Heat

HEAT When a hot body is placed in contact with a cold one, the former gets colder and the latter warmer. From this observation it is natural to conclude that a certain quantity of heat has passed from the hot body to the cold one. Heat is a form of energy. Heat is felt by its effects. Some of the effects of heat are : (a) Change in the degree of hotness (b) Change in length, surface area and volume (c) Change in state of a substance (d) Change in the resistance of a conductor (e) Thermo e.m.f. effect SI Unit : J (joule). Also measured in calorie. ● Calorie It is defined as the amount of heat required to raise the temperature of 1 g water by 1ºC or 1 K. ● International calorie International calorie is the amount of heat required to raise the temperature of 1 g water from 14.5 ºC to 15.5 ºC rise of temperature at pressure of 1 atm. ● kilo Calorie kilo calorie is defined as the amount of heat required to raise the temperature of 1 kg water from 14.5 ºC to 15.5 °C. (1 kcal = 1000 calorie) at pressure of 1 atm. ● British thermal unit (B.T.U.) It is the amount of heat required to raise the temperature of one pound water by 1 ºF. (1 B.T.U. = 252 calorie). Mechanical equivalent of heat According to Joule, work may be converted into heat and vice-versa. The ratio of work done W to the heat produced is always constant. = constant (J)  W = J H H W must be in joule, irrespective of nature of energy or work and H must be in calorie. J is called mechanical equivale1 of heat. It is not a physical quantity but simply a conversion factor. It converts unit of work into that of heat and vice-versa. J = 4.186 joule/cal or 4.186 × 103 joule/kcal. For rough calculations we take J = 4.2 joule/cal Specific heat (s or c ) The amount of energy required to raise the temperature of unit mass of a substance by 1ºC (or 1K) is called its specific heat. It is represented by s or c. If the temperature of a substance of mass m changes from T to T + dT when it exchanges an 1 dQ m dT The specific heat depends on the pressure, volume and temperature of the substance. For liquids and solids, specific heat measurements are most often made at a constant pressure as functions of temperature, because constant pressure is quite easy to produce experimentally. Unit : joule/kg-K, cal/ g – ºC Specific heat of water : cwater = 1 cal/g–ºC = 1 cal/g–K = 1 kcal/kg-K = 4200 joule/kg-K When a substance does not undergo a change of state (i.e., liquid remains liquid or solid remains solid), then the amount of heat required to raise the temperature of mass m of the substance by an amount  is Q = ms Variation in specific heat of water is less than 1% over the interval from 0 to 100ºC. Such a small variation is typical for most solids and liquids, so their specific heats can generally be taken to be constant over fairly large temperature ranges. amount of heat dQ with its surroundings then its specific heat is c = ● There are many possible processes to give heat to a gas. A specific heat can be associated to each such process which depends on the nature of process. ● Value of specific heats of gas can vary from zero (0) to infinity. ● Generally two types of specific heats are defined for a gas (a) Specific heat at constant volume (Cv) (b) Specific heat at constant pressure (Cp) ● These specific heats can be molar or gram. Molar heat capacity The amount of energy needed to raise the temperature of one mole of a substance by 1ºC (or 1K) is called molar heat capacity. The molar heat capacity is the product of molecular weight and specific heat i.e., 1  dQ  Molar heat capacity C = Molecular weight (M) × Specific heat(c)  C =  ·   dT  If the molecular mass of the substance is M and the mass of the substance is m then number of m m  dQ  moles of the substance = C=   M  dT  M SI Unit : J/mol-K Thermal capacity (Heat capacity) The quantity of heat required to raise the temperature of the whole substance through 1ºC is called thermal capacity. The thermal capacity of mass m of the whole substance of specific heat (s) = ms Thermal capacity = mass × specific heat Thermal capacity depends on property of material of the body and mass of the body. SI Unit : cal/°C or cal/K, Dimensions: ML2T–2K–1. Water equivalent of a body As the specific heat of water is unity so the thermal capacity of a body (ms) represents its water equivalent also. ● Mass of water having the same thermal capacity as the body is called the water equivalent of the body ● The water equivalent of a body is the amount of water that absorbs or gives out the same amount of heat as is done by the body when heated or cooled