to gas, and this gives us the specific latent heat of vaporisation, 𝐿𝑣.

𝑄 = 𝑚𝐿𝑣 (26) Where, 𝑄 is the amount of heat added to boil substance completely (J) 𝑚 is the mass of the substance (kg)

𝐿𝑣 is the specific latent heat of vaporisation (J·kg−1) Latent heats are also sometimes called latent heat coefficients or heats of transformation.

The SI unit of specific latent heat is J·kg−1; however, the energy involved in changing the phase of 1 kg of a substance is usually very high, so units of kJ·kg−1 are often used instead.

The word latent means ‘hidden’. These heats are called such because in phase changes, energy enters or leaves the system without any temperature change, so in effect, the energy is hidden.

NOTE: in general, 𝑳𝒗 > 𝑳𝒇. This means it is generally easier to melt a substance than to boil it.



The table below shows some representative values for 𝐿𝑓 and 𝐿𝑣 in units of kJ·kg−1, together with the substances

melting and boiling points. Note that this table shows that the energy involved in phase changes can easily be comparable to or greater than those involved in temperature changes.

SUBSTANCE MELTING

POINT (°C) 𝑳𝒇 (kJ·kg−1)

BOILING POINT (°C)

𝑳𝒗 (kJ·kg−1)

Hydrogen −259,3 58,6 −252,9 452 Nitrogen −210,0 25,5 −195,8 201 Oxygen −218,8 13,8 −183,0 213 Ethanol −114 104 78,3 854

Ammonia −75 332 −33,4 1 370 Mercury −38,9 11,8 357 272

Water 0,00 334 100,0 2 256 Sulfur 19 38,1 444,6 326 Lead 327 24,5 1 750 871

Aluminium 660 380 2 450 11 400 Silver 961 111 2 193 2 336 Gold 1 063 64,5 2 660 1 578

Copper 1 083 134 2 595 5 069 Uranium 1 133 84 3 900 1 900 Tungsten 3 410 184 5 900 4 810

Example 16 1. Determine the amount of heat needed to be transferred in order to raise the temperature of 243,5 g of

water at 33 °C to steam at 122 °C. 2. Liquid helium has a very low boiling point (4,2 K), as well as a low latent heat of vaporization

(2,09 × 104 J·kg−1). If energy is transferred to a container of liquid helium at the boiling point from an immersed electric heater at a rate of 10,0 W, how long does it take to boil away 2,00 kg of the liquid?

Solutions:

NOTE: unlike heat capacity, latent heat is not temperature dependent. This is because, these values are always measured at the phase transition temperature.



The First Law of Thermodynamics and Heat Transfer The First Law of Thermodynamics, also known as the Law of Conservation of Energy, is something that you are familiar with already, as you have used this law in both Physics and Chemistry in previous grades. In its simplest form, it states that energy can neither be created nor destroyed. For heat, this takes on a slightly different meaning than before, in that we are now transferring energy between objects and/or systems. Heat (𝑄) and work (𝑊 ) are two different ways of transferring energy between objects/systems. So, another way of expressing the first law of thermodynamics is that any change in internal energy (Δ𝑈) of a system is given by the sum of heat (𝑄) that flows across its boundaries and the work (𝑊) done on the system by its surroundings:

Δ𝑈 = 𝑄 + 𝑊 (27) Although we will occasionally do so, we will seldom look at energy entering or leaving our system, i.e. we will most commonly consider situations in which the total energy of our system remains constant. If that is true, then, the heat exchanges within the system must add up to zero. This is a technique known as calorimetry.

Σ𝑄 = 0 (28) The result of this is that the total energy lost by warmer objects will be equal to the total energy gained by cooler objects, until thermal equilibrium is reached. Mathematically, we can say:

𝑄ℎ𝑜𝑡 + 𝑄𝑐𝑜𝑙𝑑 = 0 (29)

System and Surroundings In order to avoid confusion, scientists discuss thermodynamic values in reference to a system and its surroundings. Everything that is not part of a system constitutes its surroundings. The system and surroundings are separated by a boundary.

