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TOPPER’S HANDWRITTEN NOTES, , EXPLORE AND LEARN

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CLICK HERE TO DOWNOAD COMPLETE, STUDY MATERILAS IN FREE OF COST, CLICK HERE FOR HANDWRITTEN NOTES

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Chapter 12, Thermodynamics, , 1 Marks Questions, 1.If a air is a cylinder is suddenly compressed by a piston. What happens to the, pressure of air?, Ans.Since the sudden compression causes heating and rise in temperature and if the piston, is maintained at same Position then the pressure falls as temperature decreases., , 2. What is the ratio of find volume to initial volume if the gas is compressed, adiabatically till its temperature is doubled?, Ans.Since for an adiabatic Process,, PVY= constant, Since PV = RT, P=, , So,, , constant, , Or TV y - 1 = constant T1, V1 = Initial temperature and Initial Volume, ∴ T1 V1 y - 1 = T2 V2 y - 1 T2, V2 = Final temperature and Final volume., , Since T2 = 2 T1(Given)

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So,, , Since, , > 1,, , is less than, , ., , 3.What is the ratio of slopes of P-V graphs of adiabatic and isothermal process?, Ans.The slope of P-V graph is, , For an isothermal process, (PV = constant), So,, , For an adiabatic process ( PVY = constant), , Divide 2) by 1), So, the ratio of adiabatic slope to isothermal slope is Y., , 4.What is the foundation of Thermodynamics?, Ans.The foundation of thermodynamics is the law of conservation of energy and the fact the, heat flows from a hot body to a cold body., , 5.Differentiate between isothermal and adiabatic process?

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Ans., Isothermal process, , Adiabatic process, , 1) In this, temperature remains constant, , 1) In this, no heat is added or removed., , 2) It occurs slowly, , 2) It occurs suddenly., , 3), , Here, system is thermally conducting to, surroundings, , 4) State equation :→ PV = constant, , 3), , 4), , Here, system is thermally insulated from, surroundings., State equation : → PVY = constant., , 6.A Carnot engine develops 100 H.P. and operates between 270C and 2270C. Find 1), thermal efficiency; 2) heat supplied3) heat rejected?, Ans. Here, energy = W = 100 H. P., = 100 × 746 W ( 1 H.P. = 746W), , =, , High temperature, TH = 2270C = 227 + 273 = 500K, Low temperature, Th = 270C = 27 + 273 = 300K, , 1) Thermal efficiency,, , 2) The heat supplied QH is given by:-

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3) The heat rejected QL is given by:-, , or, , 7.Draw a p – v diagram for isothermal and adiabatic expansion?, Ans., , 8.State zeroth law of thermodynamics?, Ans. Acc. to this, when the thermodynamic system A and B are separately in thermal, equilibrium with a third thermodynamic system C, then the system A and B are in thermal, equilibrium with each other also., , 9.Can a gas be liquefied at any temperature by increase of pressure alone?, Ans. No, a gas can be liquefied by pressure alone, only when temperature of gas is below its, critical temperature.

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10.Can you design heat energy of 100% efficiency?, , Ans. Since efficiency of heat engine = 1-, , , so, efficiency will be 100% or 1 if T2 = OK or T1 =, , α. Since both these conditions cannot be practically attained, so heat engine cannot have, 100% efficiency., , 11.If air is a bad conductor of heat, why do we not feel warm without clothes?, Ans.This is because when we are without clothes air carries away heat from our body due to, convection and we feel cold., , 12.A body with large reflectivity is a poor emitter why?, Ans.This is because a body with large reflectivity is a poor absorber of heat and poor, absorbers are poor emitters., , 13.Animals curl into a ball, when they feel very cold?, Ans.When animals curl, they decrease their surface area and since energy radiated varies, directly to surface area hence loss of heat due to radiation is also reduced., , 14.Why is the energy of thermal radiation less than that of visible light?, Ans.The energy of an electromagnetic ware is given by :- E = hf, h = Planck’s constant; f = frequency of wave. Since the frequency of thermal radiation is less, than that of visible light, the energy associated with thermal radiation is less than associated, with visible light., , 15.Two rods A and B are of equal length. Each rod has its ends at temperature T1 and T2, (T1 > T2). What is the condition that will ensure equal rates of flow through the rods A, and B?

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Ans. Heat flow,, , K = Thermal conductivity, A = Area, T1 = Temperature of hot body, T2 = Temperature of cold body, d = distance between hot and cold body., Q = heat flow, When the rods have the same rate of conduction,, Q1 = Q2, , K1, K2 → Thermal conductivity of first and second region, A1, A2 → Area of first and second region, or, K1 A1 = K2 A2, or, , 16.A Sphere is at a temperature of 6oo k. Its cooling rate is R in an external, environment of 200k. If temperature falls to 400k. What is the cooling rate R1 in terms, of R?, Ans.Acc. to Stefan’s law;

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increases by what factor?, Ans.By Stefan’s law : →, Rate of energy radiated α T4, T = Temperature, E1 = constant T14, E2 = constant T24, T1 = Initial temperature, T2 = Final temperature, T2 = 2T1, T24 = (2)4 T14, T24 = 16T14, E2 = constant (16 T14), E2 = 16 (constant T14), E2 = 16 E1, , 18. On a winter night, you feel warmer when clouds cover the sky than when sky is, clear. Why?, Ans.We know that earth absorbs heat in day and radiates at night. When sky is covered, with, clouds, the heat radiated by earth is reflected back and earth becomes warmer. But if sky is, clear the heat radiated by earth escapes into space.

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19. If a body is heated from 270 C to 9270C then what will be the ratio of energies of, radiation emitted?, Ans.Since, By Stefan’s law:→, E = Energy radiated, T = Temperature., E 1, T 1, , Initial energy and temperature, , E 2, T 2, , Final energy and temperature., , T1 = 270C = 27+273 = 300K, T2 = 9270C = 927+273K = 1200K., E = constant T4, So, E1 = constant T14, , Equating equation 1) &2)

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or E1 : E2 = 1 : 256, , 20. Which has a higher specific heat ; water or sand?, Ans.Water has higher specific heat than sand as, , where T = Temperature, Q = Heat, m = Mass,, , C = Specific heat; Since for water temperature increases less slowly than sand hence the, result., , 21. Why is latent heat of vaporization of a material greater than that of latent heat of, fusion?, Ans .When a liquid changes into a gas, there is large increase in the volume and a large, amount of work has to be done against the surrounding atmosphere and heat associated, with change from solid to gas is latent heat of vaporization and hence the answer., , 22. Draw a P – V diagram for Liquid and gas at various temperatures showing critical, point?, Ans.

