Heat, State Change, and Calorimetry 3rd Class Edition 3 PDF

3rd Class Edition 3 • Part A 1 CHAPTER 8 Heat, State Change, and Calorimetry LEARNING OUTCOME When you complete this chapter, you should be able to: Explain terminology regarding heat, and perform calculations regarding heat during changes of state and calorimeter tests. LEARNING OBJECTIVES Here...

3rd Class Edition 3 • Part A 1 CHAPTER 8 Heat, State Change, and Calorimetry LEARNING OUTCOME When you complete this chapter, you should be able to: Explain terminology regarding heat, and perform calculations regarding heat during changes of state and calorimeter tests. LEARNING OBJECTIVES Here /s what you should be able to do when you complete each objective: 1. Define and explain internal energy, heat, specific heat, heat units, temperature, and the relationship between the different temperature scales. 2. Define sensible heat, and use the sensible heat equation to calculate the mass of a substance, the temperature change of a substance, and the amount of heat required to change the temperature of a substance, if no change of state occurs. Explain the changes of state, and define latent heat, latent heat of fusion, and latent heat of evaporation. Given start and end conditions, calculate the heat required to change the states of water and other substances. Determine the final temperatures and the original masses for mixtures of ice, water, steam, and other substances. Explain the working principle of a simple calohmeter, and use the calorimeter equation to ii'^^vs^. ..it$8ii ;.%:gs% determine specific heat and final temperature. m "^%^%iS€;'^^'. •.^*§>'^s~<S:',-:', ^^^ic;^ :^'^^f& :s^^^ l.'v%:l?1w:i^t^ -•^•w^^m 381 Heat, State Change, and Calorimetry • Chapter 8 f& INTRODUCTION TO HEAT, STATE CHANGE, AND CALORIMETRY A considerable amount of Power Engineering centres around the application or removal of heat. In a Power Engineering plant, heat is frequently used to increase or decrease temperature within a process or to induce a change of state in different fluids. Understanding how heat affects these processes requires knowledge of basic heat terminology, such as internal energy, specific heat, and latent heat. Calculations can be used to determine how much heat is required to affect changes of state, or to determine final temperatures ofmuctures. These calculations provide a greater understanding of processes that involve these two conditions. Calorimetry, which is used to determine the heating value of fuels, is especially important to understand. Fuels of differing heating values have a considerable impact on the performance of combustion systems. The term water equivalent, commonly referred to in calorimetry, is also useful. 3rd Class Edition 3 • Part A 1 383 ?& Chapter 8 • Heat, State Change, and Calorimetry OBJECTIVE 1 Define and explain internal energy, heat, specific heat, heat units, temperature, and the relationship between the different temperature scales. INTERNAL ENERGY Any substance, whether it is solid, liquid, or gas, is comprised of atoms and molecules, which have very small mass and are in a state of continuous motion (vibration). Because they are in motion, the atoms and molecules must have a certain velocity and, therefore, must have kinetic energy, since kinetic energy of any mass is determined by KE = ^ m v2; (m is mass; v is velocity). These same atoms and molecules also contain a form of potential energy (not to be confused with gravitational potential energy). Atoms and molecules are a certain distance apart and there is a cohesive force that tends to hold them within that distance. Holding them "together" requires potential energy, with the amount depending on the distance between the molecules and the cohesion force between them. This potential energy can be shown to equal the force times the distance between the molecules. The sum of the kinetic and potential energies within a body or substance is known as the internal energy of the body or substance. HEAT Heat is a form of energy, which, when supplied to a body, will increase the internal energy of the body. Conversely, if heat is removed from the body the internal energy will decrease. Heat cannot be stored; it can only be transferred. Adding heat increases both the kinetic and potential energy within the body. The velocity of molecules increases, thus increasing the kinetic energy. At the same time, adding heat to the body (that is, heating the body) causes it to expand, which forces the molecules further apart against their cohesive force, thus increasing the potential energy. Removing heat decreases both the kinetic and potential energy within the body. The velocity of molecules decreases, thus decreasing the kinetic energy. At the same time, removing heat (that is, cooling) from the body causes it to contract, which moves the molecules closer together under their cohesive force, thus decreasing the potential energy. Heat can also be referred to as an energy transfer process or energy in transition. It is a form of energy that may be transferred from one body to another, due to a difference in the temperature of the bodies. Heat energy will only flow directly from a body or substance of higher temperature (hotter) into a body or substance of lower temperature (colder). The internal energy of the hotter body will decrease, while the internal energy of the colder body will increase until they reach the same temperature (i.e. they reach thermal equilibrium). 384 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 UNITS OF HEAT The joule (J) is the basic unit of measurement for all forms of energy, including heat. Since the joule itself is a small unit it is common to use kilojoules, megajoules and gigajoules when considering large quantities. 