Module - 1: Thermodynamics

Transcription

1 Thermodynamics: Module - : Thermodynamics Thermodynamics (Greek: thermos = heat and dynamic = change) is the study of the conversion of energy between heat and other forms, mechanical in particular. All those problems that are related to the interconversion of heat energy and work done are studied in thermodynamics. In thermodynamics, we discuss different cycles such as Carnot cycle, Rankine cycle, Otto cycle, diesel cycle, refrigerator, compressors, turbines and air conditioners. Thermal equilibrium and Temperature: The central concept of thermodynamics is temperature. Temperature is familiar to us all as the measure of the hotness or coldness of objects. We shall learn afterwards that temperature is a measure of the average internal molecular kinetic energy of an object. It is observed that a higher temperature object which is in contact with a lower temperature object will transfer heat to the lower temperature object. The objects will approach the same temperature, and in the absence of loss to other objects, they will then maintain a constant temperature. They are then said to be in thermal equilibrium. Thermal equilibrium is the subject of the Zeroth Law of Thermodynamics. The Zeroth Law of Thermodynamics: The "zeroth law" states that if two systems are in thermal equilibrium with a third system, then they must be in thermal equilibrium with each other. This law states that if object A is in thermal equilibrium with object B, and object B is in thermal equilibrium with object C, then object C is also in thermal equilibrium with object A. The message of the zeroth law is: Every body has a property called temperature. Two objects are defined to have the same temperature if they are in thermal equilibrium with each other. This law allows us to build thermometers. For example the length of a mercury column (object B) may be used as a measure to compare the temperatures of the two other objects. The zeroth law, which has been called a logical afterthought, came to light only in 90s, long after the first and second laws of thermodynamics had been discovered and numbered. Because the concept of temperature is fundamental to those two laws, the law that establishes temperature as a valid concept should have the lowest number-hence the zero. The zeroth law has a fairly straightforward statistical interpretation and this will allow us to begin to establish the equivalence between the statistical definitions and the conventional thermodynamic ones. Temperature and Heat: If you take a can of cola from the refrigerator and leave it on the kitchen table, its temperature will riserapidly at first but then more slowly until the temperature of the cola equals that of the room(the two are then in thermal equilibrium). In generalizing this situation, we describe the cola or coffee as a system (with temperature T S ) and the relevant part of the kitchen as the environment (with temperature T E ) of that system. Our observation is that if T S is not equal to T E, then T S will change until the two temperatures are equal and thus thermal equilibrium is reached. Such a change in temperature is due to the transfer of energy between the thermal energy of the system and the system s environment. It may be mentioned that thermal energy is an internal energy that consists of the kinetic and potential energies associated with the random motions of the atoms, molecules and other microscopic bodies within an object. The transferred energy is called heat and is symbolized Q. Heat is

2 positive when energy is transferred to a system s thermal energy from its environment (we say that heat is absorbed). Heat is negative when energy is transferred from a system s thermal energy to to its environment ( we say that heat is released or lost). We are then led to this definition of heat: Heat is the energy that is transferred between a system and its environment because of a temperature difference that exists between them. Recall that energy can also transferred between a system and its environment as work W via a force acting on a system. Heat and work, unlike temperature, pressure, and volume, are not intrinsic properties of a system. They have meaning only as they describe the transfer of energy into or out of a system. Let us now look into the Molecular Theory of Matter for an explanation of heat and temperature. Molecular Theory of Matter states that matter is made up of tiny particles called molecules. These particles are in constant motion within the bounds of the material. Since the relationship between kinetic energy of an object and its velocity is: KE = ½ mv, which means that the more energy an object has, the faster it is traveling (or vice versa). Thus, when you provide extra energy to an object, you cause its molecules to speed up. Those molecules, in turn, can cause other molecules to speed up. The sum effect of the speed or energy of these molecules is the phenomenon we call heat. Molecules can go into high-energy motion, causing heat, from various energy sources such as Light, Chemical reactions, Electrical resistance, Friction and nuclear reactions. Heat n i ( K. E) i Heat is defined as "The total kinetic energy of all the molecules of a body" and temperature is a measure of the average internal molecular kinetic energy of an object. Units of Heat: Before scientists realized that heat is transferred energy, heat was measured in terms of its ability to raise the temperature of water. Thus, the calorie (cal) was defined as the amount of heat that would raise the temperature of g of water from.5 0 C to C. In 98, the scientific community decided that since heat (like work) is transferred energy, the SI unit for heat should be the one we use for energy, namely, the joule. The calorie is now defined to be.860 J (exactly). The calorie used in nutrition is really is kilocalorie.. Calorie (cal): It is the amount of heat required to increase the temperature of g of water from.5 o C to 5.5 o C. ( cal=.86 J). Kilocalorie (kcal): It is the amount of heat required to raise the temperature of kg of pure water through o C.. British Thermal Units (BTU): It is defined as the quantity of heat required to raise the temperature of pound (lb) of pure water through o F. It is also referred to as pound-degree Fahrenheit unit. Conversion: B.T.U = Cal ( or 5 Cal) calorie =.86 Joule ( or. joule) Flow of Heat: Natural flow of heat always takes place from a region of high temperature to a region of lower temperature. Scales of Temperature: There are three main scales of temperature:

3 () Centigrade or Celsius Scale () Kelvin Scale () Fahrenheit Scale Conversion: C F 5 K Heat Capacity (The Storage of Thermal Energy): The heat capacity C, of a substance is defined as the amount of heat required to increase the temperature of that substance by one Celsius degree. For such a system Q C T The unit is cal/ o C or J/ o C. Specific Heat: The specific heat of a material is the amount of heat energy required to raise the temperature of unit mass of a substance by one Kelvin or C is known as specific heat of the substance." Mathematical Expression: If Q is the amount of heat required to raise the temperature of "m" kg of the substance through T Kelvin, Then Q m, (i) Q T, (ii) or Q = m S T where S is the constant called specific heat of substance. Q C Or, S mt m Unit of Specific Heat: Unit of specific heat is J/kg K in S.I. system. The specific heat of water is calorie/gram C =.86 joule/gram C which is higher than any other common substance. As a result, water plays a very important role in temperature regulation. The specific heat per gram for water is much higher than that for a metal, hence it takes a lot of heat to change the temperature of water. Water Equivalent: The water equivalent of a body is defined as the amount of water, which will absorb same quantity of heat as the substance for the same rise of temperature. The symbol representing water equivalent is W and unit gm. Mechanical Equivalent of Heat: The mechanical equivalent of heat (J) is defined as the amount of work that is necessary to produce one unit (calorie) of heat. J =.86 joules/cal J =.86 x 0 7 ergs/cal J = 778 ft-lbs/ B.Th.U

4 Latent Heat: Latent heat describes the amount of energy in the form of heat that is required for a material to undergo a change of phase. A solid consists of molecules that are tightly bound to each other by forces acting between them. Energy must be supplied in the form of heat to overcome these forces. As the bonding forces weaken, groups of molecules break free, and as more heat is supplied the solid changes into a liquid, in which small groups of molecules constantly break apart, and reform and groups slide freely past each other. While this is happening, the temperature of the substance remains unchanged. All of the heat energy is used in loosening the bonds between molecules. If still more energy is applied, it has the effect of making the groups of molecules move faster. When they move faster they strike harder against any object with which they come into contact. It is the speed of motion of molecules that a thermometer measures as temperature. Once a solid has melted to become a liquid, the application of additional heat raises the temperature of the liquid. When a liquid is cooled, its molecules lose energy and move more slowly, and the temperature of the liquid falls. When their energy falls to a certain level the molecules start bonding together. This requires less energy than moving about, and so heat energy is released as the liquid solidifies. The temperature of the substance remains unchanged, but the surrounding medium is warmed by the release of energy. For a phase change, the heat liberated or absorbed is given by Q ml () where l is the latent heat of fusion (latent heat of melting) or latent heat of vaporization. How much heat does it take to get water to change state? If the water is at a temperature of 00 degrees C (that is, the 80 cal / gram absorbed 50 cal / gram absorbed boiling point, or degrees F) it Gas Solid takes an additional 50 calories Liquid Water vapor Ice Water of heat to convert one gram of water from the liquid state to the vapor state. When the vapor converts to the liquid state, cal / gram released 50 cal / gram released calories of energy will be released per gram of water. If you are converting solid water (ice) to liquid water at 0 degrees C, it will require about 80 calories of heat to melt one gram of ice, and the 80 calories will be released when the liquid water is frozen to the solid state. Latent Heat of Fusion: Fusion is the change of state from solid to liquid. In the process of fusion, the molecule absorbs energy. This energy is latent heat. When a solid substance changes from the solid phase to the liquid phase, energy must be supplied in order to overcome the molecular attractions between the constituent particles of the solid. This energy must be supplied externally, normally as heat, and does not bring about a change in temperature.