through 1ºC. Water equivalent = mass of body × specific heat of material  (w = ms). Latent heat or Hidden heat When phase of a body changes, change of phase takes place at constant temperature [melting point or boiling point] and heat released or absorbed is Q = mL where L is latent heat. Heat is absorbed if solid converts into liquid (at melting point) or liquid convert into vapours (at boiling point) and heat is released if liquid converts into solid or vapours convert into liquid, ● Latent heat of fusion It is the quantity of heat (in kilocalories) required to change 1 kg mass of a substance from solid to liquid state at its melting point. Latent heat of fusion for ice : 80 kcal/kg = 80 cal /g. ● Latent heat of vaporization The quantity of heat required to change its 1 kg mass from liquid to vapour state at its boiling point. Latent heat of vaporisation for water : 536 kcal/kg = 536 cal/g  540 cal/g ● Change of State Melting Conversion of solid into liquid state at constant temperature is known as melting. ● Boiling Evaporation within the whole mass of the liquid is called boiling. Boiling takes place at a constant temperature known as boiling point. A liquid boils when the saturated vapour pressure on its surface is equal to atmospheric pressure. Boiling point reduces on decreasing pressure. ● Evaporation Conversion of liquid into vapours at all temperatures is called evaporation. It is a surface phenomenon. Greater the temperature, faster is the evaporation. Smaller the boiling point of liquid, more rapid is the evaporation. Smaller the humidity, more is the evaporation. Evaporation increases on decreasing pressure that is why evaporation is faster in vacuum. Heat of evaporation Heat required to change unit mass of a liquid into vapour at a given temperature is called heat of evaporation at that temperature. ● Sublimation Direct conversion of solid into vapour state is called sublimation. ● Heat of sublimation Heat required to change unit mass of solid directly into vapours at a given temperature is called, heat of sublimation at that temperature. ● Camphor and ammonium chloride sublimates on heating in normal conditions. ● A block of ice sublimates into vapours on the surface of moon because of very-very low pressure on its surface ● Condensation The process of conversion from gaseous or vapour state to liquid state is known as condensation . These materials again get converted to vapour or gaseous state on heating. ● Hoar frost Direct conversion of vapours into solid is called frost, This process is just reverse of the process of sublimation. Ex. : Formation of snow by freezing of clouds. ● Regelation Regelation is the melting of ice caused by pressure and its resolidification when the pressure is removed. Ice shrinks when it melts, and if pressure is applied, deliberately promoting shrinkage, it is found that melting is thereby assisted. In other words, melting of cold ice is ordinarily effected by using the temperature, but if pressure is present to help with the shrinkage the temperature need not be raised so much. Ice heals up after being cut through by the wire. Melting takes place under the wire because pressure lowers the melting temperature Refreezing (regelation) occurs above the wire when the water escapes to normal pressure again. Increase of pressure lowers the melting (or freezing) point of water. Conversely, if a substance expands on melting, the melting point is raised by pressure. Phase of a substance The phase of a substance is defined as its form which is homogeneous, physically distinct and mechanically separable from the other forms of that substance. ● Phase diagram A phase diagram is a graph in which pressure (P) is represented along the y-axis and temperature (T) is represented along the x-axis. ● Characteristics of Phase diagram (i) (ii) (iii) (iv) Different phases of a substance can be shown on a phase diagram. A region on the phase diagram represents a single phase of the substance, a curve represents equilibrium between two phases and a common point represents equilibrium between three phases. A phase diagram helps to determine the condition under which the different phases are in equilibrium. A phase diagram is useful for finding a convenient way in which a desired change of phase can be produced. Phase diagram for water The phase diagram for water consists of three curves AB, AC and AD meeting each other at the point A, these curves divide the phase diagram into three regions. Region to the left of the curve AB and above the curve AD represents the solid phase of water (ice). The region