1 kilojoule (kj) = 1000 Joules = 103 J 1 megajoule (MJ) = 1 000 000 Joules = 106 J 1 gigajoule (GJ) = 1 000 000 000 Joules = 109 J TEMPERATURE We commonly think of temperature as an indication of the degree of hotness or coldness in a body. A more accurate definition would be ca measure of the level, or intensity of internal energy." For example, consider a quantity of water with a large amount of internal energy (that is, the water is hot ). As the molecules move about at high velocity, they collide with each other very frequently and quite violently. If a mercury-in-glass thermometer is placed in the water, the water molecules collide with the glass molecules, which, in turn, begin to move more rapidly as energy is transferred to them. The glass molecules collide with each other and then with the mercury molecules inside the thermometer. This energy transfer to the mercury molecules causes them to move more randomly and the mercury expands up into the thermometer tube. Eventually, the temperature of the water, glass, and mercury reach the same level and energy transfer between them ceases. At that point the thermometer is indicating the temperature of the water, which is really an indication of the internal energy of the water. Temperature Scales There are four measurement scales used to define temperature. The most common, for everyday use, are the Fahrenheit and Celsius scales, while the Rankine and Kelvin scales are used largely for specific scientific, thermodynamic calculations. Each scale has its own importance and the relationships between the scales must be understood to successfully complete many calculations. Figure 1 shows the relationship between these scales and the key temperatures of consideration for each. Three temperatures are used as the key reference points for these scales. • Absolute zero • Freezing point (temperature) of water at atmospheric pressure • Boiling point (temperature) of water at atmospheric pressure 1. Kelvin The kelvin scale indicates temperature in kelvins (K). It is called an absolute scale because it begins at absolute zero, which is the temperature at which there is absolutely zero molecular motion. Water, at atmospheric pressure, will freeze at 273 K and will boil at 373 K. Widely used in metric system calculations, the kelvin (K) temperature of a substance is commonly referred to as its absolute temperature. 2. Celsius The Celsius scale indicates temperature in degrees Celsius (°C, formerly called Centigrade). It is a metric scale, not starting at absolute zero, in which one °C is exactly equal to one K. However, the Celsius scale specifies the freezing point of water as 0°C and the boiling point as 100°C. Absolute zero is equal to -273°C. 3rd Class Edition 3 • Part A1 385 ?&• Chapter 8 • Heat, State Change, and Calorimetry 3. Rankine The Rankine scale is the absolute scale in the Imperial and US Standard systems of measurement and indicates temperature in degrees Rankine (°R). It is based on absolute zero, which it calls 0°R, but establishes the freezing point of water as 492°R and the boiling point as 672°R. 4. Fahrenheit The Fahrenheit scale is used in the Imperial and US Standard systems to indicate temperature in degrees Fahrenheit (°F). It parallels the Rankine scale so that an increment of one °F is exactly equal to an increment of one °R. However, the Fahrenheit scale specifies the freezing point of water as 32°F and the boiling point as 212°F. Absolute zero is equal to -460°F. Figure 1 - Temperature Scales Celsius Kelvin Fahrenheit Rankine °C Boiling point (water) K 100 373 0 273 °F °R 212 672 Freezing point (water) Absolute zero -273 0 32 492 0_ 460 1-460 _( Converting Between Scales It will often be required to convert temperature from one unit to another, in order to perform calculations. The following conversion formulae apply. • Converting between Fahrenheit and Celsius, use: °F = 9/5°C+32 °C - 5/9 (°F-32) • Converting between Fahrenheit and Rankine, use: °R = °F+460 °F = °R-460 • Converting between Celsius and Kelvin (absolute), use: K = °C+273 °C = K-273 • Converting between Rankine and Kelvin, use: °R = 9/5 xK K = 5/9 x°R 386 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 ^ SPECIFIC HEAT The specific heat of a substance may be defined as the 'quantity of heat required to change the temperature of a unit mass of the substance by one degree." We can think of it as heat that transfers into or out of the substance, or heat that is absorbed or rejected by a unit mass of the substance, to cause the temperature to change by one degree. The base unit for specific heat is the J/g°C (joules per gram degree Celsius). Using larger units this can also be expressed as kJ/kg°C (kilojoules per kilogram degree Celsius). However, since one °C is equal to one K, specific heat is usually expressed as kJ/kgK. For example, the specific heat of water is given as 4.183 kJ/kgK. This means that 4.183 kj of heat must be added to 1 kg of water in order to raise its temperature by 1 K (or by 1°C). Likewise, to cool 1 kg of water by 1 K would require the removal of 4.183 kj of heat from the water. Specific Heat of Solids and Liquids The specific heat of a particular solid or liquid may vary slightly if temperature change is drastic. Most quoted specific heat values are, therefore, an average for that substance. Table 1 shows the specific heats of some common substances. Table 1 - Typical Specific Heats Substance Specific heat (kJ/kgK) aluminum 0.909 brass 0.383 carbon 0.795 copper 0.388 gold 0.130 ice 2.135 cast iron 0.544 silver 0.235 mild steel 0.494 ammonia (liquid) 4.73 mercury 0.139 gasoline 2.093 water 4.183 Specific Heat of Gases Heat may be transferred into or out of a gas under an infinite number of pressures and temperatures. The specific heat of a gas is very dependent on the conditions under which it is operating. Therefore, the value of the specific heat of a gas will vary with the conditions. 3rd Class Edition 3 • Part A1 387 Chapter 8 • Heat, State Change, and Calorimetry OBJECTIVE 2 Define sensible heat, and use the sensible heat equation to calculate the mass of a substance, the temperature change of a substance, and the amount of heat required to change the temperature of a substance, if no change of state occurs. SENSIBLE HEAT When heat is added to or removed from a substance, causing a change in temperature, but no change in state, then the heat is known as sensible heat. For example, only sensible heat is added to bring the temperature of ice from -10°C up to 0°C or the temperature of water from 0°C up to 100°C. Likewise, only sensible heat is removed to reduce the temperature of water from 100°C to 0°C or to reduce the temperature of ice from 0°C down to -10°C. Sensible Heat Equation The amount of sensible heat that is added to or removed from any solid or liquid, causing a temperature change, but no state change, depends on the following three factors: • the mass of the substance • the specific heat of the substance • the temperature change of the substance Given a particular heat transfer situation, the following formula can be applied to determine any of the unknown factors. Q = mc^-Ti) where Q = the heat added or removed (kj) m = mass of the substance (kg) c = specific heat of the substance (kJ/kg°C) TI = initial temperature of the substance (°C) T^ = final temperature of the substance (°C) Note: AT (pronounced "delta T") is often used to