5 The units of heat of fusion are usually expressed as joules per mole (the SI units) or calories per gram or Btu per pound-mole l fusion 80cal/gm or kj/kg The specific latent heat of fusion is defined as "The specific latent heat of fusion of a substance is the amount of heat required to convert unit mass of the solid into the liquid without a change in temperature." The specific latent heat of fusion of ice at 0 ºC, for example, is kj.kg -. This means that to convert kg of ice at 0 ºC to kg of water at 0 ºC, kj of heat must be absorbed by the ice. Conversely, when kg of water at 0 ºC freezes to give kg of ice at 0 ºC, kj of heat will be released to the surroundings. Latent Heat of aporization: Evaporation is the change of state from liquid to vapor. In the process of evaporation, the molecule absorbs energy. This energy is latent heat. How did you make the water evaporate? Probably you added heat. You might have set it out in the sun, or possibly put it over a fire. To make water evaporate, you put energy into it. The individual molecules in the water absorb that energy, and get so energetic that they break the hydrogen bonds connecting them to other water molecules. They become molecules of water vapor. l vaporization 50cal/gm or 60kJ/kg The definition of the specific latent heat of vaporization is 'The specific latent heat of vaporization is the amount of heat required to convert unit mass of a liquid into the vapour without a change in temperature." For water at its normal boiling point of 00 ºC, the latent specific latent heat of vaporization is 60 kj.kg -. This means that to convert kg of water at 00 ºC to kg of steam at 00 ºC, 60 kj of heat must be absorbed by the water. Conversely, when kg of steam at 00 ºC condenses to give kg of water at 00 ºC, 60 kj of heat will be released to the surroundings. Heating / Cooling Curves: The diagram on the left shows the uptake of heat by kg of water, as it passes from ice at -50 ºC to steam at temperatures above 00 ºC, affects the temperature of the sample. A: Rise in temperature as ice absorbs heat. B: Absorption of latent heat of fusion. C: Rise in temperature as liquid water absorbs heat. D: Water boils and absorbs latent heat of vaporization. E: Steam absorbs heat and thus increases its temperature. The above is an example of a heating curve. One could reverse the process, and obtain a cooling curve. The flat portions of such curves indicate the phase changes. 5

6 First Law of Thermodynamics: The energy can be neither created nor destroyed; it can only change forms. This principle is based on experimental observations and is known as the first law of thermodynamics or the conservation of energy principle. Mathematical Representation: Let a system absorbs Q amount of heat energy. Addition of heat energy increases the internal energy of system from U to U and some useful work is also performed by the system. Increase in internal energy is given by: U = U U and work done is W According to the first law of thermodynamics: Q = U + W () Sign Convention: Q = positive if heat is added to a system Q = negative if heat is released from a system W = positive if work is done by the system W = negative if work is done on the system SYSTEM (+) (-) Q (-) W (+) For a gas, the internal energy, U, is directly proportional to its temperature measured in Kelvin. This means that U only reflects a change in the kinetic energy of the gas molecules. Remember that the potential energy can not change except when there is a phase change: liquid to solid or liquid to vapor. Work Done by Gas (Pressure-olume work): Consider a cylinder fitted with a frictionless and weightless non-conducting piston of area of cross-section "A" as shown in Fig. An ideal gas is enclosed in the cylinder. Let the volume of gas at initial state is " ". An external pressure "P" is exerted on the piston. If we supply "Q" amount of heat to the system then it will increase its internal energy by U. "After a certain limit gas exerts pressure on the piston. If piston is free to move, it will be displaced by "h" & volume of system increases from to. We know that pressure is the force per unit area i.e. P = F/A Or F = PA... (i) We also know that the work done by the gas on the W = F d Where d = displacement of piston = h Putting the value of F and d, we get Or W = (PA) h W = P (Ah) But Ah = change or increase in volume = Hence Application of the First Law to Gas Processes, Four Special Cases: The variables that give the state of the gas P,, and T are seen to be important, along with the thermodynamic variables Q, U, and W. In principle, all six of these could change, but there are some 6

