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https://docs.google.com/document/d/11lxKFSHDb70C9aXNZEIGjqf2lBjZweYn/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true 12-Transference of heat SYLLABUS Thermal expansion of solids, liquids and gases; Calorimetry, latent heat; Heat conduction in one dimension; Elementary concepts of convection and radiation; Newton’s law of cooling; bulk modulus of gases; Blackbody radiation: absorptive and emissive powers; Kirchhoff’s law, Wien’s displacement law, Stefan’s law, Specific heat of a liquid using calorimeter (M and E) Thermal Expansion When a body is heated, it expands in terms of length, area & volume and temperature rises. In a solid, molecules can only have thermal agitation (random vibrations). As temperature of a body increases, the vibrations of molecules will become fast and due to this the rate of collision among neighbouring molecules increases, it develops a thermal stress in the body and due to this the intermolecular separation increases which results in thermal expansion of body. At ordinary temperatures, the atoms in a solid oscillate about their equilibrium position with an amplitude of approximately 10–11 m. The average spacing between the atoms is about 10–10 m. As the temperature of solid increases, the atoms oscillate with greater amplitudes, as a result the average separation between them increases. Consequently the object expands. In the similar way the block diagram shown in Fig. explains the way how thermal expansion takes place. Thermal expansion of a substance can be classified in three broad categories, these are (i) Linear Expansion (ii) Superficial Expansion and (iii) Cubical Expansion or Volume Expansion (i) Linear Expansion: Consider a rod of length l1 at a temperature T1. Let it be a heated to a temperature T2 and the increased length of the rod be l2 then l2= l1 (1+ αΔt) α = Coefficient of linear expansion and Example 1. The density of substance at 0°C is 10 g cm-3 and at 100°C, its density is 9.7g cm-3. The coefficient of linear expansion of substance is Solution: (ii) Superficial Expansion (expansion in surface area): If A1 is the area of solid at T1 °C and A2 is the area at T2°C . Then A2 = A1(1 + Δt ) Coefficient of superficial (areal) expansion and (iii) Volume expansion (Part A: expansion in solids): If V1 is the volume of solid at T1°C and V1 is the volume at T2 °C then coefficient of cubical (volume) expansion and Note: For isotopic solids: Relation between expansion coefficient are As temperature increases, density of solid decreases. If d1 is the density at T1°C, d2 is the density at T2°C then (iv) Expansion of gases (Part B: expansion in gases): Pressure coefficient of a gas is the ratio of increase in pressure for 1C rise in temperature to the pressure at 0C, provided the volume of the gas is kept constant. p = Where p = pressure coefficient Pt = pressure at tC Po= pressure at 0C Volume coefficient of a gas is similarly defined as (pressure being kept constant) v = Where v = volume coefficient Vt = volume at tC Vo= volume at 0C Experiments have shown the value of p (or v) is the same for all gases and equal to 1/273 per degree celsius, i.e. p = v = perC or per K, Where K stands for absolute of kelvin temperature. Example 2. A grid iron pendulum consists of 5 iron rods and 4 brass rods. What will be the length of each brass rod if the length of each iron rod is 1 m ? (αFe= 12 × 10-6/°C, αbrass = 20 × 10-6/°C) Solution: There are 5 iron rods and 4 brass rods in the pendulum. The number of iron in one half of the pendulum including the central rod is 3. The number of brass rods is 2. Since the pendulum is compensated, ERROR in measurament DUE TO THERMAL EXPANSION HEATING A METALLIC SCALE A metallic scale (linear) expands in length when heated. As a result all the markings are displaced from their usual (correct) positions. A reading of l unit on a heated scale is equivalent to an actual length of l , where is coefficient of linear expansion of material of scale, and Δt is the temperature of the heated scale. ⇒ If the reading is x, actual length = x(1 + α Δt) actual length = reading (1 + α Δt) DIFFERENCE OF LENGTHS OF TWO RODS Consider two rods 1 and 2 of lengths l1 and l2. Let they be heated through a temperature . If l’1 and l’2 are their expanded lengths, then: l’2 = l2 where is coefficient of linear expansion of rod 2 l’1 = l1 where is coefficient of linear expansion of rod 1 Since the difference of length of two rods is constant Example3. Two rods of length l1 and l2 are made of materials whose coefficients of linear expansion are α1 and α2 respectively. If the difference between the two lengths is independent of temperature, then relation between length and linear expansion coefficients. Solution: TIME PERIOD OF PENDULUM Time period (T) of a sample pendulum of length l is given by If there is a rise in temperature by Δt, length of the pendulum increases and hence the time period increase. As a result the clock slows down. If l0 be the length of the pendulum and corresponding time period be T0 then If the pendulum be heated by Δt (rise in temperature), the new time period T1 is: ⇒ as α is very small ⇒ The above relation gives the time lost per second by the pendulum clock. If Δt is the fall in temperature, same equation will give the time gained by the clock per second as its oscillation will become faster due to reduction in its length. Example 4: A clock with a metallic pendulum is 5 second fast each day at a temperature of 15and 10 seconds slow each day at a temperature of 30. Find coefficient of linear expansion for the metal. Solution: Time lost or gained per second by a pendulum clock is given by ( Here is difference of temperature) Here temperature is higher then graduation temperature thus clock will loose time and if it is lower then graduation temperature will gain time. If graduation temperature of clock is T0 then we have. At 15°C, clock is gaining time, thus 5 = ……(1) At 30°C clock is loosing time, thus 10 = ……(2) Dividing equation (2) by (1), we get 2(T0 – 15) = (30 – T0) or T0 = 20°C Thus from equation (1) STRESS IN OBJECTS DUE TO THERMAL EXPANSION If cross–sectional area of wire is A and F be the tension developed in wire due to stretching. Thus the stress developed in the wire due to this tension F is given as Strees = = F / A …..(1) Strain produced in wire due to its elastic properties is Strees = …..(2) If young’s modulus of the material of wire is Y, we have or or …..