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23. Why is temperature gradient required for flow of heat from one body to another?, Ans. Heat flows from higher temperature to lower temperature. Therefore, temperature, gradient (i.e. temperature difference) is required for the heat to flow one part of solid to, another., , 24. Why are Calorimeters made up of metal only?, Ans. Calorimeters are made up of metal only because they are good conductor of heat and, hence the heat exchange is quick which the basic requirement for the working of, calorimeter., , 25. If a body has infinite heat capacity? What does it signify?, Ans. Infinite heat capacity means that there will be no change in temperature whether heat, is taken out or given to the substance., , 26. Define triple point of water?, Ans. Triple point of water represents the values of pressure and temperature at which water, co-exists in equilibrium in all the three states of matter., , 27. State Dulong and petit law?, Ans.Acc. to this law, the specific heat of all the solids is constant at room temperature and is

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equal to 3R., , 28. Why the clock pendulums are made of invar, a material of low value of coefficient, of linear expansion?, Ans.The clock pendulums are made of Inver because it has low value of α (co-efficient of, linear expansion) i.e. for a small change in temperature, the length of pendulum will not, change much., , 29. Why does the density of solid | liquid decreases with rise in temperature?, Ans.Let P = Density of solid | liquid at temperature T, P1 = Density of solid | liquid at Temperature T+∆T, Since Density =, , So, P =, , →(1) P1 =, , (2), , V1 = Volume of solid at temperature T + ∆T, V = Volume of solid at temperature T, Since on increasing the temperature, solids | liquids expand that is their volumes increases,, so by equation, i) & 2) Density is inversely proportional to volumes, so if volume increases on increasing the, temperature, Density will decrease., , 30. Two bodies at different temperatures T1, and T2 are brought in thermal contact do, not necessarily settle down to the mean temperature of T1 and T2?, Ans.Two bodies at diff temperatures T1 and T2 when in thermal contact do not settle always, at their mean temperature because the thermal capacities of two bodies may not be always

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equal., , 31. The resistance of certain platinum resistance thermometer is found to be 2.56 Ω at, 00c and 3.56 Ω at 1000c. When the thermometer is immersed in a given liquid, its, resistance is observed to 5.06 Ω. Determine the temperature of liquid?, Ans.Ro = Resistance at00c = 2.56Ω, Rt = Resistance at temperature T = 1000c = 3.56Ω 100, Rt = Resistance at unknown temperature t ;, Rt = 5.06Ω, Since,, , t=, , =, , =, , =, , t = 2500c, , 32. Calculate Cp for air, given that Cv =0.162 cal g-1 k-1 and density air at N.T. P is, 0.001293 g|cm3?, Ans.Specific heat at constant pressure = Cp= ?

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Specific heat at constant volume = Cv = 0.162 Cal g-1 k-1, Now, Cp – Cv =, , Or CP – Cv =, , Cp – Cv =, , =, , =, , = 6.8×10-4+2, Cp – Cv = 0.068, Cp = 0.162+0.068, Cp = 0.23 Cal g-1 k-1, , 33. Develop a relation between the co-efficient of linear expansion, co-efficient, superficial expansion and coefficient of cubical expansion of a solid?, Ans.Since, co-efficient of linear expansion = α =, , ∆L = change in length, L = length, ∆T = change in temperature

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Similarly, co-efficient of superficial expansion = β =, , ∆S = change in area, S = original area, ∆T = change in temperature, Co-efficient of cubical expansion, = Y =, , ∆V = change in volume, V = original volume, ∆T = change in temperature., Now, ∆L=αL ∆T, L + ∆L = L + αL ∆T, L + ∆L = L (1+α∆T) → (1), Similarly V+ ∆V = V (1+Y∆T) →(2), And S+∆S=S (1+β∆T) → (3), Also, (V+∆V) = (L+∆L)3, V+∆V =, , V+∆V = L3, Since α2, α3 are negligible, so,, V+γ V∆T= V(1+3α∆T) [as L3=V], So, V+γV∆T = V+V3α∆T

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γV∆T = 3α∆T, Y = 3α, Similarly, β = 2α [using L2 = S (Area)], So,, , 34. Calculate the amount of heat required to convert 1.00kg of ice at – 100c into steam at, 1000c at normal pressure. Specific heat of ice = 2100J|kg|k. Latent heat of fusion of ice =, 3.36x105J|kg, specific heat of water = 4200J|kg|k. Latent heat of vaporization of water =, 2.25 x106J|kg?, Ans.(1) Here, heat is required to raise the temperature of ice from – 100c to 00c., So, change in temperature = ∆T = T2-T1 = 0-(-10) = 100c, So, ∆Q1=cm∆T, C = specific heat of ice, M = Mass of ice, ∆T = 100c, ∆Q1 = 2100×1×10=21000J, (2) Heat required to melt the ice to 00c water:∆Q2 = mL, L = Latent heat of fusion of ice = 3.36×105J/kg, m = Mass of ice

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36. How would a thermometer be different if glass expanded more with increasing, temperature than mercury?, Ans. If glass expanded more with increasing temperature than mercury, the scale of the, thermometer would be upside down., , 37. Show the variation of specific heat at constant pressure with temperature?, Ans., , 38. Two thermometers are constructed in the same way except that one has a spherical, bulb and the other an elongated cylindrical bulb. Which one will response quickly to, temperature change?, Ans. The thermometer with cylindrical bulb will respond quickly to temperature changes, because the surface area of cylindrical bulb is greater than the of spherical bulb., , 39. State Carnot’s Theorem?, Ans. According to Carnot’s Theorem, no engine working between two temperatures can be, more efficient than a Carnot’s reversible engine working between the same temperatures.

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2 Marks Questions Part 1, 1.A motor car tyre has a Pressure of four atmosphere at a room temperature of 270C. If, the tyre suddenly bursts, calculate the temperature of escaping gas?, Ans. Since the tyre suddenly bursts, the change taking place is adiabatic, for adiabatic, change:-, , Or, , Hence, T1 = 273 + 27 = 300K, P1 = Initial Pressure; P2 = final Pressure, , So,, , So, Putting the above values in eq4 i), , Taking 1.4 Power, , W1=-150J→ (1)

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Work done by the gas in the process B → C is : →, , Adding equation i) & 2), Net work done by the gas in the whole process is W = W1 + W2, , W = 150 – 70 = - 22 OJ, , T2 = 201.8 K, ∴ T2 = 201.8 – 273 = - 71.20C, , 2.How does Carnot cycle operates?, Ans. A Carnot cycle operates a follows:-

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1) It receives thermal energy isothermally from some hot reservoir maintained at a constant, high temperature TH., 2) It rejects thermal energy isothermally to a constant low–temperature reservoir (T2)., 3) The change in temperature is reversible adiabatic process., Such a cycle, which consist of two isothermal processes bounded by two adiabatic processes,, is called Carnot cycle., , 3.Calculate the work done by the gas in going from the P-V graph of the thermodynamic, behavior of a gas from point A to point B to point C?, Ans.Work done by the gas in the process A → B is, , W1 = - (area under curve A B), = =P A B = 500 Pa, = 5×105 N|m2

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4.Why does absolute zero not correspond to zero energy?, Ans.The total energy of a gas is the sum of kinetic and potential energy of its molecules., Since the kinetic energy is a function of the temperature of the gas. Hence at absolute zero,, the kinetic energy of the molecules ceases but potential energy is not zero. So, absolute zero, temperature is not the temperature of zero energy., , 5.State the Second law of thermodynamics and write 2 applications of it?, Ans.According to second law of thermodynamics, when a cold body and a hot body are, brought into contact with each other, heat always from hot Body to the cold body. Also, that, no heat engine that works in cycle completely converts heat into work., , Second law of thermodynamics is used in working of heat engine and of refrigerator., , 6.At 00C and normal atmospheric pressure, the volume of 1g of water increases from, 1cm3 to 1.091 cm3 on free zing. What will be the change in its internal energy? Normal, atmospheric pressure is 1.013x105 N|m2 and the latent heat of melting of ice is 80 cal/g?, Ans.Since, heat is given out by 1 g of water in freezing is, m = Mass of water = 1 g, Q = - (mLf) Lf = Latent heat of melting of ice = 80 cal|g, , During freezing, the water expands against atmospheric pressure. Hence, external work, done (W) by water is :- W = P × ∆ V, P = 1.013×105 N|m2; ∆ V = 1.091 – 1 = 0.091 cm3 = o.o91 × 10-6 m3, ∆ V = V2 – V1; V2 = final volume = 1.91 cm3, V1 = Initial volume = 1 cm3