7 interrelations that reduce the freedom of independent adjustment. With one or more variables being kept constant, we select four to illustrate the versatility of thermodynamics: Quantity held constant P T Q Type of process Constant volume, Isochoric Constant pressure, Isobaric Constant temperature, Isothermal No heat flow, Adiabatic In discussing each of these cases, we shall make reference to the trends of pressure and volume on the P- diagram.. Constant volume, = 0, Isochoric: Heat supplied at constant volume is also known as "ISOCHORIC SYSTEM". An isochoric process is one in which the volume of system during the supply of heat does not change. This is achieved only when the piston of cylinder is fixed. Consider a cylinder fitted with a frictionless piston. An ideal gas is enclosed in the cylinder. The piston is fixed at a particular position so that the volume of cylinder remains constant during the supply of heat. Let Q amount of heat is added to the system. Addition of heat causes the following changes in the system: Internal energy increases from U to U. olume of the system remains unchanged. Temperature increases from T K to T K. Pressure increases from P to P. No work is performed. Applying First Law of Thermodynamics But Thus, As, Q U W W P Q U P 0 Q U P(0) Q U This expression indicates that the heat supplied under isochoric process is consumed in increasing the internal energy of the system but no work is performed.. Constant Pressure P = 0, Isobaric Process: A thermodynamic process in which pressure of the system remains constant during the supply of heat is called an ISOBARIC PROCESS. Explanation: Consider a cylinder fitted with a frictionless piston. The piston is free to move in the cylinder. An ideal gas is enclosed in the cylinder. 7

8 Let the initial volume of the system is and initial internal energy is U. Let Q P the gas is heated from T K to T K. Addition of heat causes the following changes in the system: Internal energy increases from U to U. olume of the system increases from to. Temperature increases from T K to T K. Work (W) is done by the gas on the piston. According to the first law of thermodynamics: Q U W But Thus, As, W P Q U P ( ). Constant Temperature T = 0, Isothermal Process: A thermodynamic process in which the temperature of the system remains constant during the supply of heat is called an ISOTHERMAL PROCESS. Isothermal Compression: Consider a cylinder of non-conducting walls and good heat conducting base. The cylinder is fitted with a frictionless piston. An ideal gas is enclosed in the cylinder. In the first stage pressure on the piston is increased and the cylinder is placed on a cold body. Due to compression, the temperature of the system increases but at the same time Q amount of heat is removed from the system and the temperature of the system is maintained. According to the first law of thermodynamics: Q U W Since temperature is constant, therefore, there is no change in internal energy of the system. i.e. ΔU = 0 As the work is done on the system, therefore, W is negative, Q 0 W But W P Thus, Q W Isothermal Expansion: In another situation the cylinder is placed over a hot body and the pressure on the system is decreased. Due to expansion, the temperature of the system is decreased but at the same time Q amount of heat is According to the first law of thermodynamics: Q U W Since temperature is constant, th ΔU = 0 As the work is done by the system, therefore, ΔW is positive, Q 0 W Thus, Q W 8