represent the change in temperature (T^ - Ti), in which case the formula may be written as: Q = mcAT Note: Occasionally temperatures T^ and Ti are quoted as K rather than °C. Realize that a change in K has the same numerical value as a change in °C. 388 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 Example 1 Calculate the heat added when the temperature of 6.3 kg of silver is increased from 18°C to 47°C. The specific heat of silver is 0.235 kJ/kgK. Solution 1 Q = mcdz-Ti) = 6.3 kg x 0.235 kJ/kgK x (47°C - 18°C) = 6.3 kg x 0.235 kJ/kgK x 29°C = 42.9345 kj (Ans.) Example 2 How much heat is lost by 20 kg of water if its temperature drops from 100°C to 15°C? Take the specific heat of water as 4.183 kJ/kgK. Solution 2 Q = mcAT = 20kgx4.183kJ/kgKx(100°C-15°C) = 7111kJ(Ans.) Example 3 Example 2 could be asked in a different way, as follows. What is the amount of heat transferred if 20 kg of water at 100°C is cooled to 15°C? Take the specific heat of water as 4.183 kJ/kgK. Solution 3 Q = mcOz-Ti) = 20kgx4.183kJ/kgKx(15°C-100°C) = -7111 kj (Ans.) the negative sign indicates heat is lost Example 4 How much heat is required to raise the temperature of 3 L of water from 18°C to 71°C? Assume 1 L of water has a mass of 1 kg. Solution 4 Q = mcA T = (3 L x 1 kg/L) x 4.183 kJ/kgK x (71°C - 18°C) = 3kgx4.183kJ/kgKx53°C = 665.1 kj (Ans.) 3rd Class Edition 3 • Part A1 389 Chapter 8 • Heat, State Change, and Calorimetry Example 5 The temperature of a block of ice is increased from -18°C to -5°C when 500 kj of heat are transferred to it. Calculate the mass of the ice, if specific heat for ice is 2.135 kJ/kgK. Solution 5 Q - m c (T2 - Ti) To transpose to isolate m: m= Q c(T2-Ti) 500 kj 2.135 kJ/kgK x (-5°C - (-18°C)) 500 kj 27.755 kj/kg 18.01 kg (Ans.) Example 6 A 2 tonne slab of cast iron, at a temperature of 14°C, has 20 MJ of heat transferred to it. Calculate its final temperature. Specific heat for cast iron is 0.544 kJ/kgK. Solution 6 Q transpose At m cA T Q mc 20 000 kj 2000 kg x 0.544 kJ/kgK •c 18.38°( and T2-Ti TZ -14°C TZ AT •c 18.38°( 'c +14°C 18.38°( 3C (Ans.) 32.38°( 390 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 Example 7 After 36.86 kj of heat is taken from 3.8 kg of copper, its temperature has decreased to 380 K. Calculate the initial temperature of the copper (to the nearest degree, in °C). Specific heat for copper is 0.388 kJ/kgK. Solution 7 Q = mcAT -36.86 kj = 3.8 kg x 0.388 kJ/kgK x AT Note: The negative sign indicates heat is transferred away from the copper. ^ ^ -36.86 kj 3.8 kg x 0.388 kJ/kgK -25 K T^-Ti = -25 K -TI = -25K-T2 TI = 25 K+380 K = 405 K - 405 - 273°C = 132°C(Ans.) 3rd Class Edition 3 • Part A1 f& ?& Chapter 8 • Heat, State Change, and Calorimetry Self-Test 2 One cubic metre of mercury has 28.356 MJ of heat transferred from it. Calculate the temperature change in °C. Mercury has a density of 13 600 kg/m3 and its specific heat is 0.139 kJ/kgK. Self-Test 3 A material has a specific heat of 0.414 kJ/kgK. What is the mass of the material if 16.5 MJ of heat, added or removed, causes the temperature to change by 19°C. Self-Test 4 An 85 kg block of ice at -25°C receives 1600 kJ of heat. A second block of ice, at -1°C and 58 kg, has 2100 kJ of heat removed. Which block ends up at the coldest temperature and by how much is it colder? Specific heat of ice = 2.135 kJ/kgK. 392 3rd Class Edition 3 ' Part A1 Heat, State Change, and Calorimetry • Chapter 8 OBJECTIVE 3 Explain the changes of state, and define latent heat, latent heat of fusion, and latent heat of evaporation. CHANGE OF STATE Many substances can exist, at any given time, in one of three possible states: solid, liquid, or gas (vapour). Substances can be changed from one state to another if a significant change in their internal energy occurs. The internal energy is changed by the addition or removal of heat. Note: The terms "state" and "phase" are often used interchangeably. Therefore, a change of state may also be referred to as a change of phase. There are four basic state changes that can occur. I.Solid.to-Liquid If enough heat is supplied to a solid, the internal energy of the molecules (that is, their speed of vibration) will increase to such an extent that the cohesion force between the molecules will be overcome, causing the molecules to move farther apart. The state changes from solid to liquid and is often referred to as melting or fusion." 2. Liquid-to-Gas (Vapour) The addition of even further heat, to the liquid, will eventually increase the vibration and weaken the cohesive force enough that the molecules will separate and begin to escape from the surface of the liquid. Here the state changes from liquid to gas (or vapour) and is often referred to as boiling" or "evaporation". 3. Gas (Vapour)-to-Liquid If enough heat is removed from a vapour, the internal energy of the molecules will decrease to such an extent that the cohesion force between the molecules will be restored, and the molecules will move closer together. The state changes from vapour to liquid and is often referred to as condensing. 4. Liquid-to-Solid The removal of even further heat, from the liquid, will eventually decrease the internal energy and strengthen the cohesive force enough that the molecules will again come together. The state changes from liquid to solid and is often referred to as "freezing" or "solidification". Water as an Example A simple example of state change involves water. At atmospheric pressure, water is a solid (ice) at temperatures ofO°C and lower. If heat energy is supplied to the ice, at 0°C, the change in internal energy causes the ice to melt, which is a state change from solid to liquid. Between 0°C and 100°C, water is a liquid. However, if heat energy is supplied to the water at 100°C the cohesive force between molecules weakens enough that the water boils, changing from liquid to vapour state (steam). 3rd Class Edition 3 • Part A1 393 ?