SYSTEM

SURROUNDINGS BOUNDARY



Example 17 Three ice cubes are used to chill a soda, initially at 20 °C with mass 𝑚𝑠𝑜𝑑𝑎 = 0,25 kg. The ice is at 0 °C and each ice cube has a mass of 6,0 g. Assume that the soda is kept in a foam container so that heat loss to the environment can be ignored and that the soda has the same specific heat capacity as water. Find the final temperature when all ice has melted. Solution:



MODULE 2: THERMAL PHYSICS TUTORIAL For this tutorial, the following data will be useful. [NOTE: only if these values are used will you arrive at the answers given.]

Substance 𝜶

(K−1) 𝜷

(K−1) 𝒄

(J·kg−1·K−1) 𝑳𝒇

(J·kg−1) 𝑳𝒗

(J·kg−1) Density, 𝝆 (kg·m−3)

Steel/Iron 1,18 × 10−5 3,54 × 10−5 452

Aluminium 2,5 × 10−5 7,5 × 10−5 900 3,80 × 105 1,14 × 107

Brass 1,9 × 10−5 5,7 × 10−5 390 8 730 Copper 1,7 × 10−5 5,1 × 10−5 387 1,34 × 105 5,069 × 106

Mercury (liquid)

6,1 × 10−5 1,82 × 10−4 142 1,18 × 104 2,72 × 105 1,369 × 104

Mercury (solid)

390

Invar 1,6 × 10−6 4,8 × 10−6 Glass 8,5 × 10−6 25,5 × 10−6 840 Gold 1,43 × 10−5 129

Turpentine 9,72 × 10−4 Pyrex 9,6 × 10−6

Motor Oil 9,0 × 10−4 Water 6,9 × 10−5 2,07 × 10−4 4 186 3,34 × 105 2,256 × 106 1 000

Ethanol 1,10 × 10−3 2 430 Ice 2 050 917

Steam 2 100 Oxygen (at STP)

651 2,43 × 105 1,429

Lead 2,9 × 10−5 8,7 × 10−5 127 2,45 × 104 8,71 × 105

Silver 235 1,11 × 105 2,336 × 106 Wood 1 700

Concrete 1,45 × 10−5 4,35 × 10−5 840

Marble 7,1 × 10−5 2,11 × 10−4

1 cal = 4,186 J 1 kcal = 4,186 kJ 1 Cal = 1 kcal = 4,186 kJ



EXERCISE 3: HEAT 1. Two identical mugs contain hot chocolate from the same pot. One mug is full, while the other is only one-

quarter full. Sitting on the kitchen table, the full mug stays warmer longer. Explain. 2. How is heat transfer related to temperature?

3. When heat transfers into a system, is the energy stored as heat? Explain briefly.

4. What three factors affect the heat transfer that is necessary to change an object’s temperature?

5. Two materials – a floor tile and a floor rug – have been on the floor for some time. When you touch both, the

tile feels cold and the rug does not. Is the tile at a much lower temperature than the rug? Explain fully. 6. Why does a paper wrapping keep hot things hot and cold things cold?

7. Which quantity must be the same for two bodies if they are to be in thermal equilibrium? [Multiple Choice]

A. Internal energy. B. Potential energy. C. Temperature. D. Mass.

8. On a hot day, the temperature of an 80,0 kL swimming pool increases by 1,50 °C. What is the net heat transfer

during this heating? Ignore any complications, such as loss of water by evaporation. {𝟓, 𝟎𝟐 × 𝟏𝟎𝟖 J} 9. When we add 4,2 × 106 J of thermal energy to a block of steel, its temperature increases by 12 °C.

a) What is the mass of the above block? {770 kg} b) Suppose we now have 220 kg of this same kind of steel. How much heat must be added to raise its

temperature by 12 °C? {1,2 × 106 J}

10. A 520 g steel block is heated up to 95 °C and placed into water, which is originally at a temperature of 15 °C. The mass of the water is 350 g. The container the water is in is extremely light plastic and acts as a good insulator (it doesn’t absorb any of the heat). a) What is the final temperature of the combination, assuming no heat is lost to the container or the

surroundings? {26 °C} b) Which substance loses heat?