to the right of the curve AB and above the curve AC represents the liquid phase of water. The region below the curves AC and AD represents the gaseous phase of water (i.e. water vapour). A curve on the phase diagram represents the boundary between two phases of the substance. ● ● ● Along any curve the two phases can coexist in equilibrium Along curve AB, ice and water can remain in equilibrium. This curve is called fusion curve or ice line. This curve shows that the melting point of ice decreases with increase in pressure. Along the curve AC, water and water vapour can remain in equilibrium. The curve is called vaporisation curve or steam line. The curve shows that the boiling point of water increases with increase in pressure. Along the curve AD, ice and water vapour can remain in equilibrium. This curve is called sublimation curve or hoar frost line. Triple point of water The three curves in the phase diagram of water meet at a single point A, which is called the triple point of water. The triple point of water represents the co-existence of all the three phases of water ice water and water vapour in equilibrium. The pressure corresponding to triple point of water is 6.03 × 10–3 atmosphere or 4.58 mm of Hg and temperature corresponding to it is 273.16K. ● ● ● Significance of triple point of water Triple point of water represents a unique condition and it is used to define the absolute temperature. While making Kelvin's absolute scale, upper fixed point is 273.16 K and lower 1 fixed point is 0 K. One kelvin of temperature is part of the temperature of triple point 273.16 of water. Effect of change in pressure on M.P. and B.P. for water then result  M.P. & B.P.  If P   then result  M.P. & B.P.  If P   Example : If to a given mass (m) of a solid, heat is supplied at constant rate and a graph is plotted between temperature and time, as shown in figure, it is called heating curve. ● In the region OA Rate of heat supply P is constant and temperature of solid is changing with time. So, Q = mcS T  P t = mcS T [ Q = P t] 1 T  = The slope of temperature-time curve so specific heat of solid cS  slope of line OA t specific heat (or thermal capacity) is inversely proportional to the slope of temperature-time curve. ● In the region AB Temperature is constant, so it represent so it represents change of state, i.e., melting of solid at melting point T1. At point A melting starts and at point B all soli is converted into liquid. So between A and B substance is partly solid and partly liquid. If LF is the latent heat of fusion then P(t 2  t1 ) Q = mLF  LF = [as Q = P(t2 – t1)]  LF  length of line AB m i.e., Latent heat of fusion is proportional to the length, of line of zero slope. 1 [In this region specific heat  = ] tan º ● In the region BC Temperature of liquid increases so specific heat (or thermal capacity) of liquid will be 1 inversely proportional to the slope of line BC, cL  slope of line BC ● In the region CD Temperature is constant, so it represents change of state, i.e., liquid is boiling at boiling point T2. At C all substance is in liquid state while at D is vapour state and between C and D partly liquid and partly gas. The length of line CD is proportional to latent heat of vaporisation, i.e., LV  Length of line CD. 1 [In this region specific heat  = ] tan 0º ● In the region DE The line DE represents gaseous state of substance with its temperature increasing linearly with time. The reciprocal of slope of line will be proportional to specific heat or thermal capacity of substance in vapour state. ● ● Law of Mixtures : Principle of calorimetry When two bodies at different temperatures are mixed, heat will be transferred from body at higher temperature to a body at lower temperature till both acquire same temperature. The body at higher temperature releases heat while body at lower temperature absorbs it, so that Heat lost = Heat gained (Principle of calorimetry) Principle of calorimetry represents the law of conservation of heat energy. Temperature of mixture (T) is always lower temperature (TL) and higher temperature (TH), TL T TH The temperature of mixture can never be lesser than lower temperature (as a body cannot be cooled below the temperature of cooling body) and greater than higher temperature (as a body cannot be heated above the temperature of heating body). Further more usually, rise in temperature of one body may not be equal to the fall in temperature of the other body though heat gained by one body is equal to the heat lost by the other. Illustrations Illustration 13. 