& Chapter 8 • Heat, State Change, and Calorimetry LATENT HEAT When heat is supplied to or removed from a substance, causing a change in state, but no change in temperature, then the heat is known as latent heat. During a state change there is no change in temperature until the entire substance has changed state. Latent Heat of Fusion The latent heat of fusion is the heat required to change a unit mass of solid to a unit mass of liquid, with no change in temperature. That is, the entire heat energy (kj) supplied to each kg of the solid is used to change it to a liquid. For example, at atmospheric conditions, the latent heat of fusion of ice is 335 kj/kg, which means 335 kj of heat must be added to change each kg of ice at 0°C, into water at 0°C. Note: It also requires 335 kj of heat removal to change each kg of water at 0°C, into ice at 0°C Table 2 shows the latent heats of fusion for some typical substances. "Melting point" is the temperature at which the substance will change from solid to liquid. This is also the temperature at which the reverse change, liquid to solid, will occur (called the "freezing point"). Table 2 - Latent Heat of Fusion Substance Melting point (°C) Latent heat of fusion (kJ/kg) oxygen -219 13.8 nitrogen -210 25.5 mercury -39 11.8 ice 0 335 lead 327 24.5 aluminum 660 390 silver 961 110 gold 1063 copper 1083 _65^ 210 Latent Heat of Evaporation The latent heat of evaporation (or vaporization) is the heat required to change a unit mass (1 kg) of liquid to a unit mass of vapour (or gas), at the same temperature and pressure. That is, the entire heat energy (kj) supplied to each kg of the liquid is used to change it to a vapour. For example, at atmospheric pressure the latent heat ofevaporation of water is 2257 kj/kg, which means it requires 2257 kj of heat to change every kg of water at 100°C, into steam at 100°C. Also, when steam at 100°C is cooled it changes back to water at 100°C. This state change, from vapour to liquid, also requires the removal of 2257 kj/kg, the latent heat of evaporation. 394 3rd Class Edition 3 9 Part A1 Heat, State Change, and Calorimetry • Chapter 8 ^ Table 3 shows the latent heats of evaporation (at atmospheric pressure) for several common substances. Table 3 - Latent Heat of Fusion Substance Melting point (°C) Latent heat of fusion (kJ/kg) nitrogen -195.81 201.0 oxygen -182.97 213.0 water 100.0 2257 mercury 357.0 297.0 copper 2927 4720 lead 1750.0 870.7 silver 2160 2335 aluminum 2460 10907 gold 2860 3rd Class Edition 3 • Part A1 1697 _ 395 Chapter 8 • Heat, State Change, and Calorimetry OBJECTIVE 4 Given start and end conditions, calculate the heat required to change the states of water and other substances. HEAT REQUIRED FOR CHANGES OF STATE If a substance exists in one state, and heat is either added or removed, such that a change of state occurs, the substance ends up at a final temperature in that new state. It is possible to calculate the total amount of heat that was added to, or removed from, the substance. The total heat will be the sum of the total sensible heat and the total latent heat for the state change(s). For a given mass of a substance (m), this heat added/removed can be shown as: Total heat == total sensible heat + total latent heat = me AT ^ + mx latent heat (of fusion or evaporation) IfcLp? represents the latent heat of fusion and tL^ represents the latent heat of evaporation, then the respective state changes can be shown as: Total sensible and latent, involving fusion = m c AT + mLp Total sensible and latent, involving evaporation = me AT + m LE Example 8 How much heat must be added to 4.5 kg of ice at 0°C to change it into water at 0°C? The latent heat of fusion for ice is 335 kj/kg. Solution 8 Note: Since the ice is already at 0°C, there is no sensible heat involved. Q(kJ) = mcAT+mLp = 0+4.5 kg x 335 kj/kg = 1507.5 kj (Ans.) Example 9 How much heat must be removed from 4.5 kg of water at 0°C to change it to ice at 0°C? Solution 9 Note: This example simply illustrates that the latent heat applies in both "directions" of a state change, resulting in exactly the same amount of heat transfer. Again, there is no sensible heat involved. Q(kJ) = mcAT+mLp = 0+4.5 kg x 335 kj/kg = 1507.5 kj (Ans.) 396 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 Example 10 A block of ice having a mass of 2.2 kg, at a temperature of -8°C, is completely changed to water at 0°C. How much heat is required? Use specific heat of ice =2.135 kJ/kg°C. Solution 10 Note: In this case, sensible heat must first be added to the ice to increase it from -8°C to 0°C. Then latent heat must be supplied to change the ice to water at 0°C. Q (kj) = sensible heat + latent heat = me AT +mLj7 = 2.2 kg x 2.135 kJ/kg°C x (0°C - (-8°C)) + 2.2 kg x 335 kj/kg = 37.576 kj + 737 kj = 774.6 kJ(Ans.) Example 11 Ten litres of water at 278 K are changed to ice at 0°C. How much heat has been removed from the water? Assume 1 litre of water has a mass of 1 kg, and the specific heat of water is 4.183 kJ/kg°C. Solution 11 Note: Sensible heat must first be removed from the water to lower it to 0°C, then latent heat removed to change the water to ice. Temperature must be converted to °C. Initial temperature = 278 K= (278 - 273)°C = 5°C Q (kj) = sensible heat + latent heat of fusion = mcAT-^-mLp = (10 L x 1 kg/L) x 4.183 kJ/kg°C x (5°C - 0°C) + 10 kg x 335 kj/kg = 209.15 kj+3350 kj = 3559.15kJ(Ans.) 3rd Class Edition 3 • Part A1 397 Chapter 8 • Heat, State Change, and Calorimetry Example 12 Heat is supplied to 30 kg of ice at -18°C. The heat is sufficient to completely melt the ice and result in water at a temperature of 14°C. How much total heat was supplied? Use specific heat for ice = 2.135 kJ/kg°C, Lp = 335 kj/kg, and specific heat for water = 4.183 kJ/kg°C. Solution 12 Note: First, consider the sensible heat plus latent heat required to change the ice to water at 0°C; then add the sensible heat required to raise the water to 14°C. Sensible heat added to the ice: Q = mcAT = 30kgx2.135kJ/kg°Cx(0-(-18°C) = 30kgx2.135kJ/kg°Cxl8°C = 1152.9 kj Latent heat to melt the ice: Q = mLp = 30 kg x 335 kj/kg = 10 050 kj Sensible heat added to the water: Q = mcAT = 30kgx4.183kI/kg°Cx(14°C-0°C)) = 1756.86 kj Therefore Total heat = 1152.9 kj + 10 050 kj + 1756.86 kj - 12 959.76 kj (Ans.) 398 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 Example 13 How many megajoules of heat are removed while converting 12 kg of steam, at 100°C, into ice at -15°C? Assume atmospheric pressure and use latent heat of fusion = 335 kj/kg, latent heat of evaporation = 2257 kj/kg, specific heat of water is 4.183 kJ/kgK, and specific heat of ice is 2.135kJ/kgK. Solution 13 There are two state changes. We must consider the latent heat required to change the steam to water at 100°C, the sensible heat to lower the water temperature to 0°C, the latent heat to freeze the water at 0°C, and the sensible heat to lower the ice temperature to -15°C. Latent heat removed from the steam at 100°C: Q = mLE = 12 kg x 2257 kj/kg = 27 084 kj Sensible heat removed from the water: Q = mcAT = 12kgx4.183kJ/kg°Cx(100°C-0°C) = 5019.6 kj Latent heat to freeze the water at 0°C: Q = mLp = 12 kg x 335 kj/kg = 4020 kj Sensible heat removed from the ice: Q = mcAT = 12 kg x 2.135 kJ/kg°C x (0°C - (-15°C)) = 12 kg x 2.135 kJ/kg°C x 15°C = 384.3 kj therefore Total heat = 27 084 kj + 5019.6 kj + 4020 kj + 384.3 kj = 36 507.9 kj = 36.51 MJ (Ans.) 3rd Class Edition 3 • Part A1 399 ^ Chapter 8 • Heat, State Change, and Calorimetry Example 14 50 kg of gold, at a temperature of 70°C, is placed in a furnace and raised to its melting point of 1063°C. How much heat is required to totally liquefy the gold? Specific heat of gold is 0.130 kJ/kgK and latent heat of fusion is 65 kj/kg. Solution 14 Q (kj) = sensible heat + latent heat = mcAT-^-mLp = 50 x 0.130 kJ/kgK x (1063°C - 70°C) + 50 kg x 65 kj/kg = 6454 kj + 3250 kl = 9704kJ(Ans.) Self-Test 5 500 kg of steam at atmospheric pressure and 100°C is condensed to a final temperature of 65°C. How much heat is removed from the steam by the condenser? Self-Test 6 150 kg of ice at -20°C is converted to water at 100°C. How much heat is absorbed by each kg during this change? 