9 . No Heat Flow Q = 0, Adiabatic Process: A thermodynamic process in which no heat flows to or from the system and hence Q AB 0 Sometimes we say that the system is thermally isolated but this condition does not mean that temperature is constant. In fact all the thermodynamic variables P,, and T change in an adiabatic process. Adiabatic System Insulation Q = 0 Applying First Law of Thermodynamics: ΔQ = ΔU+ ΔW U AB W AB This tells us that if work is done by the system (that is, if W is positive), the internal energy of the system decreases by the amount of work. Conversely, if work is done on the system (that is, if W is negative), the internal energy of the system increases by that amount. In an idealized adiabatic process, heat cannot enter or leave the system because of the insulation. Thus, the only way energy can be transferred between the system and its environment is by work. If we remove shot from the position and allow the gas to expand, the work done by the system (the gas) is positive and the internal energy of the gas decreases. If, instead, we add shot and compress the gas, the work done by the system is negative and the internal energy of the gas increases. Reversible and Irreversible Process: Reversible Process: A process that, once having taken place, can be reversed and in doing so leaves no change in either the system or the surroundings. Net work and net heat transfer must be zero. A reversible process is an ideal that we never achieve all processes are irreversible. Why do we study reversible processes? They are ideals that we can shoot for. They are standards to which we can compare different processes. Irreversible Process: An irreversible process is such that it cannot be traced in the opposite direction by reversing the controlling factors. Causes of irreversibility: - Friction - Unrestrained expansion - Heat transfer through a finite temperature difference - Mixing of different substances - Others (chemical reactions, inelastic deformation of solids, electric resistance) During an irreversible process heat energy is always used to overcome friction. Energy is also dissipated in the form of conduction and radiation. This loss takes place whether the engine works on one direction or the reverse direction. Such energy cannot be regained. In actual practice all engine is irreversible. If electric current is passed through a wire, heat is produced. If the direction of current is reversed heat is again produced. This is also an example of an irreversible process. 9

10 Carnot Reversible Engine Heat engine are used to convert heat into mechanical work. Sadi Carnot (French) conceived a theoretical engine which is free from all defects of practical engines. Its efficiency is maximum and it is an ideal heat engine. For any engine, there are three essential requisites: (i) Source (ii) Sink and (iii) Working substance Conducting Piston Working Substance Cylinder Conducting Source Insulating Conducting Sink Stand At T Fig: Carnot s Reversible Engine At T (i) Source: The source should be at a fixed high temperature T from which the heat engine can draw heat. It has infinite thermal capacity and any amount of heat can be drawn from it at constant temperature T. (ii) Sink: The sink should be at a fixed lower temperature T to which any amount of heat can be rejected. It has infinite thermal capacity and its temperature remains constant at T. (iii) Working Substance: A cylinder with non-conducting sides and conducting bottom contains the perfect gas as the working substance. A perfect non-conducting and frictionless piston is fitted into the cylinder. The working substance undergoes a complete cyclic operation. A perfectly non-conducting stand is also provided so that the working substance can undergo adiabatic operation. A Working Principle of a Carnot Engine An ideal cycle of reversible engine operations in which a substance maintain the following four processes: - Reversible isothermal expansion - Reversible adiabatic expansion - Reversible isothermal compression - Reversible adiabatic compression adiabatic Q = 0 T isothermal Q in Net work done T isothermal Q out adiabatic Q = 0 E F G H Step-: Place the engine containing the working substance over the source at temperature T. The working substance is also at Fig: Carnon Cycle temperature T. Its pressure is P and volume is as shown by the point A in Fig. Decrease the pressure. The volume of the working substance increases. Work is done by the working substance. As the bottom is perfectly conducting P D B C 0

11 to the source at temperature T, it absorbs heat. The process is completely isothermal. The temperature remains constant. Let the amount of heat absorbed by the working substance be Q at the temperature T. The point B is obtained. Consider one gram molecule of the working substance work done from A to B (Isothermal Process) W P. d () RT W d d RT RT ln () P RT RT P Step-: Place the engine on the stand having an insulated top. Decrease the pressure on the working substance. The volume increases. The process is completely adiabatic. Work is done by the working substance at the cost of its internal energy. The temperature falls. The working substance undergoes adiabatic change from B to C. At C the temperature is T. Work done from B to C (adiabatic process) W P. d () W cons tant P Cons tant d P P P P P d cons tant R( T T) P P P P RT Here Constant Constant P C C P p v cons tant W R(T T ) () (iii) Step : Place the engine on the sink at temperature T. Increase the pressure. The work is done on the working substance. As the base is conducting to the sink, the process is isothermal. A quantity of heat Q is rejected to the sink at temperature T. Finally the point D is reached. Work done from C to D (isothermal process) W P. d (5) P RT P RT