(3) Equation - (3) gives the expression for tension in the wire due to decrease in its temperature by ΔT. This result gives the tension in wire if initially wire is just taught between clamps. If it already has some tension in it then this expression will give the increment in tension in the wire. Example 5. A uniform metal rod of 2 m m2 cross-section is heated from 0°C to 20°C. The coefficient of liner expansion of the rod is 12 × 10-6 per °C, Y=1011 N/m2. The energy stored per unit volume of the rod is solution: Energy per unit volume SPECIFIC HEAT When heat energy flows into a substance, the temperature of the substance usually rises. The heat required to raise the temperature of unit mass of a body through 1 oC or ( 1 oK ) is called specific heat capacity or simply specific heat of the material of the body. If ΔQ heat changes the temperature of mass m by ΔT. … (1) The SI unit of specific heat is J/kg K. Heat is so frequently measured in calories, therefore the practical unit cal/g C is also used quite often. The specific heat capacity of water is approximately 1 cal/g °C. From Eq. (1), we can define the specific heat of a substance as “the amount of energy needed to raise the temperature of unit mass of that substance by 1°C (or 1 K)”. A closely related quantity is the Molar heat capacity C. It is defined as, … (2) where n is the number of moles of the substance. If M is the molecular mass of the substance, then n = were m is the mass of the substance and, C = … (3) The SI units of C is J / mole K. Key points: (a) It depends on nature of material of body. Dulong and petit has found formula for elemental solids that (with few exceptions such as carbon) Atomic weight × Specific heat = 6 cal / oC So, heavier the element lesser will be the specific heat, i.e., CHg < CCu< CAl (b) Specific heat of a substance also depends on temperature (particularly at low temperatures) the variation of specific heat with temperature for wateris shown in Fig (A) for metals in Fig (B). This temperature dependence of specific heat is usually neglected. Fig. (A) Water Fig. (B) Metals (c) Specific heat also depends on the state of substance, i.e., solid, liquid or gas. e.g., specific heat of solid copper will be different from that of liquid copper. In case of water Specific heat of In cal/g °C In SI units, i.e., J/kg K Ice (solid) 0.5 2100 Water (liquid) 1 4200 Steam (gas) 0.47 1970 (d) If a substance is undergoing change of state which takes place at constant temperature (called isothermal change) , specific heat = = [as ΔT = 0] i.e., specific heat of a substance at its melting pint or boiling point or isothermal change is infinite. (e) Specific heat is found to be maximum for hydrogen (3.5 cal/gm °C) then for water (1 cal/gm °C = 4200 J/kg K). For all other substances specific heat is lesser than 1 cal/gm °C and is minimum for radon and actinium (= 0.22 cal/gam °C) (f) If the temperature of a body changes without transfer of heat with the surroundings (adiabatic change) as in shaking a liquid or compressing a gas, c = = 0 [as Q = 0] i.e., specific heat of a substance, when it undergoes adiabatic change, is zero. (g) Specific heat of a substance can also be negative. Negative specific heat means that in order to raise the temperature, a certain quantity of heat is to be withdrawn from the body. Specific heat of saturated water vapours is negative. (h) When specific heats are measured, the values obtained are also found to depend on the conditions of the experiment. In general measurements made at constant pressure are different from those at constant volume. For solids and liquids this difference is very small and usually neglected. The specific heat of gases are quite different under constant pressure condition (cp) to constant volume condition (cv). (i) As by definition c = (Q/m ΔT), heat required to change the temperature of m gm of a substance through ΔT: Q = mc ΔT and as ΔT = (Q/mc), greater the specific heat of a substance lesser will be the change in temperature for a given mass when same amount of heat is supplied. Now as specific heat of water is very large (1 cal/g °C), by absorbing or releasing large amount of heat its temperature changes by small amounts. This is why, it is used in hot water bottles or as coolant in radiators. This is also how the sea moderates the climate of nearby coastal land. WATER EQUIVALENT Water–equivalent of a body is the mass of water which when given same amount of heat as to the body, changes the temperature of water through same range as that of the body, i.e., W = (m × c) gm The unit of water equivalent W is gm while its dimensions [M]. Units and dimensions of some physical–quantities used in heat are given below in a tabular form. S. No. Physical quantity Symbol Dimensions Units SI CGS 1. Heat Q [ML2T–2] Joule calorie 2. Specific–heat c [L2T–2θ–1] J/kg K cal/gm C° 3. Molar sp. heat C [ML2T–2θ–1μ–1] J/mol K cal/mol C° 4. Latent heat L [L2T–2] J/kg cal/gm 5. Thermal capacity Tc [ML2T–2θ–1] J/K cal/C° 6. Water–equivalent W [M] kg gm PHASE CHANGES AND LATENT HEAT Suppose that we slowly heat a cube of ice whose temperature is below 0°C at atmospheric pressure, what changes do we observe in the ice? Initially we find that its temperature increases according to equation Q = mc(T2 – T1). Once 0°C is reached, the additional heat does not increase the temperature of the ice. Instead, the ice melts and temperature remains at 0°C. The temperature of the water then starts to rise and eventually reaches 100°C, whereupon the water vaporizes into steam at this same temperature. During phase transitions (solid to liquid or liquid to gas) the added heat causes a change in the positions of the molecules relative to one another, without affecting the temperature. The heat necessary to change a unit mass of a substance from one phase to another is called the latent heat (L). Thus, the amount of heat required for melting and vaporizing a substance of mass m are given by, Q = mL … (1) For a solid-liquid transition, the latent heat is known as the latent heat of fusion (Lf) and for the liquid-gas transition, it is known as the latent heat of vaporization (Lv). PRINCIPLE OF CALORIMETRY When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat will be transferred from body at higher temperature to 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, i.e. principle of calorimetry represents the law of conservation of heat energy. While, using this principle always keep in mind that: (a) Temperature of mixture (T) is always ≥ lower temperature (TL) and ≤ higher temperature (TH), i.e., TL ≤ T ≤ TH i.e. the temperature of mixture can never be lesser than lower temperatures (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 when there is no chemical reaction). (b) When temperature of a body changes, the body releases heat if its temperature falls and absorbs heat when its temperature rises. The heat released or absorbed by a body of mass m is given by: Q = mc ΔT Where c is specific heat of the body and ΔT change in its temperature in C° or K. (c) When state of a body changes, change of state takes place at constant temperature [m.pt. or b.pt.] and heat released or absorbed is given by Q = mL Where L is latent heat. Heat is absorbed if solid converts into liquid (at m.pt.) or liquid converts into vapours (at b.pt.) and is released if liquid converts into solid or vapours converts into liquid. HEAT TRANSFER Heat can be transferred from one place to the other by any of three possible ways: conduction, convection and radiation. In the conduction, convection processes, a medium is necessary for the heat transfer. Radiation, however, does no have this restriction. This is also the fastest mode of heat transfer, in which heat is