400 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 3rd Class Edition 3 • Part A 1 -^ I^VVBd • £ uowps sseio PJ£ AjfOLUuoieo pue 'Q6ueqo Q^S l)^H • 8 ^deHO ^ Heat, State Change, and Calorimetry • Chapter 8 f£ OBJECTIVE 5 Determine the final temperatures and the original masses for mixtures of ice, water, steam, and other substances. HEAT TRANSFER IN A MIXTURE When two substances at different temperatures are mixed, heat will be transferred between the substances until a final, common temperature is reached. This process of establishing a common temperature is often referred to as "reaching thermal equilibrium". Since heat will only flow directly from a hotter to a colder substance, it follows that heat will be transferred from the substance at higher temperature to the substance at lower temperature, until both (that is, the mixture) are at the same temperature. Therefore, the temperature of the hotter substance will decrease and the temperature of the colder substance will increase, until they are equal. If we assume that no heat is lost to the surroundings, we can further conclude that the amount of heat gained by the colder substance must be equal to the amount of heat lost by the hotter body. The following relationship can be stated: Heat gained by colder substance (kj) = heat lost by hotter substance (kj) Mixtures with No Change of State The heat gained or lost by a solid or liquid substance, provided there is no change of state, is determined by the equation, Q= m c AT. The heat transferred in this case is sensible heat only and the simple heat transfer equation becomes: me AT (for colder) = me AT (for hotter) Given the appropriate information, we can now calculate the final temperature of the mixture, the masses of the substances, or the temperature changes. 3rd Class Edition 3 • Part A1 403 Chapter 8 • Heat, State Change, and Calorimetry Example 15 4 kg of solid lead at 150°C is dropped into 8 kg of water at 30°C. Find the final temperature of the lead and water, given that specific heat of lead = 0.1275 kJ/kgK and specific heat of water = 4.183 kJ/kgK. Solution 15 Since the lead is at a higher temperature, heat will be transferred from the lead to the water. The water will gain heat and the lead will lose heat. Final temp ofnuxture = t°C Heat gained by water = me AT = 8 kg x 4.183 kJ/kgK x (t - 30)°C = (33.46^-1003.92) kj Heat lost by lead = me At = 4 kg x 0.1275 kJ/kgKx (150 - t)°C = (76.5- 0.5 lt)k] Heat gained by water = heat lost by lead 33.46^-1003.92 = 76.5-0.5U 33.97^ = 1089.48 t = 32.1°C(Ans.) Example 16 A piece of brass, at a temperature of 125°C, is placed into 4 kg of turpentine at 7°C. Thermal equilibrium is reached at 48°C. Calculate the mass of the brass, given that specific heat of brass is 0.383 kJ/kgK and specific heat ofturpentine is 1.8 kJ/kgK. Solution 16 Heat lost by brass = heat gained by turpentine me AT (brass) = me AT (turpentine) m x 0.383 kJ/kg°C x (125 - 48)°C = 4kg x 1.8kJ/kg°C x (48 - 7)°C m x 0.383 kJ/kg°C x 77°C = 4 kg x 1.8 kJ/kg°C x 41°C m x 29.491 kj/kg = 295.2k] m= 295.2 kj 29.49 IkJ kg = 10 kg (Ans.) 404 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 Example 17 20 kg of aluminum at 265°C and 5 kg of copper at 152°C are dropped into a container, which contains 200 kg of oil at 10°C. What will the final temperature of the oil be? The specific heats are: aluminum = 0.909 kJ/kgK, copper = 0.388 kJ/kgK, oil = 2.12 kJ/kgK. Solution 17 Note: In this case there are 3 substances. They will all end up at the final temperature, t, with the aluminum and copper losing heat and the oil gaining heat. The total heat gained by the oil will equal the total heat lost by the aluminum plus the copper. Heat gained by the oil = me AT = 200kgx2.12kJ/kg°Cx(^-10)°C = (424 (-4240) kj Heat lost by the aluminum = me AT - 20 kg x 0.909 kJ/kg°C x (265 - t)°C = (4817.7- 18.18 t)k] Heat lost by the copper = me AT = 5kgx0.388kJ/kg°Cx(152-OOC = (294.88 - 1.94 t) kj Heat gained by the oil = heat lost by copper + heat lost by aluminum (4241- 4240) kj = (294.88 - 1.941) kj + (4817.7 - 18.18 t) kj 424 (+1.94 ^+18.18^ = 294.88+4817.7+4240 444.12^ = 9352.58 t = 21.06°C (Ans.) Mixtures Involving Change of State If the substances that are being mixed are in different states, one or both of the substances will go through a state change before the final, common temperature is reached. The heat gained by the colder substance will equal the heat lost by the hotter substance. However, since a state change occurs, the heat will consist of sensible heat plus latent heat. If tV represents latent heat, which can be either latent heat of fusion (Lp) or latent heat of evaporation (Lg), the mixture equation can be stated as follows. (sensible + latent) gained by colder = (sensible + latent) lost by hotter (m c4T+mL), for colder = (m c4T+m L), for hotter The simplest examples of such mixtures involve ice, water and steam. If ice and water are mbced, the process will usually involve the ice changing to water, with the resultant mixture being water at some final temperature. The ice must gain latent heat of fusion (Lp) as it changes state. If water and steam are mixed, the process will usually involve the steam changing to water, so the steam must lose latent heat ofevaporation (Lg). If steam and ice are mixed, they will both go through a state change, usually resulting in water at a final temperature. 