11. A pot of 10,0 kg of 15,0 °C water is put on a stove and brought to a boil. How much heat was needed for this?

{𝟑, 𝟓𝟔 × 𝟏𝟎𝟔 J} 12. What is the increase in temperature of a 5,0 g steel nail hit into a piece of wood by a hammer with force of

500 N? The length of the nail is 6,0 cm. [You are reminded that the work done on (energy transferred to) an object by a force is 𝑊 = 𝐹Δ𝑥.] {1,3 °C}

13. To sterilize a 50,0 g baby bottle made of glass, the temperature must be raised from 22,0 °C to 95,0 °C. How

much heat transfer is required? {𝟑, 𝟎𝟕 × 𝟏𝟎𝟑 J} 14. A 0,250 kg block of an unknown, pure material is heated from 20,0 °C to 65,0 °C by the addition of 4,35 kJ of

energy. Calculate its specific heat capacity and identify the substance of which it is most likely composed. {387 J·kg−1·K−1}



15. If the price of electrical energy is R2,85 per kWh, what is the cost of using electrical energy to heat the water in a swimming pool (12,0 m × 9,00 m × 1,5 m) from 15 to 27 °C? {R6 300}

16. 3,0 kg of gold at a temperature of 20 °C is placed into contact with 1,0 kg of copper at a temperature of 80 °C.

At what temperature do the two substances reach thermal equilibrium? {42,5 °C}

17. 1,0 kg of cold water at 5,0 °C is added to a container of 5,0 kg of hot water at 65 °C. What is the final temperature of the water when it arrives at thermal equilibrium? {55 °C}

18. A steel block of mass 0,60 kg is put into a vessel containing 5,65 kg of water at 7,2 °C. The equilibrium

temperature comes to 13,2 °C. Calculate the initial temperature of the steel block. Take the heat capacity of the container to be 20 J·K−1. {536,45 °C}

19. During a bout of flu, a 60 kg physicist ran a temperature of 39 °C. Assuming that the human body is mostly

water, how much heat (in kJ) is required to raise her body temperature to 39 °C? {500 kJ}

20. Suppose 200 g of water at 40 °C are mixed with 120 g water at 15 °C. Assuming no loss of heat to the surroundings, calculate the resulting temperature of the mixture. {31 °C}

21. A 100 g steel ball is heated for some time in a Bunsen burner flame. It is then rapidly plunged into 250 g of water, and causes a rise in temperature from 12 °C to 30 °C. Find the temperature of the ball as it was plunged into the water. Assume no water was converted into vapour. {447 °C}

22. Truck brakes are used to control speed on a downhill and do work, converting gravitational potential energy

into increased internal energy (higher temperature) of the brake material. This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck, i.e. prevents the truck from speeding up. Since the mass of the truck is much greater than the mass of the brake material absorbing energy, the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment; in other words, the brakes may overheat. Calculate the temperature increase of 10,0 kg of brake material with an average specific heat capacity of 800 J·kg−1·°C−1, if the material retains 10 % of the energy from a 10 000 kg truck descending 75,0 m in height at a constant speed. {91,9 °C}

23. Temperature and phase changes can be caused by heat transfer (i.e. heating and/or cooling). What else (besides heating and/or cooling) can cause these changes?

24. How much heat transfer (in kilocalories) is required to thaw a 0,450 kg package of frozen vegetables

originally at 0 °C if their specific latent heat of fusion is the same as that of water? {35,9 kcal} 25. How much heat is needed to transform a 1,0 kg block of ice at −5,0 °C to a puddle of water at 10,0 °C?

{𝟑, 𝟔𝟓 × 𝟏𝟎𝟓 J} 26. Suppose 100 g of water at 60,0 °C and 10,0 g of ice at −10,0 °C are mixed together. What is the final

temperature of the mixture? {47,3 °C} 27. A copper calorimeter having a mass of 24,40 g when empty is two-thirds filled with water at 21 °C, and the

total mass is found to be 130,56 g. Small chips of ice at 0 °C are added and the mixture is stirred. At one stage, when all the ice had just melted, the lowest temperature reached was found to be 10,5 °C and the total mass of the calorimeter and its contents was now 143,16 g. From these data, calculate the specific heat of fusion of ice. {𝟑, 𝟑𝟒 × 𝟏𝟎𝟓 J·kg−1}