transferred from one place to the other in the form of electromagnetic radiation. (i) Conduction Figure shows a rod whose ends are in thermal contact with a hot reservoir at temperature T1 and a cold reservoir at temperature T2. The sides of the rod are covered with insulating medium, so the transport of heat is along the rod, not through the sides. The molecules at the hot reservoir have greater vibrational energy. This energy is transferred by collisions to the atoms at the end face of the rod. These atoms in turn transfer of heat through a substance in which heat is transported without direct mass transport is called conduction. Most metals use another, more effective mechanism to conduct heat. The free electrons, which move throughout the metal can rapidly carry energy from the hotter to cooler regions, so metals are generally good conductors of heat. The presence of ‘free’ electrons also causes most metals to be good electrical conductors. A metal rod at 5°C feels colder than a piece of wood at 5°C because heat can flow more easily from your hand into the metal. Heat transfer occurs only between regions that are at different temperature, and the rate of heat flow is . This rate is also called the heat current, denoted by H. Experiments show that the heat current is proportional to the cross-section area A of the rod and to the temperature gradient , which is the rate of change of temperature with distance along the bar. In general H = … (1) The negative sign is used to make a positive quantity since is negative. The constant k, called the thermal conductivity is measure of the ability of a material to conduct heat. A substance with a large thermal conductivity k is a good heat conductor. The value of k depends on the temperature, increasing slightly with increasing temperature, but k can be taken to be practically constant throughout a substance if the temperature difference between its ends is not too great. Let us apply Eq. (1) to a rod of length L and constant cross sectional area A in which a steady state has been reached. In a steady state the temperature at each point is constant in time. Hence, Therefore, the heat ΔQ transferred in time Δt is …(2) Thermal Resistance (R) Eq. (2) in differential form can be written as …(3) Here, ΔT = temperature difference (T.D) and = thermal resistance of the rod. (ii) Convection Although conduction does occur in liquids and gases also, heat is transported in these media mostly by convection. In this process, the actual motion of the material is responsible for the heat transfer. Familiar examples include hot-air and hot-water home heating systems, the cooling system of an automobile engine and the flow of blood in the body. You probably have warmed your hands by holding them over an open flame. In this situation, the air directly above the flame is heated and expands. As a result, the density of this air decreases and then air rises. When the movement results from differences in density, as with air around free, it is referred to as natural convection. Air flow at a beach is an example of natural convection. When the heated substance is forced to move by a fan or pump, the process in called forced convection. If it were not for convection currents, it would be very difficult to boil water. As water is heated in a kettle, the heated water expands and rises to the top because its density is lowered. As the same time, the denser, cool water at the surface sinks to the bottom of the kettle and is heated. Heating a room by a radiator is an example of forced convection. Ingen Hausz Experiment: Ingen Hausz provided a method to compare the thermal conductivities of different materials. He took wax coated rods of different materials but of the same area. One end of the rods is kept in a hot water bath and the other end is kept at the temperature of surroundings. If 1, 2, 3 . . . represent the lengths upto which the wax has melted and K1, K2, K3 . . . are their thermal conductivities, then = . . . = constant or Example-6. A wall is made of two equally thick layers A and B of different materials. The thermal conductivity of A is twice that of B. In the steady state, the temperature difference across the wall is 36°C. The temperature difference across the layer A will be Solution: (iii) Radiation The third means of energy transfer is radiation which does not require a medium. The best known example of this process is the radiation from sun. All objects radiate energy continuously in the form of electromagnetic waves. The rate at which an object radiates energy is proportional to the fourth power of its absolute temperature. This is known as the Stefan’s law and is expressed in equation form as P = σAeT4 Here P is the power in watts (J/s) radiated by the object, A is the surface area in m2, e is a fraction between 0 and 1 called the emissivity of the object and σ is a universal constant called Stefan’s constant, which has the value σ = 5.67 × 10–8 W/m2-K4 BLACK BODY RADIATION (i) Perfectly black body A BODY THAT ABSORBS ALL THE RADIATION INCIDENT UPON IT AND HAS AN EMISSIVITY EQUAL TO 1 IS CALLED A PERFECTLY BLACK BODY. A BLACK BODY IS ALSO AN IDEAL RADIATOR. IT IMPLIES THAT IF A BLACK BODY AND AN IDENTICAL ANOTHER BODY ARE KEPT AT THE SAME TEMPERATURE, THEN THE BLACK BODY WILL RADIATE MAXIMUM POWER AS IS OBVIOUS FROM EQUATION P = EAσT4 ALSO. BECAUSE E = 1 FOR A PERFECTLY BLACK BODY WHILE FOR ANY OTHER BODY E < 1. MATERIALS LIKE BLACK VELVET OR LAMP BLACK COME CLOSE TO BEING IDEAL BLACK BODIES, BUT THE BEST PRACTICAL REALIZATION OF AN IDEAL BLACK BODY IS A SMALL HOLE LEADING INTO A CAVITY, AS THIS ABSORBS 98% OF THE RADIATION INCIDENT ON THEM. (ii) Absorptive power ‘a’ “It is defined as the ratio of the radiant energy absorbed by it in a given time to the total radiant energy incident on it in the same interval of time.” As a perfectly black body absorbs all radiations incident on it, the absorptive power of a perfectly black body is maximum and unity. (iii) Spectral absorptive power ‘aλ’ The absorptive power ‘a’ refers to radiations of all wavelengths (or the total energy) while the spectral absorptive power is the ratio of radiant energy absorbed by a surface to the radiant energy incident on it for a particular wavelength λ. It may have different values for different wavelengths for a given surface. Let us take an example, suppose a = 0.6, aλ = 0.4 for 1000 Å and aλ = 0.7 for 2000 Å for a given surface. Then it means that this surface will absorbs only 60% of the total radiant energy incident on it. Similarly it absorbs 40% of the energy incident on it corresponding to 1000 Å and 70% corresponding to 2000 Å. The spectral absorptive power aλ is related to absorptive power a through the relation (iv) Emissive power ‘e’ (Don’t confuse it with the emissivity e which is different from it, although both have the same symbols e). “For a given surface it is defined as the radiant energy emitted per second per unit area of the surface.” It has the units of W/m2 or J/s–m2. For a black body e=σT4. (v) Spectral emissive power ‘eλ’ “It is emissive power for a particular wavelength