3rd Class Edition 3 • Part A1 405 Chapter 8 • Heat, State Change, and Calorlmetry Example 18 Five kg of ice at -10°C are mixed with 50 kg of water at 60°C. What will be the final temperature of the nuxture if all of the ice melts? Take the specific heat of ice as 2.04 kJ/kgK, the specific heat of water as 4.183 kJ/kgK, and the latent heat of fusion of ice as 335 kj/kg. Solution 18 Note: The heat added to the ice consists of sensible heat to raise it to 0°C, latent heat to change it to water, and sensible heat to raise the water to the final temperature. Heat lost by the water consists of sensible heat to lower it from 60°C to the final temperature. Final temp of mixture = t°C Sensible heat to raise ice 0°C = me AT = 5 kg x 2.04 kJ/kg°C x 10°C = 102 kj Latent heat to melt ice = mx Lp = 5kgx335kJ/kg = 1675 kj Heat to raise melted ice to { = me AT = 5 kg x 4.183 kJ/kg°C x (t - 0)°C = 20.92 tk] Total heat gained by ice = 102k] + 1675 kj + 21 tk] = (1777+21()kJ Heat lost by water = me AT = 50kgx4.183kJ/kg°Cx(60-0°C = (12 549- 209.15 t)k] Heat gained by ice = heat lost by water 1777 + 211 = 12549-209.15^ 230.15^ - 10772 t = 46.8°C(Ans.) 406 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 Example 19 Two kg of steam at atmospheric pressure and 100°C are blown into 100 litres of water at 20°C. If all the steam condenses, find the temperature of the final mixture. Latent heat of evaporation for steam is 2257 kj/kg and specific heat of water is 4.183 kj/kg K. Solution 19 Note: Since all steam condenses, we know the final mixture will be water. The steam will lose latent heat when it condenses to water, plus sensible heat when the condensate falls to the final temperature. The water will gain sensible heat. Use density of water = 1000 kg/m3. 100 litres of water = 100 kg. First, find the mass of water: 1000 litres = 1 m3 so, 100 litres = O.lm3 = density x volume = 1000 kg/m3 x 0.1 m3 = 100kg Final temp of mixture = t°C Heat gained by water = me AT = 100 kg x 4.183 kJ/kg°C x (t - 20)°C = (418.3^-8366) kj Heat lost by condensing steam = mx LE = 2 kg x 2257 kj/kg = 4514 kj Heat lost by steam condensate = me AT = 2kgx4.183kJ/kg°Cx(100-0°C = (836.6 - 8.366 0 kj Total heat lost by steam = (4514 + 836.6 - 8.366 t) kj = (5350.6 - 8.366 f) kj Total heat gained by water = total heat lost by steam 418.3^-8366 = 5350.6-8.366^ 426.671 = 13716.6 t = 32.15°C (Ans.) 3rd Class Edition 3 • Part A1 407 Chapter 8 • Heat, State Change, and Calorimetry Example 20 Find the final temperature when 30 kg of ice at 268 K are mixed with 5 kg of steam at 373 K, at atmospheric pressure. Solution 20 Note: We will assume that both the ice and the steam will change state, to water, at some final temperature, Y. The ice must gain sensible heat to raise it to 0°C, latent heat of fusion at 0°C, and then sensible heat as water to the final temp. The steam must lose latent heat ofevaporation at 100°C and then sensible heat as water, down to the final temp. Atmospheric pressure tells us that evaporation occurs at 100°C. Temperature of ice = 268 K = (268 - 273)°C =-5°C Temperature of steam = 373 K = (373 - 273)°C = 100°C For ice, specific heat = 2.135 kJ/kg°C and latent heat of fusion = 335 kj/kg For water, use specific heat = 4.183 kJ/kg°C For steam, latent heat of evaporation = 2257 kJ/kgK Final temperature = t°C Sensible heat to raise ice to 0°C = me AT = 30kgx2.135kJ/kg°Cx5°C = 320.25 kj Latent heat to melt the ice = mx Lp = 30 kg x 335 kj/kg 10 050 kj Heat to raise melted ice to Y = me AT = 30kgx4.183kJ/kg°Cx^-0)OC = 125.49 tk] Total heat gained by ice = (320.25 + 10 050 + 125.49 t) kj = (10 370.25 + 125.49 t) kj Latent heat lost by steam = mx L^ = 5kgx2257kJ/kg = 11 285 kj Heat lost by steam condensate = me AT = 5kgx4.183kJ/kg°Cx(100- t)°C = (2091.5 -20.195 t) kj Total heat lost by steam = (11 285 + 2091.5 - 20.915 t) kj = (13 376.5-20.915 QkJ Heat gained by ice = heat lost by steam 10 370.25 + 125.49 t = 13 376.5 - 20.915 t 146.411 = 3006.25 t = 20.53°C (Ans.) 408 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 f£ Example 21 What mass of ice, at -8°C, must be added to 1 tonne of water, at 60°C, to reduce the water temperature to 6°C? Solution 21 Note: Let mj be the mass of the ice and mW the mass of the water. Sensible heat to raise ice to 0°C = mj c A T = mi kg x 2.135 kJ/kg°C x 8°C = 17.08mjkJ Latent heat to melt the ice = mj x Lp = mjkgx335kj/kg = 335mjkJ Heat to raise melted ice to 6°C = mj c A T = m; kg x 4.183 kJ/kg°C x 6°C = 25.1m^kJ Total heat gained by ice = (17.08 mj + 335 mj + 25.1 mj) kj = 377.18mjkJ Heat lost by water = m^cAT = 1000 kg x 4.183 kJ/kg°Cx54°C = 225 882 kj Heat gained by ice = Heat lost by water 377.18m; = 225882 mi = 225 882 377.18kg = 598.9 kg (Ans.) Self-Test 10 What will the temperature of the water be, at thermal equilibrium, if 130 kg of copper, at 700°C, is dropped into 600 litres of water at 20°C? The specific heat of copper is 0.388 kl/kgK and the specific heat of water is 4.183 kJ/kgK. 3rd Class Edition 3 • Part A1 409 ;!£- Chapter 8 • Heat, State Change, and Calorimetry Self-TestH Find the final temperature if 400 kg of ice at 0°C is mixed with 90 kg of steam at 100°C, at atmospheric pressure. Use specific heat of water as 4.183 kJ/kgK. Self-Test 12 Assuming no heat loss to atmosphere, what is the final temperature if 15 tonnes of water, at 25°C, is injected with 250 kg of steam, at atmospheric pressure and 100°C? 410 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 3rd Class Edition 3 • Part A1 T® ?& Chapter 8 • Heat, State Change, and Calorimetry OBJECTIVE 6 Explain the working principle of a simple calorimeter, and use the calorimeter equation to determine specific heat and final temperature. CALORIMETRY Calorimetry is a scientific process, which involves the determination of the heating values of fuels or the specific heats of substances. The term derives from calorie", a unit of heat measurement that is replaced in the metric system by kj. One gram-calorie was the amount of heat required to raise the temperature of one gram of a substance through 1°C. There are various types of calorimeters. The bomb calorimeter", for example, burns a measured amount of fuel and determines the heating value (or calorific value) of the fuel. That is, it determines the number ofkilojoules of heat that will be produced by the combustion of 1 kilogram of the fuel. This requires a fairly complex calorimeter design. Simple Calorimeter A basic calorimeter design, called the "simple calorimeter", is primarily used to determine the specific heat of a substance. The apparatus consists of a vessel, of known mass, which contains a known quantity of water, at a known temperature. The specific heat of the mass is found by placing the mass in the water and allowing the mass, water and container to reach thermal equilibrium. Figure 2 illustrates a simple calorimeter apparatus, which consists of the following: • An outer