5 g ice at 0ºC is mixed with 5 g of steam at 100ºC. What is the final temperature? Solution: Heat required by ice to raise its temperature to 100ºC, Q1 = m1L1 + m1c11 = 5 × 80 + 5 × 1 × 100 = 400 + 500 = 900 cal Heat given by steam when condensed Q2 = m2L2 = 5 × 536 = 2680 cal As Q2 > Q1. This means that whole steam is not even condensed. Hence temperature of mixture will remain at 100°C. Illustration 14. A calorimeter of heat capacity 100 J/K is at room temperature of 30ºC. 100 g of water 40ºC of specific heat 4200 J/kg-K is poured into the calorimeter. What is the temperature of water in calorimeter ? Solution: Let the temperature of water in calorimeter is t. Then heat lost by water = heat gained by calorimeter (0.1) × 4200 × (40 – t) = 100 (t – 30)  420 × 40 – 420 t = 100 t – 3000  t = 38.07ºC Illustration 15. Find the quantity of heat required to convert 40 g of ice at –20ºC into water at 20ºC. Given Lf = 0.336 × 106 J/kg. Specific heat of ice = 2100 J/kg-K, specific heat of water = 4200 J/kg-K Solution: Heat required to raise the temperature of ice from –20°C to 0ºC = 0.04 × 2100 × 20 = 1680 J Heat required to convert the ice into water at 0ºC = mL = 0.04 × 0.336 × 106 = 13440 J Heat required to heat water from 0ºC to 20ºC = 0.04 x 4200 x 20 = 3360 J Total heat required = 1680 + 13440 + 3360 = 18480 J Illustration 16. Steam at 100ºC is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15ºC till the temperature of the calorimeter and its contents rises to 80ºC. What is the mass of steam condensed? Latent heat of steam = 536 cal/g. Solution: Heat required by (calorimeter + water) Q = (m1c1 + m2c2)  = (0.02 + 1.1 × 1) (80 – 15) = 72.8 kcal If m is mass of steam condensed, then heat given by steam Q = mL + mc  = m × 536 + m × 1 × (100 – 80) = 556 m 556 m = 72.8 72.8  Mass of steam condensed m = = 0.130 kg 556 Illustration 17. An iron block of mass 2 kg, fall from a height 10 m. After colliding with the ground it loses 25% energy to surroundings. Then find the temperature rise of the block. (Take specific heat of iron. 470 J/kg ºC) Solution: 3 3 10 10 mS = mgh   = = 159ºC 4 4  470 Illustration 18. The temperature of equal masses of three different liquids A, B, and C are 10ºC, 15ºC and 20°C respectively· The temperature when A and Bare mixed is 13ºC and when B and C are mixed, it is16ºC. What will be the temperature when A and C are mixed? Solution: when A and B are mixed mS1 × (13 – 10) = m × S2 × (15 – 13) 3S1 = 2S2 ……(1) when B and C are mixed S2 × 1 = S3 × 4 ……(2) when C and A are mixed S1( – 10) = S3 × (20 – ) ……(3) by using equation (1), (2) and (3) 140 we get  = ºC 11 Illustration 19. 5 kg of steam at 100ºC is mixed with 10 kg of ice at 0ºC. Choose incorrect alternative (Give swater = 1 cal/gºC, LF = 80 cal/g, LV = 540 cal/g) (A) Equilibrium temperature of mixture is 160°C (B) Equilibrium temperature of mixture is 100°C 1 (C) At equilibrium, mixture contains 13 kg of water 3 2 (D) at equilibrium, mixture contains 1 kg of steam 3 Ans. (A) Solution: Required heat Available heat 10 kg ice (0°C) 5 kg steam (100°C) So available heat is more than required heat therefore final temperature will be 100°C. 800  1000 10 10 40 1  kg. Total mass of water = 10 +   13 kg Mass of heat condensed = 540 3 3 3 3 10 5 2   1 kg Total mass of steam = 5 – 3 3 3 1. BEGINNER’S BOX - 2 A bullet of mass 10 g is moving with speed 400 m/s. Find its kinetic energy in calories ? 2. Calculate amount of heat required to convert 1 kg steam from 100°C to 200°C steam ? 3. Calculate heat required to raise the temperature of 1 g of water by 1°C ? 4. 420 J of energy supplied to 10 g of water will raise its temperature by ? 5. The ratio of the densities of the two bodies is 3 : 4 and the ratio of specific heats is 4 : 3. Find the ratio of their thermal capacities for unit volume ? 6. Heat releases by 1 kg steam at 150°C if it is converted into 1 kg water at 50°C. 7. 200 g water is filled in a calorimetry of negligible heat capacity. It is heated till its temperature is increase by 20°C. Find the heat supplied to the water. 8. A bullet of mass 5 gm is moving with speed 400 m/s strike a target. Then calculate rise of temperature of bullet. Assuming all the lose in kinetic energy is converted into heat energy of bullet is its specific heat is 500 J/kg°C. 9. 1 kg ice at –10°C is mixed with 1 kg water at 100°C. Then find equilibrium temperature and mixture content. 10. 540 g of ice at 0°C is mixed with 540 g of water at 80°C. The final temperature of the mixture is (Given latent heat of fusion of ice = 80 cal/g and specific heat capacity of water = 1 cal/g°C) (A) 0°C (B) 40°C (C) 80°C (D) less than 0°C