28. An ice cube of mass 50,0 g is taken from a refrigerator where its temperature was −10,0 °C and is dropped into a glass of water at 0 °C. If no heat is exchanged with the surroundings (including the glass), how much water will freeze on to the cube? {3,07 g}



29. A glass tumbler contains 190 g of water at 21 °C. An ice cube at 0 °C of mass 27 g is added and the water is stirred. Calculate the temperature of the water when all the ice has melted, assuming that the heat exchanged with the surroundings (including the glass) is negligibly small. {4,0 °C}

30. An ice cube of mass 50,0 g is taken from a refrigerator where its temperature was −15 °C and is dropped into

glass of water (mass of glass 200 g; mass of water 200 g) at 25 °C. If the only heat exchanged with the surroundings causes 0,1 g of water vapour (originally at 25 °C) to condense on the surface of the glass, determine the final equilibrium temperature of the glass and water? {6,1 °C}

31. An ice cube of mass 100 g is taken from a refrigerator where its temperature was −20 °C and is dropped into

glass of water (mass of glass 300 g; mass of water 250 g) at 20 °C, causing the temperature to decrease to an equilibrium temperature of 10 °C. If the only heat exchanged with the surroundings causes water vapour (originally at 20 °C) to condense on the surface of the glass, determine the mass of water that condenses on the glass. {12 g}

32. 500 g of ice at −10 °C are placed in a vessel containing 1 litre of water at 20 °C. If the thermal capacity of the vessel and the heat exchange with the surroundings are both negligibly small, what will be the mass of the remaining ice after equilibrium between the ice and water has been established? {280 g}

33. A block of dry ice of mass 2,0 kg at its sublimation temperature is used for making ice. What is the maximum mass of water, initially at 20 °C, which can be frozen to ice at 0 °C? Assume the vaporized carbon dioxide does not exchange heat with the water. Take the specific latent heat of sublimation, 𝐿𝑠(𝑑𝑟𝑦 𝑖𝑐𝑒) to be 5,74 × 105 J·kg−1. {2,7 kg}

34. A 400 g mass of brass which has been heated to a uniform temperature of 500 °C is lowered into a vessel of thermal capacity 100 J·K−1 and which contains 1 litre of water at 25 °C. Some steam is generated locally and escapes into the atmosphere at 100 °C. The final temperature of the water and the brass is 35 °C. What mass of water has evaporated? Assume the heat losses to the surroundings by the vessel and the water are negligibly small. {12 g}

35. Calculate the heat (in kJ) required to convert 10,0 g of ice at −20,0 °C completely into steam at 130 °C at atmospheric pressure. {31,13 kJ}

36. A casting weighing 50 kg is taken from an annealing furnace where its temperature was 480 °C and plunged into a tank containing 360 kg of oil at a temperature of 27 °C. The final temperature is 38 °C. The specific heat capacity of the oil is 2 090 J·kg−1·K−1. What is the specific heat capacity of the casting? Neglect the heat capacity of the tank itself and any heat losses. {𝟑𝟕𝟎 J·kg−1·K−1}

37. A calorimeter of thermal capacity 25 J·K−1 contains 100 g of water at 0 °C. A 200 g copper cylinder and a 300 g lead cylinder, both at 100 °C, are placed in the calorimeter. Calculate the final temperature attained, assuming that heat losses are negligible. {21 °C}

38. Suppose we have a 200,0 kg steel ingot and a 200,0 kg wooden block, both at room temperature (20,00 °C). If we add 1,143 MJ of heat (about the amount of energy in a Snickers bar) to each what will the final temperature of each block be? {𝑻𝒇(𝐬𝐭𝐞𝐞𝐥) = 𝟑𝟐, 𝟔𝟒 °C; 𝑻𝒇(𝐰𝐨𝐨𝐝) = 𝟐𝟑, 𝟑𝟔𝟐 °C}

39. You have a glass of 1 kg lukewarm water (25 °C). To make it cold you put in some ice cubes (−5 °C). After an

equilibrium temperature is reached, there is a very small amount of ice left. What is the minimum mass of the ice cubes added? (Assume no heat is lost to the environment.) {50 g}



40. In 1986, a gargantuan iceberg broke away from the Ross Ice Shelf in Antarctica. It was approximately a rectangle 160 km long, 40,0 km wide, and 250 m thick.