vessel, with a highly polished outer surface (to prevent heat loss by radiation) and an insulated cover • A smaller, inner vessel made of a known material (usually copper or aluminum), which is insulated and suspended within the outer vessel • A stirrer, which fits through the cover and extends into the inner vessel • A thermometer, which extends into the inner vessel Note: For the purpose of clarity, we will refer to the inner vessel as the calorimeter", since only this vessel and its contents enter into the calculation for specific heat. The entire assembly, including the outer vessel and the insulation, may be referred to as the "calorimeter apparatus". Figure 2 - Simple Calorimeter Apparatus Thermometer Stirrer Inner \^ vessel z Insulated lid Outer vessel Insulation 412 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 Using the Calorimeter The calorimeter is first weighed (to determine its mass), then partially filled with water. It is weighed again, with the difference equaling the mass of the water. The calorimeter is then placed into the outer vessel and the cover is installed. The apparatus is allowed to stand until all components reach the same temperature, which occurs when a constant temperature is observed on the thermometer. Meanwhile, the substance to be tested is weighed to determine its mass. The substance to be tested is then heated to a known temperature, which is greater than the calorimeter temperature. For example, it may be placed in boiling water and held until it is at that temperature throughout. When the apparatus and substance have reached their respective temperatures, the substance is placed into the water in the calorimeter. The cover plate is immediately replaced and the stirrer is activated to ensure the water mixes thoroughly, maintaining a consistent temperature during the test. Ttiermometer readings are recorded at regular intervals. Heat transfers from the substance to the water (plus calorimeter), due to the difference in their initial temperatures. The insulation prevents heat loss beyond the inner vessel. This heat transfer continues until all three components are at the same temperature. This point of thermal equilibrium is recorded. Since heat loss to the surroundings is negligible, it is assumed that all heat transfer occurs only from the substance to the water (plus the calorimeter). The expression for the amount of heat transferred can be stated as: Heat lost by substance = heat gained by water + heat gained by calorimeter Since the heat transfer for each individual component is represented by ££Q = m c AT\ the relationship can be further stated as: me A T(substance) = mcA T(^ater) + m C 4 T(calorimeter) Using this equation, it is possible to calculate the specific heat of the substance. Since all masses are known, the initial and final temperatures have been recorded, and the specific heats of water and the calorimeter material are known, the only unknown in the equation is the specific heat of the substance. 3rd Class Edition 3 • Part A1 413 Chapter 8 • Heat, State Change, and Calorimetry Example 22 0.4 kg of a substance is subjected to calorimetry to determine its specific heat. The calorimeter is made of 1.5 kg of copper. If the specific heats of copper and water are 0.39 kJ/kgK and 4.183 kJ/kgK respectively, what is the specific heat of the substance if the following measurements were noted? Initial temperature of substance = 60°C Initial temp. of calorimeter and water = 20°C Mass of water = 0.5 kg Final thermal equilibrium temperature = 27°C Solution 22 Heat lost by substance = heat gained by water + heat gained by calorimeter ^ c A -r(substance) = m c A r(water) + m C 4 T(calorimeter) 0.4 kg x c x (60°C - 27°C) = 0.5 kg x 4.183 x (27°C - 20°C) + 1.5 kg x 0.39 x (27°C - 20°C) 13.2 c = 14.64+4.095 18.735 C(substance) = —^ = 1.419kJ/kgK(Ans.) Example 23 An aluminum calorimeter, with a mass of 50 g and containing 60 g of water, is heated to 18°C. 0.6 kg of copper, at 80°C, is placed in the calorimeter. At what temperature will thermal equilibrium be reached? Use the following specific heats: aluminum = 0.909 kJ/kgK, copper = 0.388 kj/kg, water =4.183 kj/kg Solution 23 Note: Let the final temperature be t. Heat lost by copper = heat gained by water + heat gained by aluminum mc4T(copper) = mcAT(^^)-^-mcA ^aluminum) 0.6 kg x 0.388 x (80°C - t) = 0.06 kg x 4.183 x ((- 18°C) + 0.05 kg x 0.909 x (t- 18°C) 18.642-0.233^ = 0.25K-4.518 + 0.045 ^-0.818 -0.529^ = -23.978 t = 45.32°C (Ans.) 414 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 3rd Class Edition 3 • Part A1 7® ?& Chapter 8 • Heat, State Change, and Calorimetry OBJECTIVE 7 Explain water equivalent, and perform calculations involving calorimetry and wafer equivalents. WATER EQUIVALENT The water equivalent of a substance is "the equivalent mass of water that would require the same amount of heat transfer as the substance, to produce the same temperature change. The calculation for water equivalent can be explained as follows: Let the heat equation, Q = m c AT, be rewritten to represent water, W, and a certain metal, M, as follows: Water: QW = mWcW4TWand Substance: QM = mMdVMTM However, by definition of water equivalent, the heat transfers are equal. So: QW = QW or, mWcW4TW = mMcM/VTM Also, by definition, the water and metal have the same temperature change (4TW = ATM). So: mwcw = mw = HlMCM mMXCM cw kg This can be written as: water equivalent (kg) = mass of substance x specific heat of substance specific heat of water Water equivalent is often used in calorimetry and other heat transfers to simplify the calculations. The specific heat of the calorimeter material, since it undergoes the same temperature change as the water, may be replaced by its water equivalent. In other problems, it may be asked to simply determine the water equivalent of a substance. 