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So, W =, W = 0.0092 J, Since, 1 cal = 4.2J so,, W=, , Since the work has been done by ice, it will be taken positive., Acc. to first law of thermodynamics,, Q = ∆∪ + W ∆∪ = change in internal energy, , So, ∆∪ = Q – W, =, ∆∪ = - 80.0022 cal, Negative sign indicates that internal energy of water decreases on freezing., , 7.Two different adiabatic paths for the same gas intersect two thermals at T1 and T2 as, shown in P-V diagram. How does, , Compare with, , ?, , Ans.Now, A B and C D are isothermals at temperature T1 and T2 respectively and BC and AD, are adiabatic., Since points A and D lie on the same adiabatic., , T1 VA Y-1 = T2 VDY-1

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Also, points B and C lie on the same adiabatic,, , or T1VB Y-1 = T2VCY-1, , From equation 1) & 2), , 8.The internal energy of a compressed gas is less than that of the rarified gas at the, same temperature. Why?, Ans.The internal energy of a compressed gas is less than that of rarified gas at the same, temperature because in compressed gas, the mutual attraction between the molecules, increases as the molecules comes close. Therefore, potential energy is added to internal, energy and since potential energy is negative, total internal energy decreases., , 9.Consider the cyclic process A B C A on a sample 2 mol of an ideal gas as shown. The, temperature of the gas at A and B are300 K and 500K respected. Total of 1200 J of heat is, with drawn from the sample. Find the work done by the gas in part BC?

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Ans.The change in internal energy during the cyclic process is zero. Therefore, heat supplied, to the gas is equal to work done by it,, ∴ WAB + WBC + WCA = - 1200J →(1), (- ve because the cyclic process is traced anticlockwise the net work done by the system is, negative), The work done during the process AB is, WAB = PA (VB-VA) = nR(TB-TA), WAB = 2×8.3(500-300) = 3320J →2), R = Universal gas constant, N = No. of volume, Since in this process, the volume increases, the work done by the gas is positive., Now, WCA = O (, , volume of gas remains constant), , ∴ 3320 + WBC + O = - 1200 (Using equation 1) & 2), WBC = - 1200 – 3320, WBC = - 4520J, , 10.A refrigerator placed in a room at 300 K has inside temperature 264K. How many, calories of heat shall be delivered to the room for each 1 K cal of energy consumed by

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the refrigerator, ideally?, Ans.High temperature, TH = 300K, Low temperature, Th = 264K, Energy = 1K cal., Co - efficient of performance, is given by:-, , Now, COP =, , QL = heat rejected, QL = COP × W, QL =, The mechanical work done by the compressor of the refrigerator is:W = QH – QL, QH = W + QL, QH =, , QH = 8.33 K cal, , 11.If the door of a refrigerator is kept open in a room, will it make the room warm or

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cool?, Ans.Since a refrigerator is a heat engine that operates in the reverse direction i.e. it extracts, heat from a cold body and transforms it to hot body. Since it exhaust more heat into room, than it extracts from it. Therefore, the net effect is an increase in temperature of the room., , 12.The following figure shows a process A B C A per formed on an ideal gas, find the net, heat given to the system during the process?, Ans .Since the process is cyclic, the change in internal energy is zero. Therefore, the heat, given to the system is equal to work done by it. The net work done by the gas in the process, ABCA is:W = WAB + WBC + WCA, Now WAB = O, During the path BC, temperature remains constant. So it is an isothermal process. So, WBC =, nRT2 Loge, , During the CA, Vα T so that, , is constant., , ∴ Work done by the gas during the part CA is :WCA = P (V1 – V2), = nR (T1 – T2), = - nR (T2 – T1) → Using equation 1), , W= O + nR T2 Loge, , - nR (T2 – T1)

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13.A certain gas at atmospheric pressure is compressed adiabatically so that its volume, becomes half of its original volume. Calculate the resulting pressure?, Ans.Let the original volume, V1=V, ∴ final volume V2 =, , (∴volume become half), , Initial pressure P1 = o.76m of Hg column, Final pressure P2 after compression =?, As the change is adiabatic, so, , Y=, , P2 = P1, , = 0.76 ×, , P2 = 0.76 × (2)1.4, , = 1.4 for air

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P2 = 2m of Hg column, P2 = h sg, P2 = 2.672 x 105 N|m2, P2 = 2× (13.6x103)× 9.8, h = height of Hg column, s = Density of air, g = Acceleration due to gravity, , 14.Why is conversion of heat into work not possible without a sink at lower, temperature?, Ans. For converting heat energy in to work continuously a part of heat energy absorbed, from the source has to be rejected. The heat energy can be rejected only to a body at lower, temperature which is sink, so we require a sink to convert heat into work, , 15.Write the sign conventions for the heat and work done during a thermodynamic, process?, Ans. 1) When heat is supplied to a system d Q is taken positive but when heat is supplied by a, system, d Q is taken negative., 2) When a gas expands, d w is taken as positive but when a gas compresses, work done is, taken as negative., , 16.Does the working of an electric refrigerator defy second law of thermodynamics?, Ans. No, it is not against the second law; this is because external work is done by the, compressor or for this transfer of heat.

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17.A Carnot engine absorb 6×105 cal at 2270c calculate work done per cycle by the, engine if it sink is at 1270c?, Ans. Here, heat abs or bed = Q1 = 6 × 105 cal., Initial temperature = T1 = 2270c = 227+273 = 500K, Final temperature = T2 = 1270c = 127 + 273 = 400K, As, for Carnot engine;, , Q2 =, Q2 = 4.8 × 105 cal, Q2 = Final heat emitted, As w = Q1 – Q2 = 6 × 105 – 4.8 × 105, = 1.2 × 105 cal, Work = w = 1.2 ×105 × 4.2 J, Dore = 5.04 × 105 J, , 18.How does second law of thermodynamics explain expansion of gas?, Ans. Since from second law,.

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d S ≥ O d S = change in entropy, During the expansion of gas, the thermodynamic probability of gas is larger and hence its, entropy is also very large. Since form second law, entropy cannot decrease ∴ following the, second law, gas molecules move from one partition to another., , 19.Why is it hotter at the same distance over the top of the fire than in front of it?, Ans. At a point in front of fire, heat is received due to the process of radiation only, while at a, point above the fire, heat reaches both due to radiation and convection. Hence the result., , 20. A metal rod of length 20cm and diameter 2cm is covered with a non-conducting, substance. One of it ends is maintained at 1000c while the other is at 00c. It is found that, 25g of ice melts in 5 min calculate coefficient of thermal conductivity of metal?, Ans.Length of rod = ∆x = 20cm = 2 × 10-3m, Diameter = 2cm, R = 10-2m, Area of cross-section = π r2, = π (10-2)2, = 10-4 π sq. m, ∆ T = T2 – T1 = 100 – 0 = 1000c, Mass of ice melted = m = 25g, Latent heat office = 80 cal/g, Heat conducted, ∆ Q = mL, = 25 x 80