λ.” Thus, Kirchhoff’s law: “According to this law the ratio of emissive power to absorptive power is same for all surfaces at the same temperature.” Fig. Hence, but (a)black body = 1 and (e)black body = E (say) Then, Similarly for a particular wavelength λ, Here E = emissive power of black body at temperature T = σT4 From the above expression, we can see that eλ ∝ aλ i.e., good absorbers for a particular wavelength are also good emitters of the same wavelength. COOLING BY RADIATION Consider a hot body at temperature T placed in an environment at a lower temperature T0. The body emits more radiation than it absorbs and cools down while the surrounding absorb radiation from the body and warm up. The body is losing energy by emitting radiations at a rate. P1 = eAσT4 and is receiving energy by absorbing radiations at a rate P2 = aAσT04 Here ‘a’ is a pure number between 0 and 1 indicating the relative ability of the surface to absorb radiation from its surroundings. Note that this ‘a’ is different from the absorptive power ‘a’. In thermal equilibrium, both the body and the surrounding have the same temperature (say Tc) and, P1 = P2 or eAσTc4 = aAσTc4 or e = a Thus, when T > T0, the net rate of heat from the body to the surroundings is, eAσ (T4 – T04) or eAσ (T4 – T04) Rate of cooling or ∝ (T4 – T04) NEWTON’S LAW OF COOLING According to this law, if the temperature T of the body is not very different from that of the surroundings T0, then rate of cooling – is proportional to the temperature difference between them. T0 prove it let us assume that T = T0 + ΔT So that T4 = (T0 + ΔT)4 = (from binomial expansion) ∴ (T4 – ) = 4 (ΔT) or (T4 – ) ∝ ΔT (as T0 = constant) Now, we have already shown that rate of cooling and here we have shown that , if the temperature difference is small. Thus, rate of cooling or as dT = dθ or ΔT = Δθ Variation of temperature of a body according to Newton’s law Suppose a body has a temperature θi at time t = 0. It is placed in an atmosphere whose temperature is θ0. We are interested in finding the temperature of the body at time t, assuming Newton’s law of cooling to hold good or by assuming that the temperature difference is small. As per this law, rate of cooling ∝ temperature difference or or Here α = is a constant ∴ ∴ From this expression we see that θ = θi at t = 0 and θ = θ0 at t = ∞, i.e. temperature of the body varies exponentially with time from θi to θ0 (<θi). The temperature versus time graph is as shown in Fig. Note: If the body cools by radiation from θ1 to θ2 in time t, then taking the approximation and θ = θav = The equation becomes This form of the low helps in solving numerical problems related to Newton’s law of cooling. WEIN’S DISPLACEMENT LAW At ordinary temperatures (below about 600°C) the thermal radiation emitted by a body is not visible, most of it is concentrated in wavelengths much longer than those of visible light. Figure shows how the energy of a black body radiation varies with temperature and wavelength. As the temperature of the black body increases, two distinct behaviours are observed. The first effect is that the peak of the distribution shifts to shorter wavelengths. This shift if found to obey the following relationship called Wein’s displacement law λmaxT = b Here b is a constant called Wein’s constant. The value of this constant in SI unit is 2.898 × 10–3 m–K. Thus, Here λmax is the wavelength corresponding to the maximum spectral emissive power eλ. The second effect is that the total amount of energy the black body emits per unit area per unit time (= σT4) increases with fourth power of absolute temperature T. This is also known as the emissive power. We know e = = Area under eλ–λ graph = σ T4 or Area ∝ T4 A2 = (2)4A1 = 16A1 Thus, if the temperature of the black body is made two fold, λmax remains half while the area becomes 16 times. Example-7. Estimate the surface temperature of sun. Given for solar radiations, λm = 4753 A° Solution: From Wien’s displacement law λmT = b ∴ T = = 6097°K. Example-8. Experimental investigations show that the intensity of the solar radiation is maximum for wavelength in the visible region. Estimate the surface temperature of the sun. Assume the sun to be a black body. Wien’s constant (b) = . Solution : Wien’s law states that = constant = b ∴ = 6060 K Example-9: A body cools from 62°C to 50°C in 10 minutes and to 42°C in the next 10 minutes. Find the temperature of surroundings. Solution: For the first ten minutes, = –1.2°C/min and ΔT = ⇒ −1.2°C/min = −KA (56−T0)°C …(1) Similarly for the next ten minutes = −0.8 °C/min and ΔT = ⇒ −0.8°C/min = −KA (46−To)°C …(2) Dividing (1) by (2) ⇒ T0 = 26°C. measurment OF SPECIFIC HEAT CAPACITY & error analysis As shown in the figure Regnault’s apparatus to determine the specific heat capacity of a solid heavier than water, and insoluble in it. A wooden partitions P separates a steam chamber O and A calorimeter C. The steam chamber O is a double walled cylindrical vessel. Steam can be passed in the space between the two walls through an inlet A and it can escape out through an outlet B. The upper part of the vessel is closed by a cork. The given solid may be suspended the vessel is closed by a cork. The given solid may be suspended in the vessel by a thread passing through the cork. A thermometer T1 is also inserted into the vessel to record the temperature of the solid. The stem chamber is kept on a wooden platform with a removable wooden disc D closing the bottom hole of the chamber. To start with, the experimental solid (in the form of a ball or a block) is weighed and then suspended in the steam chamber. Steam is prepared by boiling water in a separate boiler and is passed through the steam chamber. A calorimeter with a stirrer is weighed and sufficient amount of water is kept in it so that the solid may be completely immersed in it. The calorimeter is again weighed with water to get the mass of the water. The initial temperature of the water is noted. When the temperature of the solid becomes constant (say for 15 minutes), the partition P is removed. The calorimeter is taken below the steam chamber, the wooden disc D is removed and the thread is cut to drop the solid in the calorimeter. The calorimeter is taken to its original place and is stirred. The maximum temperature of the mixture is noted. Calculation: Let the mass of the solid =m1 mass of the calorimeter and the stirrer =m2 mass of the water =m3 specific heat capacity of the solid =s1 specific heat capacity of the material of the calorimeter(and stirrer) =s2 specific heat capacity of water =s3 initial temperature of the solid =1 initial temperature of the calorimeter, stirrer and water =2 final temperature of the mixture = We have heat lost by the solid =m1s1(1-) heat gained by the calorimeter (and the stirrer) =m2s2(-2) Assuming no loss of heat to the surrounding the heat lost by the solid goes into the calorimeter stirrer and water. Thus m1s1(1 – ) = m2s2 (–2)+ m2s3( – 2) ……(1) or s1 = Knowing the specific heat capacity of water (s3 = 4186 J/kg-K) and that of the material of the calorimeter and the stirrer (s2=389 J/kg-K if