416 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 Example 24 The mass of a copper calorimeter is 0.5 kg and its specific heat is 0.39 kJ/kgK. Calculate its water equivalent. Use specific heat of water as 4.183 kJ/kgK. Solution 24 mCOPPER X CCOPPER ^WATER = CWATER 0.5 kg x 0.39 kJ/kg°C 4.183kJ/kg°C = 0.0466 kg (Ans.) Example 25 An aluminum calorimeter has a mass of 0.3 kg. It contains 0.5 kg of water at 18°C. How much heat must be transferred from the tested substance if the final temperature of the calorimeter is 22°C? Use specific heat of water as 4.183 kJ/kgK and of aluminum as 0.909 kJ/kgK. Solution 25 Note: First find the water equivalent of the aluminum calorimeter and then use the heat equation for the total equivalent mass of water. Water equivalent of calorimeter = mAL x CAL CWATER 0.3 kg x 0.909 kJ/kg°C 4.183 kJ/kg°C = 0.0652kg therefore, total mass of water = 0.0652+0.5 kg = 0.5652kg Heat transferred to calorimeter = m c AT = 0.5652 kg x 4.183 kJ/kg°C x (22°C - 18°C) = 0.5652 kg x 4.183 kJ/kg°Cx4°C = 9.46kJ(Ans.) 3rd Class Edition 3 • Part A1 417 ?2> Chapter 8 • Heat, State Change, and Calorimetry Example 26 A calorimeter is used to determine the specific heat of a substance. If the substance has a mass of 0.45 kg, given the following measurements, what is the specific heat of the substance? Take specific heat of water as 4.183 kJ/kgK. Initial temp. of substance = 90°C Water equivalent of calorimeter = 0.028 kg Mass of water = 0.5kg Initial temp. of water = 15°C Final thermal equilibrium temp. = 29°C Solution 26 Note: We don t know the material or the specific heat of the calorimeter. However, using the water equivalent, the calorimeter can be treated as if it is water. Heat lost by substance = heat gained by water + heat gained by calorimeter = heat gained by (water + water equivalent) mc4r(substance) = m c A T(total water) 0.45 x ex (90- 29) = (0.5 + 0.028) x 4.183 x (29 - 15) 27.45 c = 30.92 c,(substance) 30.920 27.45 kJ/kgK = 1.126kJ/kgK(Ans.) Self-Test16 The material of a calorimeter has a specific heat of 0.467 kJ/kgK and a mass of 0.6 kg. What is its water equivalent? Use 4.183 kJ/kgK for specific heat of water. 418 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 T® Self-Test17 What is the mass of a material that has a water equivalent of 7.45 kg and a specific heat of 0.875 kJ/kgK? Self-Test18 A copper calorimeter has a mass of 0.5 kg. If it contains 0.6 kg of water at 22°C when a test material is placed in it, (a) what is the water equivalent of the calorimeter, and (b) how much total heat must the calorimeter and water absorb if equilibrium temperature is 31°C? The test determines the specific heat of the material to be 0.435 kJ/kgK. 3rd Class Edition 3 • Part A1 419 rSr Chapter 8 • Heat, State Change, and Calorimetry 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 f£ SUMMARY OF HEAT, STATE CHANGE, AND CALORIMETRY In this chapter, heat, changes of state, and calorimetry were discussed. These subjects include important terms such as specific heat, sensible heat, and latent heat. This chapter also covered the relationship between potential and kinetic energy, internal energy, and temperature. Latent heat of evaporation and latent heat of fusion were explored. Calculations involving changes of state and the determination of the final temperatures of mixtures were presented step by step. Learners were given opportunities to practice these calculations. Calorimetry was explained using a simple calorimeter, which included discussion of water equivalent. Examples of calculations involving calorimetry were provided. 3rd Class Edition 3 • Part A1 421 ?& Chapter 8 • Heat, State Change, and Calorimetry CHAPTER QUESTIONS Objective 1 1. Define "specific heat55 and state the three factors that determine the amount of heat transferred to or from a substance, assuming no change of state. Objective 2 2. Show clearly, by calculation and comparing results, which of the following processes would require the greatest transfer of heat: melting 0.5 tonnes of copper from 70°C; solidifying 3 tonnes ofmolten aluminum at the fusion temperature; changing 0.75 tonnes of ice at -10°C to water at +10°C. Objective 3 3. Define the following terms: latent heat, latent heat of fusion, latent heat ofevaporation. Objective 4 4. How much heat, in MJ, is added to 1.8 tonnes of ice, at -15°C, if 60% of the ice becomes steam at atmospheric pressure and 100°C, and the remainder becomes water at 90°C? Objective 5 5. a) 4.0 kg of ice are mixed with 60 kg of water, which is at a temperature of 50°C. If the temperature of the mbcture becomes 40°C, calculate the initial temperature of the ice. b) If the mixture in '(a)5 is injected with 2.5 kg of steam, at atmospheric pressure and 100°C, what temperature will the new mbcture attain, assuming no external heat loss? 6. Find the resulting temperature when 30 kg of ice at 200 K are mixed with 10 kg of steam at 373 K and atmospheric pressure. Use specific heat of ice as 2.135 kJ/kgK and specific heat of water as 4.183 kJ/kgK. 7. 2.5 kg of brass, with specific heat 0.39 kJ/kgK and temperature of 256°C is dropped into 10 litres of water at 14°C. As soon as thermal equilibrium is achieved, the brass is removed and a piece of cast iron, specific heat = 0.544 kJ/kgK, and temperature of400°C is placed in the water. If the final temperature of the water is 24°C, what is the mass of the cast iron? Ignore heat losses to atmosphere. Objective 6 8. In an experiment to find the specific heat of iron, 2.15 kg of iron at 100°C are dropped into a calorimeter-like vessel containing 2.3 litres of water at 17°C. If the resultant mbcture is 24.4°C and the water equivalent of the vessel is 0.18 kg, what is the specific heat of the iron? 422 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 9. A calorimeter test, using a copper calorimeter, records these results: mass of tested material = 350 g; mass of calorimeter = 0.70 kg; mass of water = 0.9 kg; initial temperature of tested material = 60°C; initial temperature ofcalorimeter = 13°C; final temperature = 18°C. What is the specific heat of the tested material? Objective 7 10. a) Define water equivalent" and explain the formula used to determine the water equivalent of a material. b) A calorimeter has a water equivalent of 0.0065 kg and a mass of 1.95 kg. Calculate the specific heat of the calorimeter material. 3rd Class Edition 3 • Part A1 423 ,& Chapter 8 • Heat, State Change, and Calorimetry NUMERICAL ANSWERS TO CHAPTER QUESTIONS 2. Copper = 263.5 MJ, Aluminum = -270 MJ, Ice = 298.6 MJ; The process of changing the ice to water requires the greatest transfer of heat. 4. 3821 MJ 5. a) -58.6°C b) 62.5°C 6. 293.8 K or 21°C 7. 0.920 kg 8. 0.472 kJ/kgK 9. 1.37 kJ/kgK 10. b) 0.0139 kJ/kgK 424 3rd Class Edition 3 • Part A1 Heat, State Change, and Calorimetry • Chapter 8 7® SELF TEST SOLUTIONS 1. 0.85 kJ/kgK 2. -15°C 3. 2098 kg 4. The 58 kg block is 1.7°C colder than the 85 kg block 5. 1 201 702.5 kj or 1201.7 MJ 6. 791.5 kj 7. 2.12 GJ 8. Melting the copper requires 13 816.4 kj or 13.816 MJ more 9. 10.993 MJ 10. 33.5°C 11. 52.09°C 12. 35.07°C 13. 293.51 kg 14. 1.204 kJ/kgK 15. 43.49°C 16. 0.067kg 17. 35.62kg 18. a) 0.0464kg b) 24.33 kj 19. 0.02kg 3rd Class Edition 3 • Part A1 425 p> 426 Chapter 8 • Heat, State Change, and Calorimetry 3rd Class Edition 3 • Part A1