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= 2000 cal, = 2000 × 4.2J, ∆ t = 5 min = 300s, So,, , =, , K=, , =, , K = 1.78J |s|m|0c, K = coefficient of thermal conductivity, , 21.Calculate the temperature in Kelvin at which a perfectly black body radiates at the, rate of 5.67 w/cm2?, Ans. E = 5.67w|cm2 ; E = energy radiated, = 5.67 x 107 erg | s | cm2, , = Stefan’s constant = 5.67 × 10-5 ergs |s | cm2| K4, from Stefan’s law, E = σ T4, , T=

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T=, , 22.How do you explain the emission of long - wavelength by the object at low, temperature?, Ans.Since by Wein’s law: →, , i.e temperature is inversely proportional to the wavelength so, if temperature is less, then, wavelength will be long. If temperature is high, then wavelength will be short., , 23.If the radiation from the moon gives maxima at, , = 4700 A0 and, , conclusion can be drawn from the above information?, Ans. Acc. to wien’s displacement law,, , Now, according to the question,, , m = 4700 A0 = 4700×10-10m, , T1 = Temperature of moon,, T1 =, b = 2.9 ×10-3 mK, , Let the temperature corresponding to, , = 14x10-6m. What

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So, T2 =, , T2 =, , 24.Differentiate between conduction, convection and radiation?, Ans., Properties, 1., , Material, Medium, , 2. Molecules, 3., 4., , Conduction, , Convection, , Radiation, , Essential, , Essential, , Not Essential, , Do not leave their mean More bodily from one, , Medium does not, , position, , play any part, , Transfer of, , Can be in any direction, , heat, , along any part, , Speed of, transfer of heat, , Slow, , place to another., Only vertically upward, Rapid, , In all direction in, straight lines, Fastest with the, speed of light., , 25.The tile floor feels colder than the wooden floor even though both floor materials, are at same temperature. Why?, Ans. This happens because the tile is better heat conductor than wood. The heat conducted, from our foot to the wood is not conducted away rapidly. So, the wood quickly heats up on its, surface to the temperature of our foot. But the tile conducts the heat away rapidly and thus, can take more heat from our foot, so its surface temperature drops.

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2 Marks Questions Part 2, 26.A room has a 4m x 4m x10cm concrete roof (K1 = 1.26w|m|0C). At some instant, the, temperature outside is 460c and radius 320c., 1) Calculate amount of heat flowing per second into the room through the roof., 2) If bricks (K2 – 0.56w |m|0c) of thickness 7.5cm are laid down on roof, calculate the, new rate of heat flow under the same temperature conditions?, Ans. 1) Area of roof = 4 × 4 = 16 m2, Thickness of roof, x1 = 10 cm = 0.1m,, Thermal resistance of the roof is given by :-, , ∴ Rate of heat flow through the roof is:-, , 2) The thermal resistance of the brick is given by:-

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The equivalent thermal resistance of the roof now is :→, , ∴ Rate of heat through the roof is :→, , 27.A bar o copper of length 75cm and a bar of length 125cm are joined end to end. Both, are of circular cross – section with diameters 2cm. The free ends of copper and steel are, maintained at 1000c and 00c respectively. The surfaces of the bars are thermally, insulated. What is the temperature of copper – steel junction? Thermal conductivity of, copper = 9.2x10-2k cal |m|0c|s and that of steel is 1.1x10-2 k cal|m|0c|s?, Ans. l1 = lengths of copper bars AB, l2 = length of steel bars BC., Θ1 = temperature of free ends A, Θ2 = temperature of free ends C., Θ = temperature of copper – steel., In steady state, the heat flowing per second through two bars is the same i.e, H1 = H2

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∴ Temperature of junction = θ :→, , 28.Two rods A and B are of equal length. Each rod has its ends at temperatures T1 and, T2. What is the condition that will ensure equal rates of flow of heat through the rods A, and B?, Ans.Since θ =, , Θ = heat flow, K = co – efficient of thermal conductivity, A = Cross – Sectional Area, Θ1 = Temperature of hot body, Θ2 = Temperature of cold body, X = distance between hot and cold faces, t = time, For rod A :

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For equal rates of flow,, , KA AA = KB AB, , 29.A layer of ice 10cm thick is formed on a pond. The temperature of air is – 100C., Calculate how long it will take for the thickness of ice to increase by 1mm. Density of, ice = 1g|cm3; Thermal conductivity of ice = 0.005Cal|s|cm|0C; Latent heat of ice =, 80Cal|g?, Ans. Let t = time required to increase the thickness of ice by 1mm (=0.1cm), Mass of ice required to be formed is :m = Volume x Density, Let A = Area of upper surface, Volume = Area x Thickness, = A × 0.1, m = (A × 0.1) × 1, m = 0.1 A gram →1), Now, heat must flow from lower surface to the upper surface of ice and finally into, atmosphere., Θ = heat that flows out of pond into atmosphere., , = Latent heat of ice, m = Mass of ice, k = co – efficient of thermal conductivity, A = Cross – sectional Area, t = time, x = Distance between hot and cold surface

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K1 = Co – efficient of thermal conductivity of hot slab, K2 = Co – efficient of thermal conductivity of cold slab, , Θm = final temperature, d = Distance b/w hot and cold surface, A = Area of cross – section, t = time, Now, since is steady state, the rate of heat transfer in both the slabs is same i. e, , Θ1 – θm = because first heat flows from θ1 to the junction, Θ2 – θm = then heat flows from junction to second surface, So,

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So,, , 31.The ends of the two rods of different materials with their thermal conductivities,, radii of cross – section and length in the ratio 1:2 are maintained at the same, temperature difference. If the rate of flow of heat through the larger rod is 4 cal |s,, what is the rate of flow through the shorter rod?, Ans.K1 = thermal conductivity of first region, K2 = thermal conductivity of second region, r1 = radius of cross section of first region, r2 = radius of cross – section of second region, l1 – length of first region, l2 = length of second region, Θ1 = heat flow of first region

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Θ2 = heat flow of second region, , Now,, , Also,, , and, , Now, we know,, , So, Let, , Now, Divide eq4 1) & 2)

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Since, , 32.What are thermal radiation? Write its properties of thermal radiation?, Ans.The radiant energy emitted be a body solely on account of its temperature is called, thermal radiation., Properties of thermal Radiation:1) They travel through vacuum, 2) They obey laws of refraction, 3) They can be refracted, 4) They travel with the speed of light, 5) They do not heat the medium through which they passes.

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6) They exhibit phenomena of interference, diffraction and polarization., , 33.An indirectly heated filament is radiating maximum energy of wavelength 2.16x 107m., , Find the net amount of heat energy lost per second per unit area, the temperature, , of surrounding air is 130C. Given b = 2.88x10-3 mk, σ = 5.77x10-8 J|s|m2|k4?, Ans. By Wien’s Law:The product of wavelength, , at which maximum energy is emitted and the absolute, , temperature (T) of the black body is always constant., i.e, , T = constant = b →(1), , b = Wien’s constant = 2.9×10-3 mK, Now,, , T = 13333.3K, Now, Temperature of surrounding, To = 13 + 273 = 286K., Net amount of heat energy lost per second per unit area:-

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E = 1.824 × 108 J/s/m2, , 34. Animals in the forest find shelter from cold in holes in the snow. Why?, Ans. Animals in the forest find shelter from cold in holes in the snow because snow has, trapped air (as in ice there is no air) so, it acts as a heat insulator. Therefore, the snow, prevents the transmission of heat from the body of the animal to the outside., , 35.A brass boiler has a base area of 0.15m2 and thickness 1.0cm. It boils water at the, rate of 6kg| min when placed on a gas stove, Estimate the temperature of the part of, flame in contact with the boiler. Thermal conductivity of brass = 109J|s|m|0C, heat of, vaporization of water = 2256J|g?, Ans. Rate of boiling of water is = 6.0Kg / min, =, , = 100 g/s, ∴ Rate at which heat is supplied by the flame to water is :m = Rate of boiling of water, L = heat of vaporization of water, θ=mL, =, , Θ = 225 600J/s