12 RT W d d RT RT ln RT ln (6) The ve sign indicates that work is done on the working substance. (iii) Step : Place the engine on the insulating stand. Increase the pressure. The volume decreases. The process is completely adiabatic. The temperature rises and finally the point A is reached. Work done from D to A (adiabatic process) W P. d R( T T ) (7) W and W are equal and opposite and cancel each other. Net work done = W + W RT ln RT ln RT ln RT ln (8) Efficiency of a Heat Engine In common-sense language, efficiency is output input For a heat engine, we are giving heat energy as input and we are getting work as output. So the efficiency of a heat engine can be written as Wnet Efficiency of a heat engine (9) Q in From First Law of Thermodynamics we know that Q = U+ W But over the course of a cycle, the gas returns to its original state, so there can be no net change in the internal energy. Thus U = 0, so we have Q = 0+ W But the net heat added Q = Q in - Q out. So, the net work done by the gas "W net ". So we have W net = Q in - Q out. Qin Qout Qout Therefore, the efficiency of a heat engine (0) Q Q Efficiency in terms of temperature: in in

13 Since ln = ln Q Q out in T T, then we can write, Q in RT ln Q out RT ln ln Qout T Q in T ln () Solved Problems Problem : Given that the specific heat capacity of water is times that of copper, calculate the mass of copper at a temperature of 00 C required to raise the temperature of 50 g of water from 0.0 C to.0 C, assuming no energy is lost to the surroundings. Solution: Let specific heat capacity of copper, S c cal/g C. Then specific heat capacity of water, S w = S c cal/g C. Since no energy is lost to the surroundings, we can write Heat lost by the hot body = Heat gained by the cold body m c S c (00 ) = m w S w ( 0) or m c S c 66 = 50 S c or m c = 00 g (Ans.) Problem : A series of thermodynamic processes is shown in the p-diagram of adjacent figure. In process ab, 50 J of heat are added to the system, and in process bd, 600 J of heat are added. Find [] the internal energy change in process ab; [] the internal energy change in process abd; and [] the total heat added in process acd. Solution: [] Since, no volume change occurs during process ab, so W ab = 0 and U ab = Q ab = 50 J. P 8.0 x 0 Pa b d [] Process bd occurs at constant pressure, so the work done by the system during this expansion is W ab =p[ ]=[8.0x0 Pa][5.0x0 - m.0x0 - m ] = 0 J The total work for process abd is and the total heat is W abd = W ab + W bd =0 + 0 J = 0 J.0 x 0 Pa a c O.0 x 0 - m 5.0 x 0 - m Applying First law of thermodynamics, we get Q abd = Q ab + Q bd = 50 J+ 600 J = 750 J U abd = Q abd W abd = 750 J 0 J = 50 J