the material be copper), one can calculate s1.Specific heat capacity of a liquid can also be measured with the Regnault apparatus. Here a solid of known specific heat capacity s1 is used and the experimental liquid is taken in the calorimeter in place of water. The solid should be denser than the liquid. Using the same procedure and with the same symbols we get an equation identical to equation (1) above, that is, m1s1(1– ) = m2s2 (–2) + m3s3( – 2) in which s3 is the specific heat capacity of the liquid. We get S3 = Error analysis After correcting for systematic errors, equation (1) is used to estimate the remaining errors. OBJECTIVE 1. A piece of metal floats on mercury. The coefficients of volume expansion of the metal and mercury are 1 and 2 respectively. If their temperature is increased by T, the fraction of the volume of metal submerged in mercury changes by a (A) (B) (C) (D) Solution: The correct choice is (A). It follows from the fact that the volumes of metal and mercury increase to V0 (1 + 1 T) and (1+ 2 T) respectively; Vo being the initial volume. 2. The coefficient of expansion of a crystal in one direction (x–axis) is 2.0  10–6 K–1 and that in the other two perpendicular (y–and z–axes) direction is 1.6  10–6 K–1. What is the coefficient of cubical expansion of the crystal? (A) 1.6  10–6 K–1 (B) 1.8  10–6 K–1 (C) 2.0  10–6 K–1 (D) 5.2  10–6 K–1 Solution: (D) Coefficient of cubical expansion is  = x + y + z = x + 2 y ( y = z) = 2.0  10–6 + 2 × 1.6 × 10–6 = 5.2  10–6 K–1 Hence the correct choice is (D). 3. uniform metal rod of length L and mass M is rotating with angular speed  about an axis passing through one of the ends and perpendicular to the rod. If the temperature increases by t C, then the change in its angular speed is proportional to (A) (B) (C) 2 (D) Solution: (B) At tC, the length of the rod becomes L’= L (1 + αt), where  is the coefficient of linear expansion. From the law of conservation of angular momentum, we have I  = I’’ or ML2  = ML’2 ’ or = Now, for a given value of t, (1+ at)–2 is constant, say k. = k or or ’ – = (k – 1)  i.e. (’ – ) ∝ . Hence the correct choice is (B). 4. o cylindrical rods or lengths l1 and l2, radii r1 and r2 have thermal conductivities k1 and k2 respectively. The ends of the rods are maintained at the same temperature difference. If l1 = 2l2 and r1 = r2/2, the rates of heat flow in them will be the same if k1/k2 is: (A) 1 (B) 2 (C) 4 (D) 8 Solution: (D) The rate of heat flow in rods A and B are and Q1 = Q2, if or = 2 × (2)2 = 8 Hence the correct choice is (D). 5: The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively, are T2 and T1 (T2 > T1). The rate of heat transfer through the slab, in a steady state is f, with f equal to (see Fig.) (A) 1 (B) 1/2 (C) 2/3 (D) 1/3 Solution: (C) Let A be the area of each slab. In the steady state, the rate of heat flow through the composite slab is given by … (1) Given l1 = x, l2 = 4x, K1 = K and K2 = 4K. Using these values in (1) we get Comparing this with the given rate of heat transfer, we get f = . Hence the correct choice is (C). 6. Three identical rods A,B and C of equal lengths and equal diameters are joined is series as shown in following fig. Their thermal conductivities are 2K,K and K/2 respectively. Calculate the temperature at two junction points. (A) 85.7, 57.1°C (B) 80.85, 50.3°C (C) 77.33, 48.3°C (D) 75.8,49.3°C Solution: (A) ith 1 = ith 2 = ith 3 (100 – T1) = (T1 – T2) = T1 = (100 – T1) = r T1 = = 85.7°C and T2 = = 57.1°C 7: Three rods made of the same material ane having the same cross-section have been joined as shown in. Each rod is of the same length. The left and right ends are kept at 0°C and 90°C respectively. The temperature of the junction of the three rods (A) 45°C (B) 60°C (C) 30°C (D) 20°C Solution: I=I1+I2 or 3T = 180°C or T = 60°C 8: A solid copper sphere (density p and specific heat C) of radius r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. What is the required for temperature of the sphere to drop to 100 K? (A) (B) (C) (D) Solution: (A) According to Stefan’s law P = eAσT4 or or t = seconds 9: A double pane window used for insulating a room thermally from outside consists of two glass sheets each pf area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state the room glass interface and glass outdoor interface are at 27°C and 0°C respectively. Calculate the rate of flow of heat through the windowpane. Also find the flow of heat through the windowpane. Also find the temperature of other interface if, conductivities of glass and air are 0.8 and 0.08 Wm-1 K-1 respectively. (A) 0.72C (B) 0.52C (C) 0.192C (D) 0.32C Solution: (B) Req = = A = 1 m2 Req = = 41.5 W 41.5 = 0.8 12 or 1 = 26.48C 41.5 = or 2 = 0.52C 10: A space object has the space of a sphere of radius R. Heat sources ensure that the heat evolution at a constant rate is distributed uniformly over its volume. The amount of heat liberated by a thermodynamic temperature. In what proportion would the temperature of the object change if its radius is decreased to half? (A) 2.19 (B) 5.19 (C) 1.19 (D) 4.19 Solution: (C) Heat liberated R3 and R2T4 Or T4  R Thus = Or T2 = T1 or T2 = That is , temperature decreases by a factor of 1.19. 11: The room temperature is +20C when outside temperature is -20C and room temperature room temperature is + 10C when outside temperature is -40C. Find the temperature of the radiator heating the room. (A) 30C (B) 60C (C) 90C (D) 45C Solution: (B) Applying Newton’s law In case (1) K1(T-Tr1)= K2(Tr2-Tout 1) And in case (2) K1(T-Tr2)= K2(Tr2-Tout 2) Dividing these equations or T=60C 12: Some water at 0C is placed in a large insulated enclosure (vessel). The water vapour formed is pumped out continuously. What fraction of the water will ultimately freeze, if the latent heat vaporization is seven times the latent heat of fusion? (A) 7/8 (B) 8/7 (C) 3/8 (D) 5/8 Solution: (A) m = mass of water, f = fraction which freezes L1 = latent heat of vaporization L2 = latent heat of fusion  L1 = 7L2 Mass of water frozen = mf Heat lost by freezing water = m fL2 Mass of vapour formed = m(1 - f) Heat gained by vapour = m (1 – f) L1 mfL2 = m (1 - f) x 7L2 f = 7 – 7f or f = 7/8 13. A substance is in the solid form at 0C. The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, find from the graph, the mass of the substance is (Specific heat capacity of water = 1000 cal kg-1 K-1 ) (A) 0.02 kg (B) 2 kg (C) 0.04 kg (D) 0.05 kg 14. A substance is in the solid form at 0C. The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, find from the graph, the specific latent heat of the melting process is (Specific heat capacity of water = 1000 cal kg-1 K-1 ) (A) 6000 cal kg-1 (B) 4000 cal kg-1 (C) 1000 cal kg-1 (D) 2000 cal kg-1 15. A substance is in the solid form at 0C. The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, find from the graph, the specific heat of the substance in the liquid state is (Specific heat capacity of water = 1000 cal kg-1 K-1 ) (A) 300 cal kg –1 K-1 (B) 500 cal kg –1 K-1 (C) 700 cal kg –1 K-1 (D) 100 cal kg –1 K-1 Solution 13: (A) 800 calories of heat raise the temperature of the substance from 