228. If the temperature of air is 16.5 0C and dew point is 6.5 0C , find the relative humidity of air. ( SVP at 6, 7 , 16 and 17 0C are 7.05, 7.51, 13.62 and 14.42 mm of Hg respectively ) [ 51.9%]
229. What is the dew point on a day if the humidity is 40% on that day when the temperature is 300 C.
[12 0C ]
230. On a certain day the dew point is 8.5 0C and the room temperature is 18.4 0C. Find the RH if the maximum vapour pressure for 8, 9, 18 and 19 0C are 8.04, 8.61, 15.46 and 16.46 mm of Hg respectively. [ 52.5 %]
229. What is the dew point on a day if the humidity is 40% on that day when the temperature is 300 C.
[12 0C ]
230. On a certain day the dew point is 8.5 0C and the room temperature is 18.4 0C. Find the RH if the maximum vapour pressure for 8, 9, 18 and 19 0C are 8.04, 8.61, 15.46 and 16.46 mm of Hg respectively. [ 52.5 %]
Transfer of heat
231. A man, the surface area of whose skin is 2 m2, is sitting in a room where the air temperature is 20 0C. If the temperature of skin is 28 0C , find the rate at which his body loses heat. The emissivity of his skin is 0.97 and = 5.7 x 10-8 W/m2 K4. [ 92.2 W]
Hints: The rate of heat loss due to radiation according to Stephen’s law is given by , where is emissivity.
232. Calculate the rate of loss of heat from an unclothed person standing in air at 23 0C . Assuming that the skin temperature is 34 0C and the body surface area is 1.5 m2 , emissivity = 0.7 and = 5.7 x 10-8 W/m2 K4. [ 46.85 W]
Hints: The rate of heat loss due to radiation according to Stephen’s law is given by .
233. A closed metal vessel contains water (i) at 30 0C (ii) at 75 0C . The vessel has surface area 0.5 m2 and a uniform thickness of 4 mm. If the outside temperature is 15 0C , calculate the heat loss per minute by conduction in each case. ( Thermal conductivity of metal = 400 W / mK ) [ 4.5×107, 18 x107J]
Hints: Use the relation of rate of flow of heat due to conduction is
231. A man, the surface area of whose skin is 2 m2, is sitting in a room where the air temperature is 20 0C. If the temperature of skin is 28 0C , find the rate at which his body loses heat. The emissivity of his skin is 0.97 and = 5.7 x 10-8 W/m2 K4. [ 92.2 W]
Hints: The rate of heat loss due to radiation according to Stephen’s law is given by , where is emissivity.
232. Calculate the rate of loss of heat from an unclothed person standing in air at 23 0C . Assuming that the skin temperature is 34 0C and the body surface area is 1.5 m2 , emissivity = 0.7 and = 5.7 x 10-8 W/m2 K4. [ 46.85 W]
Hints: The rate of heat loss due to radiation according to Stephen’s law is given by .
233. A closed metal vessel contains water (i) at 30 0C (ii) at 75 0C . The vessel has surface area 0.5 m2 and a uniform thickness of 4 mm. If the outside temperature is 15 0C , calculate the heat loss per minute by conduction in each case. ( Thermal conductivity of metal = 400 W / mK ) [ 4.5×107, 18 x107J]
Hints: Use the relation of rate of flow of heat due to conduction is
234. Assuming that the thermal conductivity of woolen glove is equivalent to a layer of quiescent air 3 mm thick, determine the heat lost per minute from the man’s hand of surface area 200 cm2 0n a winter’s day when atmospheric temperature is 3 0C. The skin temperature is taken 34 0C and thermal conductivity of air is 24 x 10-3 W/mK. [ 354 J/min]
Hints: Use the relation of rate of flow of heat due to conduction is
Hints: Use the relation of rate of flow of heat due to conduction is
235. A bar of 0.2 m in length and cross-sectional area 2.5 x 10-4 m2 is ideally lagged. One end is maintained at 373 k and other is kept in melting ice. Calculate the rate at which ice melts. [ 1.47 x 10-4 Kg / s ]
Hints: At first calculate and the energy is used to melt ice. So, mL = Q.
Hints: At first calculate and the energy is used to melt ice. So, mL = Q.
236. One face of a sheet of cork is 3 mm thick is placed contact with one face of the of a sheet of glass 5 mm thick, both sheet being 20 cm square. The outer faces of this square composite sheet are maintained at 100 0C and 20 0C , the glass being at the higher temperature. Find (a) the temperature of the glass – cork interface, (b) the rate at which heat is conducted across the sheet, neglecting edge effect. Thermal conductivity of cork and glass are 6.3 x 10-2 W/mK and 7.2 x 10-1 W/mK respectively. [ 90 0C, 57.6 W ]
Hints: The rate of flow of heat through glass due to conduction, ………..(i) Similarly, rate of flow of heat through cork
………..(ii)
Rate of flow of heat through glass is equal to that of cork. So, equating (i) and (ii), can be determined.