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Now, T2 = Temperature of cold junction = 1000C, , T1 = Temperature of hot junction, T2 = Temperature of cold junction, t = time, x = Distance b/w hot and cold junction, , Now, x = 1.0 cm = 1.0 X 10-2m, K = 109 J|s|m|0C, A = 0.15 m2, t =1s

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T1 = 237.980C, , 36. How do you explain heating of rooms based on principle of convection?, Ans. Convection is the process by which heat is transmitted from one point to another due to, the movement of heated particles of the substance., During heating of the room by a heater, the air molecules in immediate contact with heater, are heated up, they acquire sufficient energy and rise upward. The cool air particles near to, the roof are dense and more down and in turn it is heated and the moves upwards. Hence by, the movement of heated air particles, the entire room heats up., , 37. If for a gas,, , = 0.67 then which gas is this:- monatomic, diatomic and tri atomic?, , Ans.Since for an ideal gas, CP – CV = R → 1), CP = Specific heat at constant pressure, CV = Specific heat at constant volume, R = Universal Gas Constant, , And given, , or

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And we know, that, , = 1.67 is for monatomic gas ; So the gas is monatomic in question., , 38.A 50g lead bullet, specific heat 0.02 cal |g | 0C is initially at 300C. It is fixed vertically, upward with a speed of 840m |s and on returning to the starting level strikes a cake of, ice at 00C. How much ice is melted? Assume that all energy is spent in melting ice only?, Ans.Speed of bullet hitting the ice = V = 840 m | s, Heat produced due to kinetic energy of the bullet:- =, , Now, m = Mass of bullet =, , Hence, , Now, heat given by bullet due to temperature difference = m c

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m = Mass of bullet, c = Specific heat of bullet, Q2 = Initial Temperature, Q1 = Final Temperature, Total heat given by bullet = 4200 + 30 = 4230 Cal., Now, entire heat of bullet is used in melting the ice only, Let M = Mass of Ice that melted, L = Latent heat of ice, Hence, , 39.A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. If, we neglect all vibration modes, find the total energy of the system?, Ans.Let NA = Avogadro’s Number, No. of degrees of freedom of O2 molecule (diatomic) = 5, No. of degrees of freedom of 2 moles of oxygen = 2 NA × 5 = 10 NA, No. of degrees of freedom of 4 moles of argon (monatomic) = 4 NA × 3, = 12 NA (, , 3 = degrees of freedom), , Total degrees of freedom of mixture = 10 NA + 12 NA = 22 NA →1)

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Energy associated with each degree of freedom | molecule =, , Total energy of mixture = 22 NA ×, , (, , Using equation1), , 40. Show that CP - CV = R Where [CP = specific heat at constant pressure ; CV = specific, heat at constant volume and R = Universal Gas constant] for an ideal gas?, Ans.Now, Let first heat the gas at constant volume and temperature increases by, , So,, , Since volume remains the same, hence no work is heating the gas then according to law of, conservation of energy, the entire heat supplied goes into raising the internal energy and, hence the temperature of the gas., Now,, , ∴ ∆U = increase in the internal energy of the gas Let heat the gas at constant pressure and if, the temperature of the gas increases by ∆T but here external work is done to expand the gas, hence, , But, , Now, form ideal gas equation :→

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Subtracting equation 3) from equation 4), , Put, , or, , 41. How do you justify that when a body is being heated at melting point, the, temperature remains Constant?, Ans.When a body is being heated below the malting point, the heat supplied increases the, potential as well as the kinetic energy of the molecules. Due to the increase in the kinetic, energy of the mole clues, the temperature increases. But at melting point, heat goes, to, increase only the potential energy of molecules and hence the temperature remains the, same., , 42. Draw and explain a P – T diagram for water showing different phases?, Ans.

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1) The l - V curve represent those points where the liquid and vapour phases are in, equilibrium., 2) The s – l curve represent the points where the solid and liquid phases exist in equilibrium., 3) The s – v is the sublimation curve where a solid changes into vapour phase without, passing through the liquid stage, 4) Triple point → Intersection of three curves is the triple point. It represents a unique, temperature and pressure and it is only at this point that the three phases can exist together, in equilibrium., , 43. From what height should a piece of ice fall so that it completely melts? Only one –, quarter of heat produced is absorbed by the ice. Given latent heat of ice is 3.4 × 105 J |, Kg and acceleration due to gravity, g = 10m | s2?, Ans.Let m = Mass of piece of ice, h = height from which it falls., ∴ Loss of Potential energy = m g h, The Potential energy of ice is converted into heat.

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Since the ice absorbs only one – quarter of this,, ∴ Heat absorbed by ice,, , If L Joules | Kg is the latent heat of ice, them, , Equating 1) & 2) for Q, , 44. A gas can have any value of specific heat depending upon how heating is carried out., Explain?, Ans.If m = Mass of gas, Q = heat supplied, = Change in temperature, then specific heat of gas,, , 1) Let gas is compressed suddenly, So no heat is supplied from outside (i.e. Q = 0) but the, temperature of the gas in the gas increases due to compression,

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2) Let the gas is heated in a way that the temperature is constant (∆T = O) then,, , Hence, depending upon conditions of heating. The value of C will be different., , 45. A 0.20 Kg aluminum block at 800C is dropped in a copper calorimeter of mass 0.05 Kg, containing 200 cm3 of ethyl alcohol at 200C. What is the final temperature of the, mixture? Given Density of ethyl alcohol = 0.81 g | cm3 ; specific heat of ethyl alcohol =, 0.6 cal | g | 0C ; specific heat of copper = 0.094 cal | g | 0C, specific heat of Al = 0.22 cal | g, | 0C?, Ans.Let θ0C = final temperature of the mixture., Mass of ethyl alcohol = volume ×Density, = 200 × 0.81, = 162 g, Heat lost by Aluminum block = Mass X specific heat X fall in temperature, , Heat gained by the ethyl alcohol and calorimeter = (Mass of ethyl alcohol × specific heat ×, change in Temperature) + Mass of copper calorimeter × specific heat X change in

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Temperature, , But, Heat gained = Heat Lost, So, from equation 1) & 2), , 46. Why is there a difference in the specific heat curve as given by Delong’s petit law, and the experimental result at low temperatures?, , Ans.Now, from Dulong & Petit law, the specific heat is independent of temperature but it is, experimentally seen that specific heat at lower temperatures is directly proportional to the

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cube of temperatures. The above dependence is because of the fact that the particles in the, crystal oscillate as if they are coupled Quantum Harmonic Oscillator., , 47. Specific heat of Argon at constant Pressure is o.125 cal | g | K and at constant, volume is 0.075 cal | g | K. Calculate the density of argon at N.T.P. Given that J = 4.2J |, cal?, Ans.Specific heat at constant and Pressure, CP = 0.125 cal | g | K, CP = 0.125 × 4.2 × 1000 J | Kg | K, CP = 525 J | Kg | K →1), Specific heat at constant volume, CV = 0.075 cal | g | K, , The gas constant, r for I kg of gas is given by:-, , Normal pressure = P = h P g = 0.76 ×13600 × 9.8 = 101292.8N | m2, Normal Temperature = T = 273K., Suppose V = Volume of argon in m3 at N. T. P.