0°C to 80C.  800 = m (1000  0.5 )  80 ( specific heat = relative sp. heat x sp. heat of water) or m = 0.02 kg Solution 14: (B) Latent heat = 200  4 = 800 cal ( 1 div reads 200 cal ) = 0.02  L  L= 4000 cal kg-1 Solution 15: (C) In the liquid state temperature rises from 80C to 120C, that is, by 40C after absorbing (2160 – 1600) cal. 0.02 s  40 = 2160 – 1600 or s = 700 cal kg –1 K-1 16: An earthenware vessel loses 1g of water per second due to evaporation. The water equivalent of the vessel is 0.5 kg and the vessel contain 9.5 kg of water. Find the time required for the water in the vessel to cool to 28C from 30C. Neglect radiation losses. Latent heat of vaporization of water in this range of temperature is 540 cal g-1 . (A) 2 min 5 s (B) 3 min 5 s (C) 5 min 3 s (D) 6 min 2 s Solution: (B) Here water at the surface is evaporated at the cost of the water in the vessel losing heat. Heat lost by the water in the vessel = (9.5 + 0.5)  1000  (30 – 20) = 105 cal Let t the required time in seconds. Heat gained by the water at the surface = (t  10-3)  540  103 ( L = 540 cal g-1 = 540  103 cal kg-1)  105 = 540 t or t = 185 s = 3 min 5 s 17: In an industrial process 10 kg of water per hour is to be heated from 20°C to 80°C. To do this, steam at 150°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 90°C. How many kg of steam are required per hour? Specific heat of steam = 1 kilo cal kg–1°C–1. Latent heat of steam = steam = 540 kilo cal kg–1. (A) 1 kg (B) 2 kg (C) 3 kg (D) 4 kg Solution: (A) Let the mass of steam required per hour be m kg. Heat gained by water in boiler per hour is = 10 kg × 1 kilo cal kg–1 °C–1 × (80 – 20)°C = 600 kilo cal … (1) Heat lost by steam per hour is = heat needed to cool m kg of steam from 150°C to 100°C + heat needed to convert m kg of steam at 100°C into water at 100°C + heat needed to cool m kg of steam at 100°C to 90°C = m × 1 × (150 – 100) + m × 540 + m × 1 × (100 – 90) = 50 m + 540 m + 10 m = 600 m kilo cal … (2) Heat lost = heat gained. Equating (1) and (2) we have 600 m = 600 or m = 1 kg 18: A closed cubical box made of a perfectly insulating material has walls of thickness 8 cm and the only way for the heat to enter or leave the box is through two solid cylindrical metal plugs, each of cross–sectional area 12 cm2 and length 8 cm fixed in the opposite walls of the box. The outer surface A of one plug is kept at a temperature of 100°C while the outer surface of the other plug is maintained at a temperature of 4°C. The thermal conductivity of the material of the plug is 0.5 cal cm–1 s–1 (°C)–1. A source of energy generating 36 cal s–1 is enclosed inside the box. Find the equilibrium temperature of the inner surface of the box assuming that it is the same at all points on the inner surface. (A) 70°C (B) 45°C (C) 65°C (D) 76°C Solution: (D) At equilibrium, the total energy generated by the source per second must equal the heat leaving per second through the two metal plugs (Fig.). Let T°C be the equilibrium temperature. Then heat leaving the box per second through surface A = cal s–1 and heat leaving the box per second through the surface B Hence (T – 100 + T –4) = 36 or 2T – 104 = == 48 or T = 76°C 19: Find the time during which a layer of ice of thickness 2.0 cm on the surface of a pond will have its thickness increased by 2 mm when the temperature conductivity of ice = 5 × 10-3 cal cm-1 s-1 (°C)-1, density of ice at 0°C = 0.91 g cm-3 and latent heat of fusion = 80 cal g-1 (A) 6 min 5 s (B) 2 min 6 s (C) 5 min 6 s (D) 3 min 5 s Solution: (C) substituting the given quantities in the expression t = (-) we have t = [(2.2)2-2.0)2] = 306 s = 5 min 6 s. 20: Water is being boiled in a flat-bottomed kettle placed on a stove. The area of the bottom is 300 cm2 and the thickness is 2 mm. If the amount of steam produced is 1 g/min, calculate the difference of temperature between the inner and other outer surfaces of the bottom. The thermal conductivity of the material of kettle = 0.5 cal cm-1 s-1 (°C)-1 and the latent heat of stem = 540 cal g-1. (A) 0.012 °C (B) 0.0.04 °C (C) 0.02 °C (D) 0.0.08 °C Solution: (A) If ΔT is the temperature difference between the inner ad outer surface of the bottom of the kettle, then the amount of heat flowing through the bottom per second is = 750 Δ cal per second … (1) But … (2) Equating (1) and (2), we have 750 ΔT = 9 or ΔT = = 0.012 °C 21: If an anisotropic solid has coefficients of linear expansion αx, αy and αz for three mutually perpendicular directions in the solid, what is the coefficient of volume expansion for the solid? (A) β ≈ αx + αy + αz (B) β ≈ α2x + αy + αz (C) β ≈ α2x + α2y + α2z (D) β ≈ α3x + α3y + α3z Solution: (A) Consider a cube, with edges parallel to X, Y, Z of dimension L0 at T = 0. After a change in temperature ΔT = (T –0), the dimensions change to Lx = L0 (1 + αxT) Ly = L0(1 + αyT) Lz = L0(1 + αzT) and the volume of the parallelepiped is V = V0 (1 + αxT)(1 + αyT)(1 + αzT) ≈ V0 [1 + (αx + αy + αz)T] Where V0 = . Therefore, the coefficient of volume expansion is given by β ≈ αx + αy + αz. 22. In aluminum sheet there is a hole of diameter 2m and is horizontally mounted on a stand. Onto this hole an iron sphere of radius 2.004 m is resting. Initial temperature of this system is 25° C. Find at what temperature, the iron sphere will fall down through the hole in sheet. The coefficients of linear expansion for aluminum and iron are 2.4 × 10–4and 8.6 × 10 –5 respectively. (A) 82°C (B) 43°C (C) 45°C (D) 15°C Solution: (B) As value of coefficient of linear expansion for aluminum is more than that for ion, it expends faster then iron. So at some higher temperature when diameter of hole will exactly become equal to that of iron sphere, the sphere will pass through the hole. Let it happen at some higher temperature T. Thus we have at this temperature T, (diameter of hole)Al = (diameter of sphere)iron 2[1 + αAl (T – 25)] = 2.004[1 + αiron(T – 25)] 2Al (T – 25) = 0.004 + 2.004 αiron (T – 25) or or or T = 43°C 23. An iron ball has a diameter of 6 cm and is 0.010 mm too large to pass through a hole in a brass plate when the ball and plate are at a temperature of 30°C. At what temperature, the same for ball and plate, will the ball just pass through the hole? (A) 23.8°C (B) 13.8°C (C) 53.8°C (D) 83.8°C Solution: (C) We let I stand for the iron ball and B stand for the brass plate. LI = 6 cm and LI – LB = 0.001 cm at t = 30°C. Since the brass plate expands uniformly, the hole must expand in the same proportion. Then heating both the ball and the plate leads to increases in the diameters of the ball and the hole, with the hole increasing more, since αB > αI. We require ΔLB – ΔLI = 0.001 cm. ΔLB = αBLB Δt; ΔLI = αILI Δt. We can approximate LB in this formula by 6 cm = LI. Then ΔLB –ΔLI = (αB – αI)LI Δt = 0.001 cm solving, Δt = 23.8°C, and finally t = 30°C + 23.8°C = 53.8°C 24. It is desired to put an iron rim on a wooden wheel. The diameter of the wheel is 1.1000m and the inside diameter of the rim is 1.0980 m. If the rim is at 20°C initially, to what temperature must it be heated to just fit onto the wheel? (A) 52°C (B) 72°C (C) 102°C (D) 152°C Solution: (D) α for iron is found to be 1.2 × 10–5 °C–1. ΔL = 1.1000 – 1.0980 = 0.0020 m = αL Δt 0.0020 = (1.2 × 10–5)(1.098) Δt Δt + 20 = 172°C Δt = 152°C 25. Find the coefficient of volume expansion for an ideal gas at constant pressure. (A) (B) (C) (D) Solution: (A) For an idea gas PV= nRT As P is constant, we have P.dV = nRdT 26. What should be the lengths of steel and copper rod so that the length of steel rod is 5cm longer then the copper rod at all the temperatures. Coefficients of linear expansion for copper and steel are 1.7 and 1.1 (A) 2.17, 14.17 cm (B) 9.17, 14.17 cm (C) 9.17, 18.17 cm (D) 3.17, 5.17 cm Solution: (B) It is given that the difference in length of the two rods is always 5 cm. Thus the expansion in both the rods must be same for all temperatures. Thus we can say that at all temperature differences, we have or [If l1 and l2 are the initial lengths of Cu and steel rods] or or …..