a) What is the mass of this iceberg? {𝟏, 𝟒𝟕 × 𝟏𝟎𝟏𝟓 kg} b) How much heat transfer would be needed to melt it completely? {𝟒, 𝟗𝟏 × 𝟏𝟎𝟐𝟎 J} c) How many years would it take sunlight alone to melt this ice, if the ice absorbs an average of 100 W·m−2,

12,00 h per day? {48,6 y} 41. Gary has designed a 525 kg ice chair. How much heat must he remove from water at 0 °C to make the ice

chair (also at 0 °C)? {𝟏, 𝟕𝟓 × 𝟏𝟎𝟖 J} 42. A 100 g steel ball is heated for some time in a Bunsen burner flame. It is then rapidly plunged into 250 g of

water at 12 °C, causing 10 g of the water to suddenly boil, and the remaining water to rise in temperature to 35 °C. Find the temperature of the ball as it was plunged into the water. Assume no water was converted into vapour. {382 °C}

43. A small, hot steel block is placed in thermal contact with a larger cool steel block.

How will the temperature changes for the two blocks compare whilst coming to thermal equilibrium? Assume no energy is lost to the surroundings. [Multiple Choice] A. The temperature changes of the two blocks will be equal. B. The temperature of the small block will change the most. C. The temperature of the large block will change the most. D. There is not enough data to compare the temperature changes of the blocks.

44. A small block of metal is taken out of boiling water, where it had come to equilibrium at 100 °C, and placed

into a calorimeter containing water at 0 °C. The mass of the block is equal to the mass of cold water. The specific heat capacity of the metal is less than that of water. Ignoring any transfer of thermal energy to the container, the equilibrium temperature of the block and the water will be: [Multiple Choice] A. less than 50 °C B. 50 °C C. more than 50 °C D. 100 °C

hot cool



45. A substance is heated at a constant rate of energy transfer (i.e. the energy transferred per unit time is constant). A graph of its temperature against time is shown below. Which regions of the graph correspond to the substance existing in a mixture of two phases? [Multiple Choice] A. KL, MN and OP B. LM and NO C. All regions D. No regions

46. A substance is heated at a constant rate of energy transfer (i.e. the energy transferred per unit time is

constant). A graph of its temperature against time is shown below.

In which region of the graph is the specific heat capacity of the substance greatest? [Multiple Choice] A. KL B. LM C. MN D. OP

Te

mp

era

ture

Time

K

L M

N O

P

Te

mp

era

ture

Time

K

L M

N O

P



47. An AP Physics student determines the specific latent heat of fusion of ice at home as follows. She takes ice from the freezer, measures its mass and mixes it with a known mass of water in an insulating jug. She stirs until all the ice has melted and measures the final temperature of the mixture. She also measured the temperature in the freezer and initial temperature of the water. She records her measurements as follows:

Mass of ice used Initial temperature of ice

𝑚𝑖 𝑇𝑖

0,12 kg −12 °C

Initial mass of water Initial temperature of water

𝑚𝑤 𝑇𝑤

0,40 kg 22 °C

Final temperature of mixture 𝑇𝑓 15 °C

a) Set up the appropriate equation, representing energy transfers during the process of coming to thermal

equilibrium, that will enable her to solve for the specific latent heat, 𝐿𝑖 , of ice. Insert values into the equation from the data above, but do not solve the equation.

b) Explain the physical meaning of each energy transfer term in your equation (but not each symbol). c) State an assumption you have made about the experiment, in setting up your equation in a). d) Explain from the microscopic point of view, in terms of molecular behaviour, why the temperature of the

ice does not increase while it is melting. 48. In an experiment to find its specific latent heat of vaporisation, water is vaporised using an immersion heater

as shown. Two sources of error in this experiment are:

Error 1: water splashed out of the container; Error 2: vapour condensing on the inside of the heater and dripping back into the container.

What effect will each of these two experimental errors have on the calculated value for the specific latent heat? [Multiple Choice]

Error 1 Error 2 A. Decrease Decrease B. Decrease Increase C. Increase Decrease D. Increase Increase

Thermometer

Immersion heater

Stirrer

Polystyrene cup

module 2: thermal physics - consumo publishers...equation (22) can be rearranged into the more...

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