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∴ Density of Argon,, , 48. How is heat loss reduced in Calorimeter?, Ans.1) Heat loss due to radiation is reduced by polishing inner and outer surfaces of the, Calorimeter., 2) Heat loss due to conduction is reduced by filling the space between the calorimeter and, insulating jacket with poor conductor of heat., 3) Heat loss due to convection is done by using a insulating lid., , 49. What is critical temperature? How will you differentiate between a gas and a, vapour depending on critical temperature?, Ans.The temperature above which a gas connot be liquefied, no matter how great the, pressure is called critical temperature. If the substance lies above the critical temperature, then it falls in the gaseous region. If the substance lies below the critical temperature than it, falls in the vapour stage., , 50. If for hydrogen CP – CV = a and for oxygen CP – CV = b where CP & CV refer to, specific heat at constant pressure and volume then what is the relation between a and, b?, Ans.For H2, CP – CV = a, CP = Specific heat at constant pressure, CV = Specific heat at constant Volume, For O2 = CP – CV = b, And r =

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M = Molecular weight, I = Mechanic cal equivalent of heat, Now, we know that, CP – CV = r, , So, for, , from equation 1), 2a =, , from equation 2), 32b =, , Equating above equations for, , 2a = 32 b, a = 16 b, , 51. A ball is dropped on a floor from a height of 2cm. After the collision, it rises up to a, height of 1.5m. Assuming that 40% of mechanical energy lost goes to thermal energy, into the ball. Calculate the rise in temperature of the ball in the collision. Specific heat

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capacity of the ball is 800J/k. Take g = 10m/s2, Ans. Initial height = h1=2m, Final height = h2=1.5m, Since potential energy = mechanical energy for a body at rest as K.E =0, Mechanical energy lost =, =, =, =5J, Now (mechanical energy lost) × 40% = heat gained by ball, , ∆T = 2.5×10-3 0C

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2 Marks Questions Part 3, 52.A thermometer has wrong calibration. It reads the melting point of ice as – 100C. It, reads 600C in place of 500C. What is the temperature of boiling point of water on the, scale?, Ans.Lower fixed point on the wrong scale = -100C., Let ‘n’ = no. divisions between upper and lower fixed points on this scale. If Q = reading on, this scale, then, , Now, C = Incorrect Reading = 600C, Q = Correct Reading = 500C, So,, , n = 140, Now,, , On, the Celsius scale, Boiling point of water is 1000C, , So,, , Q = 1300C

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53. Write the advantages and disadvantages of platinum resistance thermometer?, Ans.Advantages of Platinum Resistance thermometer:1) High accuracy of measurement, 2) Measurements of temperature can be made over a wide range of temperature i.e. from –, 2600C to 12000C., → Disadvantages of Platinum Resistance thermometer:1) High Cost, 2) Requires additional equipment such as bridge circuit, Power supply etc., , 54.If the volume of block of metal changes by 0.12% when it is heated through 200C., What is the co-efficient of linear expansion of the metal?, Ans.The co-efficient of cubical expansion y of the metal is given by:-, , Here,

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∴ Co-efficient of linear expansion of the metal is :-, , 55. The density of a solid at00C and 5000C is in the ratio 1.027 : 1. Find the co-efficient of, linear expansion of the solid?, Ans .Density at 00C = SO, Density at 5000C = S500, Now, SO = S500, Where, Y = Co-efficient of volume expansion, ∆T = Change in temperature, , ∆T = Change in temperature, ∆T = Final Temperature – Initial temperature, ∆T = 500 - 00C, ∆T = 5000C, Or

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Now, Co-efficient of linear expansion (α) is related to co-efficient of volume expansion (Y) as, :-, , 56. If one Mole of a monatomic gas is mixed with 3 moles of a diatomic gas. What is the, molecular specific heat of the mixture at constant volume?, Ans.For, a monatomic gas, Specific heat at consent volume = CV1 =, Constant, No. of moles of monatomic gas = n1 = 1 mole, No. of moles of diatomic gas = n2 = 3 moles., For, diatomic gas, specific heat at constant volume, , Applying, conservation of energy., , ; R = Universal Gas

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Let CV = Specific heat of the mixture;, , R = Universal Gas constant, , 57. Calculate the difference between two principal specific heats of 1g of helium gas at, N. T. P. Given Molecular weight of Helium = 4 and J = 4.186 J/cal and Universal Gas, constant, R = 8.314J / mole / K?, Ans. Molecular weight of Helium = M = 4, Universal Gas Constant, R = 8.31J | mole | K, CP = specific heat at constant Pressure, CV = specific heat at constant Volume

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Now,, , for 1 mole of gas., , Where R = Universal Gas Constant = 8.31J | mole | K, J = 4.186 J | cal, M = Molecular weight of Helium = 4, , 58. Why does heat flow from a body at higher temperature to a body at lower, temperature?, Ans.When a body at higher temperature is in contact with a body at lower temperature,, molecule with more kinetic energy that are in contact with less energetic molecules give up, some of their kinetic energy to the less energetic ones., , 59. A one liter flask contains some mercury. IT is found that at different temperatures,, then volume of air inside the flask remains the same. What is the volume of mercury in, the flask? Given the co-efficient of linear expansion of glass = 9 × 10-6 / 0C and coefficient of volume expansion of mercury = 1.8 × 10-4 /, 0C, , Ans.It is given that volume of air in the flask remains the same at different temperature. This, is possible only when the expansion of glass is exactly equal to the expansion of mercury,, Co-efficient of cubical expansion of glass is :-

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Co-efficient of cubical expansion of mercury is :→, , Volume of flask, V = 1 liter = 1000 cm3., Let Vm Cm3 be the volume of mercury in the flask., Expansion of flask = Expansion of Mercury, , ∴ Volume of Mercury,, , 60. A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°, C, calculate the coefficient of performance., Ans. Temperature inside the refrigerator,, , Room temperature,, , = 9°C = 282 K, , = 36°C = 309 K, , Coefficient of performance =, , = 10.44, Therefore, the coefficient of performance of the given refrigerator is 10.44.

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61. A steam engine delivers, , of work per minute and services, , of, , heat per minute from its boiler. What is the efficiency of the engine? How much heat is, wasted per minute?, Ans. Work done by the steam engine per minute, W =, Heat supplied from the boiler, H =, , Efficiency of the engine =, , Hence, the percentage efficiency of the engine is 15 %., Amount of heat wasted =, =, , =, , Therefore, the amount of heat wasted per minute is, , ., , 62. A thermodynamic system is taken from an original state to an intermediate state by, the linear process shown in Fig. (12.13), , Its volume is then reduced to the original value from E to F by an isobaric process., Calculate the total work done by the gas from D to E to F

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Ans.Total work done by the gas from D to E to F = Area of ΔDEF, Area of ΔDEF =, , Where,, DF = Change in pressure, =, = 300 N/, FE = Change in volume, =, Area of ΔDEF =, , = 3.0 m3, = 450 J, , Therefore, the total work done by the gas from D to E to F is 450 J.