(1) It is given that l2 – l1 = 5cm ….(2) or Now from equation (2) l2 = 14.17 cm 27. A steel wire of cross-sectional area 0.5 mm2 is held between two rigid clamps so that it is just taut at 20. Find the tension in the wire at 0. Given that Young’s modulus of steel is Yst = 2.1 × 1012 dynes / cm2 and coefficient of linear expansion of steel is . (A) 2.31 × 10–4 (B) 2.31 × 10–2 (C) 9.31 × 10–6 (D) 2.31 × 10–6 Solution: (D) We know that due to drop in temperature, then tension increment in a clamped wire is T = YA = 2.1 ×1012 × 0.5 × 10–2 × 1.1 × 10–5 × 20 = 2.31 × 10–6 28. Two bodies have the same heat capacity. If they are combined to form a single composite body, show that the equivalent specific heat of this composite body is independent of the masses of the individual bodies. (A) (B) (C) (D) Solution: (C) Let the two bodies have masses m1, m2 and specific heats s1 and s2 then m1s1 = m2s2 or m1/m2 = s2/s1 Let s = specific heat of the composite body. Then (m1 + m2) s = m1s1 + m2s2 = 2m1s1 s = 29. 20 gm steam at 100°C is let into a closed calorimeter of water equivalent 10 gm containing 100 gm ice at – 10°C. Find the final temperature of the calorimeter and its contents. Latent heat of steam is 540 cal/gm, latent heat of fusion of ice=80 cal/gm, specific heat of ice = 0.5 cal/°C gm. (A) 13°C (B) 63°C (C) 93°C (D) 33°C Solution: (D) Heat lost by steam = mL + ms (100 - θ) where, θ is the equilibrium temperature Heat lost by steam = 20 × 540 + 20 × 1 (100 - θ) = 10800 + 2000 - 20θ Heat gained by (ice + calorimeter) = 100 × 80 + 100 × 0.5 × 10 + 100 × θ Now Heat lost = Heat gained ∴ 10800 + 2000 – 20 θ = 8000 + 500 + 110 θ or 130θ = 4300 or θ = = 33°C. 30. Victoria Falls in Africa is 122 m in height. Calculate the rise in temperature of the water if all the potential energy lost in the fall is converted to heat. (A) 0.29 K (B) 29 K (C) 0.69 K (D) 0.99 K Solution: (A) Consider mass m of water falling. mgy = mc Δt gy = c Δt We express both sides in joules by noting c = 1 kcal/kg ⋅ K = 4184 J/kg ⋅ K Then 9.8(122) = 4184 Δt and Δt = 0.29 K 31. An electric heater supplies 1.8 kW of power in the form of heat to a tank of water. How long will it take to heat the 200 kg of water in the tank from 10 to 70°C? Assume heat losses to the surroundings to be negligible. (A) 1.75 h (B) 7.75 h (C) 4.75 h (D) 5.75 h Solution: (B) The heat added is (1.8 kJ/s)t and the heat absorbed is cm ΔT = (4.184 kJ/kg ⋅ K) (200 kg)(60 K) = 5.0 × 104 kJ Equating heats, t = 2.78 × 104 s = 7.75 h. 32. What will be the final temperature if 50 g of water at 0°C is added to 250 g of water at 90°C? (A) 15°C (B) 30°C (C) 45°C (D) 75°C Solution: (D) Heat gained = heat lost. (We assume no hat transfer to or from container.) (50 g)(1.00 cal/g ⋅ °C)(t – 0°C) = (250 g)(1.00 cal/g ⋅ °C)(90°C – t) where t is the final equilibrium temperature. 50t = 22500 – 250t or 300t = 22500 t = 75°C 33. A 500 g piece of iron at 400°C is dropped into 800 g of oil at 20°C. If c = 0.40 cal/g⋅°C for the oil, what will be the final temperature of the system. Assume no loss to the surroundings. (A) 15.7°C (B) 30.7°C (C) 45.7°C (D) 75.7°C Solution: (D) Heat lost = heat gained is written as 0.11(500)(400 – t) = 0.40(800)(t – 20) from which t = 75.7°C 34. A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100°C and the other at 0°C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the thermal resistance of the bar. (A) 15.9 K/W (B) 20.9 K/W (C) 40.9 K/W (D) 50.9 K/W 35. A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100°C and the other at 0°C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the thermal current H. (A) 1.3 watt (B) 6.3 watt (C) 12.3 watt (D) 18.3 watt 36 . A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100°C and the other at 0°C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the temperature gradient . (A) –10°C/m (B) –40°C/m (C) –50°C/m (D) –90°C/m 37 . A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100°C and the other at 0°C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the temperature 25 cm from the hot end. (A) 20.5°C (B) 40.5°C (C) 77.5°C (D) 87.5°C Solution 34 (A) Thermal resistance R = or R = = 15.9 K/W Solution 35 (B) Thermal current, H = or H = 6.3 watt Solution 36 (C) Temperature gradient = = – 50 K/m = –50°C/m Solution 37 (D) Let be the temperature at 25 cm from the hot end, then (θ – 100) = (temperature gradient) ×(distance) or θ – 100 = (– 50) (0.25) or θ = 87.5°C 38. Two metal cubes with 3 cm–edges of copper and aluminium are arranged as shown in figure. Thermal conductivity of copper is 401 W/m–K and that of aluminium is 237 W/m–K. Find the total thermal current from one reservoir to the other. (A) 0.14 K/W (B) 0.34 K/W (C) 0.74 K/W (D) 0.94 K/W 39. wo metal cubes with 3 cm–edges of copper and aluminium are arranged as shown in figure. Thermal conductivity of copper is 401 W/m–K and that of aluminium is 237 W/m–K. Find the ratio of the thermal current carried by the copper cube to that carried by the aluminium cube. (A) 0.75 K/W (B) 0.05 K/W (C) 0.15 K/W (D) 0.35 K/W Solution: 38 (A) Thermal resistance of aluminium cube R1 = or R1 = = 0.14 K/W Solution: 39 (B) Thermal resistance of copper cube R2 = or R2 = = 0.08 K/W As these two resistances are in parallel, their equivalent resistance will be R = = = 0.05 K/W 40. What temperature gradient must exist in an aluminum rod in order to transmit 8 cal/s per square centimeter of cross section down the rod? k for aluminum is 0.50 cal/s ⋅ cm ⋅ °C. (A) 36°C/cm (B) 46°C/cm (C) 86°C/cm (D) 16°C/cm Solution: (D) H = kA or 8 cal/s = (0.50 cal/s ⋅ cm ⋅ °C)(1 cm2) = magnitude of temperature gradient = 16°C/cm. The actual sign of the gradient depends on whether the heat flow is in the positive or negative x direction, being negative or positive for the two cases, respectively. 41. What is heat conduction. (A) (A ΔT)/d (B) (k ΔT)/d (C) (kA ΔT)/d (D) (ΔT)/d Solution: (C) The conduction equation, H ≡ ΔQ/Δt = (kA ΔT)/d can be reexpressed as H = ΔT/R, where R ≡ d/(kA) is called the thermal resistance of the slab. It has units of K/W in SI. 42. Consider the two insulating sheets with resistances R1 and R2 shown in Fig. What is the value of T' ? (A) (B) (C) (D) Solution: (B) For the two sheets, H1 = (T1 – T′)/R1, H2 = (T′ – T2)/R2. Noting that H1 = H2 = H, we have (T1 – T′)/R1 = (T′ – T2)/R2. Cross multiplying we get R2(T1 – T′) = R1(T′ – T2), or rearranging terms, R2T1 + R1T2 = (R1 + R2)T′, and T’ = (R2T1 + R1T2)/(R1 + R2). 