14 [] Since, U is independent of path, the internal energy change is the same for path acd as for path abd ; that is, The total work for the path acd is U acd = U abd = 50 J. W acd = W ac + W cd = p[ ] + 0 = [.0x0 Pa][5.0x0 - m.0x0 - m ]= 90 J Therefore, total heat added in the process acd : Q acd = U acd + W acd = 50 J + 90 J = 600 J Problem : One gram of water [ cm ] becomes 67 cm of steam when boiled at a constant pressure of atm [.0x0 5 Pa]. The heat of vaporization at this pressure is L =.56x0 6 J/kg. Compute [a] the work done by the water when it vaporizes; [b] its increase in internal energy. Solution: [a] For a constant pressure process, work done by the vaporizing water: W = p[ ] = [.0x0 5 Pa][67x0-6 m.0x0-6 m ] = 69 J [b] The heat added to water is the heat of vaporization: Therefore, change in internal energy: Q = ml v = [0 - kg] [.56x0 6 J/kg] = 56 J. U = Q W = 56 J 69 J = 087 J. So, to vaporize one gram of water, we have to add 56 J of heat. Most of this added energy [087 J] remains in the system as an increase in internal energy. The remaining 69 J leaves the system again as it does work against the surroundings while expanding from liquid to vapor. Problem : (i) How much heat is needed to take ice of mass m = 70 g at -0 o C to a liquid state at 5 o C? (ii) If we supply the ice with a total heat only 0 kj, what then is the final state of the water? Solution: Heat needed to reach the temperature 0 o C for ice, Q = m ice S ice ΔT = ( 0 ( -0)) cal = 600 cal = 50 J = 5. KJ Heat needed to melt ice, Q = m ice L ice = cal = cal = 90 J =.9 KJ Heat needed to raise the temperature of water to 5 o C, Q = m w S w ΔT = 70 (5 0) cal = 0800 cal Now the total heat required, Q = Q + Q + Q = = 7000 cal = 000 J = 0. KJ But the supplied heat is 0 KJ. Now we see that total ice will not melt because of supplying heat is smaller than required to melt total ice. Now the remaining heat = 0 5. = 9.88 KJ = 9880 J = 600 cal Now we consider that m g ice will melt. So we can write m L ice = 600 cal m 80 cal/g = 600 cal or m = 580 g So the remaining mass of the ice = = 0 g (Ans.)

15 Problem Sheet 0. The temperature of equal masses of three different liquids A, B and C are o C, 9 o C and 8 o C respectively. The temperature when A and B are mixed is 6 o C and when B and C are mixed is o C. What would be the temperature when A and C are mixed?. Determine the change in the internal energy of a system that (i) absorbs 580 cal of thermal energy and 60 J of work is done by the system. (ii) releases 70 cal of thermal energy while 680 J of external work is done on the system. (iii) maintains at a constant volume while 0 cal of heat is removed from the system.. Consider.0 kg of liquid water at 00 0 C by boiling at standard atmosphere pressure (which is.00 atm or Pa).The volume of that water changes from an initial value of m as a liquid to.765 m as steam. (i) How much work is done by the system during this process? (ii) How much energy is transferred as heat during the process? (iii) What is the change in the system s internal energy during the process?. A Carnot engine operates between 0 0 C and 00 0 C, absorbing.5x0 5 J per cycle at the higher temperature. (i) What is the efficiency of the engine? (ii) How much work per cycle is this engine capable of performing? 5. A Carnot engine has a power output of 0 kw. The engine operates between two reservoirs at o C and 550 o C. (i) How much thermal energy is absorbed per minutes? (ii) How much thermal energy is lost per minutes? 6. A total of 0.8 kg of water at 0 o C is placed in a kw electric kettle. How long a time is needed to raise the temperature of the water to 00 o C? 7. In an industrial process 0 kg of water per hour is to be heated from 0 0 C to 60 0 C.To do this, steam at 0 0 C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 80 0 C. How many kg of steam is required per hour? Specific heat of steam = 996 J/kg-K, Latent heat of steam =.68x0 6 J/Kg. 8. What would be the final temperature of the mixture when 5 gm of ice at -0 0 C are mixed with 0 gm of water at 0 0 C? (Sp. heat of ice = 0.5 cal/gm- 0 C; latent heat of fusion of ice = 80cal/gm.) 9. A 65 g iron block is heated to 5 0 C is placed in an insulated container (of negligible heat capacity) containing 0.0 g of water at C. What is the equilibrium temperature of this system? If your answer is 00 0 C, determine the amount of water that has vaporized. The average specific heat of iron over this temperature range is 560 J/(kg K), and the specific heat of liquid water is 86 J/(kg K). 0.An aluminium container of mass 50 gm contains 50 gm of ice at - 0 C.Heat is added to the system at the rate of 0 joules per seconds. What is the temperature of the system after.5 minutes? Specific heat of aluminium = 80 J/(kg o C). 5