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3 Marks Questions, 1.Calculate the work done during the isothermal Process?, Ans.Let an ideal gas is allowed to expand very slowly at constant temperature. Let the, expands from state A (P1, V1) to state B (P2, V2), The work by the gas in expanding from state A to B is, , For ideal gas, PV = N R T, , or P =, , Use 2) in i), , W=

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Since n, R and T are constant so,, , W=, , W isothermal = nRT Loge V, , W isothermal =, , W isothermal – nRT Loge, , W isothermal = 2.303 nRT Log 10, , If M = Molecular Mass of gas then for 1 gram of ideal gas,, , W isothermal = 2.303, , W isothermal 2.303 r T Log 10, , r = Gas constant for 1 gm of an ideal gas,, , Since P1 V1 = P2 V2, , So W isothermal = 2.303 r T log 10

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2.Five moles of an ideal gas are taken in a Carnot engine working between 1000C and, 300C. The useful work done in 1 cycle is 420J. Calculate the ratio of the volume of the gas, at the end and beginning of the isothermal expansion?, Ans.High temperature, TH = 1000C = 100+ 273 = 373K, Low temperature, TL = 300C, = 30 +273 = 303K, Amount of the gas, n = 5 moles, Useful work done per cycle, W = QH - QL, Now, W = 420 J, So, QH – QL = 420J → 1), , Now,, , Or QH =

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3.Deduce the work done in the following complete cycle?, , Ans.1) Work done during the process from A to B = WAB, WAB = area ABKLA (∴ because area under p-v curve gives work done), = area of ∆ ABC + area of rectangle, , =, , BC = KL = 4-1 = 3l = 3x10-3m3, AC = 4-2 = 2N|m2, LC = 2-0 = 2N|m2, , WAB =, , = 3×10-3+6×10-3, WAB = 9×10-3J

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Since gas expands during this process, hence WAB = 9×10-3J, 2) Work done during the process from B to C(compression) is WBC = -area BCLK, ( - ve because gas compresses during BC), = - KL × LC, WBC = -3×10-3×2, = - 6×10-3J, 3) Work done during the process from C to A :As there is no change in volume of gas in this process, WCA = O, So, net work done during the complete cycle = WAB + WBC +WCA, = 9×10-3-6×10-3+0, Net work done = 3×10-3J, , 4.One kilogram molecule of a gas at 400k expands isothermally until its volume is, doubled. Find the amount of work done and heat produced?, Ans.Initial volume, V1= V, Final volume, V2= 2V, Initial temperature T = 400k, Find temperature = 400k (∴ process is isothermal), Gas constant, R = 8. 3kJ |mole |k=8.3x10-3J|mole|k, , Work done during is thermal process=w=2.3026RT Log 10

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W = 2.3026×8.3×10-3×400×log 10, , W = 2.3026×8.3×10-3×400×Log 10 (2), W = 2.3016J, If H is the amount of heat produced than,, , 5.Calculate difference in efficiency of a Carnot energy working between:1) 400K and 350K, 2) 350K and 300K, , Ans.Efficiency of heat engine = n = 1 -, , T2 = final temperature, T1 = Initial temperature, , 1) 400K and 350K,, T2 = 350, T1 = 400, n=1-, , =

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n1 =, , 2) 350K and 300K, T2 = 300K; T1 = 350K, , n1 = 1 -, , =1-, , =, , n1 =, Change in efficiency = n2 – n1 = 14.3% - 12.5% = 1.8%, , 6.How do you derive Newton’s law of cooling from Stefan’s law?, Ans.Acc. to Newton’s law of cooling, the rate of loss of heat of a liquid is directly proportional, to the difference in temperature of the liquid and the surrounding, provided the difference, in temperature is very small., , Let a body be maintained at T K. Let To be the temperature of the surroundings. Let T ≫ To., There will be loss of heat be the body, Acc. to Stefan’s law, amount of heat energy lost per second per unit area of the body is, E=ϵσ

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σ = Stefan’s constant, ε = Emissivity of the body and surroundings, E=, , Incase of Newton’s cooling, T ≈ To, E=εσ, E=εσ, , Hence, → Hence the Newton’s law of cooling, Eα, , 7.Define the terms reflectance, absorptance and transmittance. How are they related?, Ans.1) Reflectance – Ratio of amount of thermal radiations reflected by the body in a given, time to total amount of thermal radiations incident on body It is represented by r, 2), Absorptance – is the ratio of the amount of thermal to the total amount of thermal radiations, incident on it. It is represented by a, 3) Transmittance – It is the ratio of the amount of thermal radiations transmitted by body in, a given time to the total amount of thermal radiations incident on the body in a given time. It, is represented by t., Let Q = Amount of the radiations incident by the body in a given time, Q1 = Amount of thermal radiations reflected by the body in a given time.

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Q2 = Amount of thermal radiations absorbed by the body in a given time., Q3 = Amount of thermal radiations transmitted by the body in a given time,, ∴ By definition,, , New, r + a + t =, , R+a+t=, , R+a+t=1, , If t = 0, A=1–r, that is good reflectors are bad absorbers, , 8. If half mole of helium is contained in a container at S. T. P. How much heat energy is, needed to double the pressure of the gas, keeping the volume of the gas constant? Given, specific heat of gas = 3J | g |K., Ans.Number of moles of Helium gas =

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Specific heat of Helium gas =, , Molecular weight = M = 4, Temperature, T1 = 273 K., ∴ Molar specific heat at constant volume = CV = MCV, CV = 4 × 3, CV = 12 J | mol | K, Since, Volume is constant, Pα T or, , ∴, , Or, , P2 = Final Pressure = 2 P, P1 = Initial Pressure = P, , = Constant

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Now, Heat required,, , Heat required = 1638 J, , 9. Calculate the amount of heat necessary to raise the temperature of 2 moles of HE gas, from 200C to 500C using:1) Constant – Volume Process 2) Constant Pressure Process, Here for, He; CV = 1.5 R and CP = 2.49R, Ans .1) The amount of heat required for constant – volume process is :-, , Here, n = 2 moles, CV = 1.5 R = 1.5 X 8.314 J | mol | 0C, T2 = final Temperature, T1 = Initial Temperature, , 2) The amount of heat required for constant – Pressure process is :-, , Here, n = 2 moles,

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Since the temperature rise is same in both the cases, the change in internal energy is the, same i.e. 748J. However, in constant – pressure Process excess heat is supplied which is used, in the expansion of gas., , 10. An electric heater supplies heat to a system at a rate of 100W. If system performs, work at a rate of 75 Joules per second. At what rate is the internal energy increasing?, Ans.Heat is supplied to the system at a rate of 100 W., ∴Heat supplied, Q = 100 J/s, The system performs at a rate of 75 J/s., ∴Work done, W = 75 J/s, From the first law of thermodynamics, we have:, Q=U+W, Where,, U = Internal energy, ∴U = Q–W, = 100 –75, = 25 J/s, = 25 W, Therefore, the internal energy of the given electric heater increases at a rate of 25 W.

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4 Marks Questions, 1. A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If, the geyser operates on a gas burner, what is the rate of consumption of the fuel if its, heat of combustion is, , J/g?, , Ans.Water is flowing at a rate of 3.0 litre/min., The geyser heats the water, raising the temperature from 27°C to 77°C., Initial temperature,, , Final temperature,, , = 27°C, , = 77°C, , ∴Rise in temperature, ΔT =, , = 77–27= 50°C, Heat of combustion =, , Specific heat of water, c = 4.2, , Mass of flowing water, m = 3.0 litre/min = 3000 g/min, Total heat used, ΔQ = mc ΔT, =, =, , = 15.75 g/min, , Rate of consumption =, , 2. What amount of heat must be supplied to, , kg of nitrogen (at room, , temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of, .), Ans. Mass of nitrogen,, , Rise in temperature, ΔT = 45°C

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Molecular mass of, , , M = 28, , Universal gas constant, R =, Number of moles,, , Molar specific heat at constant pressure for nitrogen,, , The total amount of heat to be supplied is given by the relation:, , = 933.38 J, Therefore, the amount of heat to be supplied is 933.38 J., , 3. Explain why, (a) Two bodies at different temperatures, , necessarily settle to the mean temperature, , and, , if brought in thermal contact do not, , ., , (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the, different parts of a plant from getting too hot) should have high specific heat.