43. A spherical blackbody of 5 cm radius is maintained at a temperature of 327°C. What is the power radiated? (A) 231 W (B) 431 W (C) 531 W (D) 631 W Solution: (A) The surface area of a sphere is 4πr2. In this case, then, the area is 4π(25 × 10–4) = 0.01πm2. The power radiated is given by Stefan’s law: P= σT4A = (5.67 × 10–8 W/m2.K4)(600 K)4(0.01π m2) = 231 W. 44.. A spherical blackbody of 5 cm radius is maintained at a temperature of 327°C. What wavelength is the maximum wavelength radiated. (A) 2.82 μm (B) 4.82 μm (C) 3.82 μm (D) 5.82 μm Solution: (B) The surface area of a sphere is 4πr2. In this case, then, the area is 4π(25 × 10–4) = 0.01πm2. The power radiated is given by Stefan’s law: P= σT4A = (5.67 × 10–8 W/m2.K4)(600 K)4(0.01π m2) = 231 W By Wien’s law, λm(600 K) = 2898 μm.K and λm = 4.82 μm. 45. A sphere of 3 radius acts like a blackbody. It is equilibrium with its surroundings and absorbs 30 kW of power radiated to it from the surroundings. What is the temperature of the sphere? (A) 2300 K (B) 2800 K (C) 3000 K (D) 2600 K Solution: (D) The power absorbed by a blackbody is Pa = σA, or (30 × 103 W)= (5.67 × 10–8 W/m2.K4) 4π(0.03 m)2. = 4.68 × 1013; Ta = temperature of surroundings = 2600 K. Since the body is in equilibrium with its surroundings, it is at the same temperature, 2600 K. 46 . A blackbody is at a temperature of 527°C. To radiate twice as much energy per second, what temperature must be increased ? (A) 951 K (B) 451 K (C) 551 K (D) 651 K Solution: (A) Since P ∝ T4, the temperature must be increased to 21/4(800 K) = 951 K. 47. Use Stefan’s law to calculate the total power radiated per square meter by a filament at 1727°C having an absorption factor of 0.4. (A) 0.16 MW/m2 (B) 0.26 MW/m2 (C) 0.36 MW/m2 (D) 0.46 MW/m2 Solution: (C) Stefan’s law gives R = ∈σT4 = 0.4(5.67 × 10–8)(2000)4 = 0.36 MW/m2 48. A blackbody is at a temperature of 527°C. To radiate twice as much energy per second, how many times of increase in radiated power when the temperature of a blackbody is increased from 7 to 287°C. (A) 5 times (B) 16 times (C) 6 times (D) 20 times Solution: (B) Since P ∝ T4, the temperature must be increased to 21/4(800 K) = 951 K. And times 49. The initial and final temperature of water as recorded by an observer are (40.6 ± 0.2) °C and (78.3 ± 0.3) °C. Calculate the rise in temperature with proper error limit. (A) (27.7 ± 0.5)°C (B) (17.7 ± 0.5)°C (C) (37.7 ± 0.9)°C (D) (37.7 ± 0.5)°C Solution: (D) Let θ1 = 40.6°C, Δθ1 = ± 0.2 °C θ2 = 78.3°C, Δθ2 = ± 0.3 °C ⇒ θ = θ2 – θ1 = 78.3 – 40.6 = 37.7°C & Δθ = ± (Δθ1 + Δθ2) = ± (0.2 + 0.3) = ± 0.5°C Hence rise in temperature = (37.7 ± 0.5)°C 50. A cylinder of radius R made of a material of thermal conductivity k1 is surrounded by a cylindrical sheet of inner radius R and outer radius 2R made of material of thermal conductivity . The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. Calculate the effective thermal conductivity of the system. (A) 4K = 2K1 + 3K2 (B) 4K = 5K1 + 3K2 (C) 4K = 6K1 + 4K2 (D) 4K = K1 + 3K2 Solution : (D) Two cylinders are in parallel, therefore equivalent thermal resistance R is given by But Here and i.e. 4K = K1 + 3K2 51. A lake is covered with ice 2 cm thick. The temperature of ambient air is −15oC. Find the rate of thickening of ice. For ice k = 4 × 10−4k-cal-m−1s−1(°C)−1. Density =9×103 kg/m3and latent heat L = 80 Kilo Cal/Kg. (A) 2.5 cm/ hour (B) 1.5 cm/ hour (C) 3.5 cm/ hour (D) 4.5 cm/ hour Solution : (B) Heat energy flowing per sec is given by . . . (I) if dm mass of ice is increased in time dt, then Since, . . . (ii) From eq. (I) and (II) Rate of thickening of ice = dx/dt == = 4.166 × 10−6 m/s = 1.5 cm/ hour. 52. A body cools from 60°C to 50°C in 10 minutes. If the room temperature is 25°C and assuming Newton’s law of cooling to hold good, find the temperature of the body at the end of the next 10 minutes. (A) 42.85°C (B) 12.85°C (C) 20.85°C (D) 52.85°C Solution : (A) According to Newton’s law of cooling rate of loss of heat α(T − To) where T = mean temperature of the body ∝ . . . (1) Also, ∝ . . . (2) where T is the required temperature Solving both equation, we get, T = 42.85°C 53. Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are same. The two bodies emit total radiant power at the same rate. The wavelength λB corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.00 μm. If the temperature of A is 5802 K, calculate (a) the temperature of B and (b) wavelength λB. (A) 2000K, 1.6 μ m (B) 3000K, 1.8 μ m (C) 1934K, 1.5 μ m (D) 4000K, 1.9 μ m Solution : (C) (a) According to stefan’s law the power radiated by a body is give by According to the given problem, , with So that i.e, or (b) According to Wien’s displacement law. ; i.e, i.e, and also λB - λA = 1μm(given) so i.e. 54 A body which has a surface area 2.00 cm2 and a temperature of 727oC radiates energy at the rate of 5 W. What is its emissivity? (Stefan boltzman constant ). (A) 0.14 (B) 0.44 (C) 0.34 (D) 0.24 solution: (B) ⇒ 55. Two rods of equal cross-sectional area A and length L are joined and what is the temperature of junction as shown in the figure. The temperature of one end of the composite rod is C and the other end of the rod is at . Draw a graph showing the variation of the temperature with the distance x from the end maintained at . Assume that the flow is longitudinal and there is no loss of heat from the lateral surface. (A) (B) (C) (D) solution: (A) ∴ when x ≥ L ∴ The graph is as shown 56. Two spherical soap bubbles coalesce to form a bigger bubble without any leakage of air. If V is the total change in volume of the contained air and S the total change in surface area, what is the value of 3PV + 4ST, where T is the surface tension of the soap solution and P is the atmospheric pressure. (A) 0 (B) 1 (C) 2 (D) 3 solution: (A) Let r1 and r2 be the initial radii of two bubbles and r is the final radius of the bubble after they coalesce. Then ⇒ …(1) also = change in total volume …(2) and = change in surface area …(3) from (1), (2) & (3) we get 57. The graphs gives the variation of temperature of two bodies having the same surface area with time where Ax and Ay represent absorptivity and εx and εy represent emissivity then (A) εx > εy and Ax < Ay (B) εx < εy and Ax > Ay (C) εx > εy and Ax > Ay (D) εx < εy and Ax < Ay solution: (D)According to Kirchoff’s law good emitter is a good absorber ∴ Ey > Ex and Ay > Ax 58. The temperature gradient in a rod of 0.5 m length is 80oC/m. If the temperature of hotter end of the rod is 30oC, then the temperature of the cooler end is (A) 40oC (B) –10oC (C) 10oC (D) 0oC. Solution: (B) 59. A body at 300oC radiates 10 J cm–2 s–1. If Sun radiates 105 J cm–2 s–1, then its temperature is (A) 3000oC (B) 5457oC (C) 300 × 104 oC (D) 5730oC Solution : (B) 60. The ratio of energy of radiation emitted by a black body at 27oC and 927oC is (A) 1 : 4 (B) 1 : 16 (C) 1 : 64 (D) 1 : 256. Solution:(B) E=T4