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(c) Air pressure in a car tyre increases during driving., (d) The climate of a harbour town is more temperate than that of a town in a desert at, the same latitude., Ans.(a) When two bodies at different temperatures, , and, , are brought in thermal, , contact, heat flows from the body at the higher temperature to the body at the lower, temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become, equal. The equilibrium temperature is equal to the mean temperature, , only, , when the thermal capacities of both the bodies are equal., (b) The coolant in a chemical or nuclear plant should have a high specific heat. This is, because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice, versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or, chemical plant. This would prevent different parts of the plant from getting too hot., (c) When a car is in motion, the air temperature inside the car increases because of the, motion of the air molecules. According to Charles' law, temperature is directly proportional, to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will, also increase., (d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold), than a town located in a desert at the same latitude. This is because the relative humidity in a, harbour town is more than it is in a desert town., , 4. A cylinder with a movable piston contains 3 moles of hydrogen at standard, temperature and pressure. The walls of the cylinder are made of a heat insulator, and, the piston is insulated by having a pile of sand on it. By what factor does the pressure of, the gas increase if the gas is compressed to half its original volume?, Ans.The cylinder is completely insulated from its surroundings. As a result, no heat is

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exchanged between the system (cylinder) and its surroundings. Thus, the process is, adiabatic., Initial pressure inside the cylinder =, , Final pressure inside the cylinder =, , Initial volume inside the cylinder =, , Final volume inside the cylinder =, , Ratio of specific heats, Y = 1.4, For an adiabatic process, we have:, , The final volume is compressed to half of its initial volume., ∴, , Hence, the pressure increases by a factor of 2.639., , 5. In changing the state of a gas adiabatically from an equilibrium state A to another, equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas, is taken from state A to B via a process in which the net heat absorbed by the system is, 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal =, 4.19 J)

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Ans.The work done (W) on the system while the gas changes from state A to state B is 22.3 J., This is an adiabatic process. Hence, change in heat is zero., ∴ ΔQ = 0, ΔW = –22.3 J (Since the work is done on the system), From the first law of thermodynamics, we have:, ΔQ = ΔU + ΔW, Where,, ΔU = Change in the internal energy of the gas, ∴ ΔU = ΔQ– ΔW = –(–22.3 J), ΔU = + 22.3 J, When the gas goes from state A to state B via a process, the net heat absorbed by the system, is:, ΔQ = 9.35 cal = 9.35, , 4.19 = 39.1765 J, , Heat absorbed, ΔQ = ΔU + ΔQ, ∴ΔW = ΔQ – ΔU, = 39.1765 – 22.3, = 16.8765 J, Therefore, 16.88 J of work is done by the system., , 6. Two cylinders A and B of equal capacity are connected to each other via a stopcock., A contains a gas at standard temperature and pressure. B is completely evacuated. The, entire system is thermally insulated. The stopcock is suddenly opened. Answer the, following:

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(a) What is the final pressure of the gas in A and B?, (b) What is the change in internal energy of the gas?, (c) What is the change in the temperature of the gas?, (d) Do the intermediate states of the system (before settling to the final equilibrium, state) lie on its P-V-T surface?, Ans.(a) 0.5 atm, (b) Zero, (c) Zero, (d) No, Explanation:, (a) The volume available to the gas is doubled as soon as the stopcock between cylinders A, and B is opened. Since volume is inversely proportional to pressure, the pressure will, decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the, pressure in each cylinder will be 0.5 atm., (b) The internal energy of the gas can change only when work is done by or on the gas. Since, in this case no work is done by or on the gas, the internal energy of the gas will not change., (c) Since no work is being done by the gas during the expansion of the gas, the temperature, of the gas will not change at all., (d) The given process is a case of free expansion. It is rapid and cannot be controlled. The, intermediate states do not satisfy the gas equation and since they are in non-equilibrium, states, they do not lie on the P-V-T surface of the system.

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5 Marks Questions, 1.Derive the equation of state for adiabatic change?, Ans.Let P = pressure, V = volume and T = Temperature of the gas in a cylinder fitted with a, perfectly, frictionless piston., , Suppose a small amount of heat d Q is given to the system. The heat is spent in two ways:1) In increasing the temperature of the gas by la small range d T, at constant volume, 2) In expansion of gas by a small volume d v, So, d Q = CV d T + P d V, In adiabatic change, no heat is supplied from outside, So, d Q = O, CV d t + P d V = O →(1), Acc. to standard gas equation, PV = RT, Diff both sides

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PdV+VdP=RdR, R d T = P d V +V d P (d R=O as R is a constant), dT=, , Using this in equation i), , Cv, , CV P d V + CV V d P + R P d V = O, (CV + R) P d V + CV V d P = O →2), As, CP – CV = R, or CP = R + CV, So equation 2) becomes, CP P d V + CV V d P = O, Dividing above equation by CV PV

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Integrating both sides, , Loge V + Loge P = constant, + Loge P = constant, , Loge, Loge P, , = constant, , = antilog (constant), , K = another constant, , 2.Derive an expression for the work done during isothermal expansion?, Ans.Consider one gram mole of ideal gas initially with pressure, volume and temperature as, P, V, T, Let the gas expand to a volume V2, when pressure reduces to P2 and at the same, temperature T, If A = Area of cross – section of piston, Force = Pressure × Area, F=PxA, If we assume that piston moves a displacement d x,, the work done : → d w = F d x, dw=P×A×dx, dw=P×dv, Total work done in increasing the volume from V1 to V2

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W=, Since, PV = RT (from ideal gas equation), P=, , W=, , W = RT, , W = R T Loge, , W=RT, , W = R T Loge, , W = 2.3026 R T Log 10, , As P1 V1 = P2 V2

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So W = 2.3026 R T Log 10, , 3.Briefly describe a Carnot cycle and derive an expression for the efficiency of Carnot, cycle?, Ans.The construction of a heat engine following Carnot cycle is :1) Source of heat :- It is maintained at higher temperature T1, 2) Sink of heat – It is maintained at lower temperature T2, 3) Working base :- A perfect ideal gas is the working substance., Theory :- Carnot cycle consist of four stages:1) I so thermal expansion, 2) Adiabatic expansion, 3) I so thermal compression, 4) Adiabatic compression., , 4. Discuss briefly energy distribution of a black body radiation. Hence deduce wien’s, displacement law?, Ans.For a black body, the monochromatic emittance, wavelength, , of the radiation emitted., , So, at a given temperature of black body :→, , of the black body and the

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a) The energy emitted is not distributed uniformly amongst all wavelengths., b) The energy emitted in maximum corresponding to a certain wavelength, , and its, , falls on either side of it., As temperature of black body is increased., a) The total energy emitted rapidly increases for any given wavelength., b) The wavelength corresponding to which energy emitted is maximum is shifted towards, shorter wavelength side i. e,, , or, , m decreases with rise in temperature, , m T = constant, , Thus is the wein’s displacement law.