237. Estimate the rate of heat loss from a room through glass window of area 2 m2 and thickness 3 mm when temperature of room is 20 0C and that of air outside is 5 0C. Thermal conductivity of glass is 0.72 W/mK respectively. [ 7200 W]
Hints: Use the relation of rate of flow of heat due to conduction
Hints: The rate of flow of heat through glass due to conduction, ………..(i) Similarly, rate of flow of heat through cork
………..(ii)
Rate of flow of heat through glass is equal to that of cork. So, equating (i) and (ii), can be determined.
237. Estimate the rate of heat loss from a room through glass window of area 2 m2 and thickness 3 mm when temperature of room is 20 0C and that of air outside is 5 0C. Thermal conductivity of glass is 0.72 W/mK respectively. [ 7200 W]
Hints: Use the relation of rate of flow of heat due to conduction
238. The silica cylinder of a radiant wall heater is 0.6 m long and has a radius of 5 mm. If it is rated at 1.5 kW, estimate the temperature when operated. State two assumptions you have made in making your statement. Given Stephen constant is 6 x 10-8 W/m2K4. [1070 K]
Hints: Use relation . The assumptions are (a) The heater is considered as a perfectly black body, (b) The temperature of surrounding is considered to absolute zero.
Hints: Use relation . The assumptions are (a) The heater is considered as a perfectly black body, (b) The temperature of surrounding is considered to absolute zero.
239. What is the ratio of the energy per second radiated by the filament of a lamp at 2500 K to that radiated at 2000 K, assuming the filament is a black body radiator. The filament of a lamp can be considered as a 90% black body radiator. Calculated the energy per second radiated when its temperature is 2000 K and surface area is 10-6 m2. [ 2.44, 0.82] Hints: Apply relation for both lamps and divide them to get result. In second case, take efficiency 90 % or 0.9.
240. The sun is a black body of surface temperature 6000 K. If the sun’s radius is 7 x 108 m, calculate the energy per second radiated from its surface. The earth is about 1.5 x 1011 m from the sun. Assuming all the radiation from the sun falls on a sphere of this radius, estimate the energy per second per meter square ( solar constant ) received by the earth. Stephen constant is 5.7 x 10-8 W/m2K4. [ 1600 W/m2 ] Hints: Apply relation to obtain energy radiated by sun per second. The energy radiated by sun spherically distributes all over and solar constant is the energy received by earth per unit area per unit time. So divide the power radiated by sun ‘P’ by surface are of the sphere made by the radius 1.5 x 1011m. i.e Solar constant S =
241. A sphere of radius 2 cm with a black surface is cooled and then suspended in a large evacuated enclosure the black walls of which are maintained at 27 0C. If the rate of change of thermal energy of the sphere is 1.85 J/s, when its temperature is -73 0C, calculate the value for the Stephen constant. [5.7 x 10-8 W/m2K4 ]
Hints: The rate of heat loss by black body is given by .
240. The sun is a black body of surface temperature 6000 K. If the sun’s radius is 7 x 108 m, calculate the energy per second radiated from its surface. The earth is about 1.5 x 1011 m from the sun. Assuming all the radiation from the sun falls on a sphere of this radius, estimate the energy per second per meter square ( solar constant ) received by the earth. Stephen constant is 5.7 x 10-8 W/m2K4. [ 1600 W/m2 ] Hints: Apply relation to obtain energy radiated by sun per second. The energy radiated by sun spherically distributes all over and solar constant is the energy received by earth per unit area per unit time. So divide the power radiated by sun ‘P’ by surface are of the sphere made by the radius 1.5 x 1011m. i.e Solar constant S =
241. A sphere of radius 2 cm with a black surface is cooled and then suspended in a large evacuated enclosure the black walls of which are maintained at 27 0C. If the rate of change of thermal energy of the sphere is 1.85 J/s, when its temperature is -73 0C, calculate the value for the Stephen constant. [5.7 x 10-8 W/m2K4 ]
Hints: The rate of heat loss by black body is given by .
242. The element of an electric fire, with an output of 1.0 kW, is a cylinder 25 cm long and 1.5 cm in diameter. Calculate the temperature when in use of it behaves as a black body. [ 1105 K ]
Hints: Use relation , where A is the curved surface area of cylindrical electric element, equal to 2 r h
243. A solid copper sphere of diameter 10 mm is cooled to a temperature of 150 K and is then placed in an enclosure maintained at 290 K. Assuming that all interchange of heat is by radiation, calculate the initial rate of rise of temperature of the sphere. The sphere may be treated as a black body. Density of copper 893 kg /m3, specific heat capacity of copper 3.7 x 102 J / kgK and Stephen constant 5.7 x 10-8 W/m2K4. [ 0.068 K/s]
Hints: Use relation . This energy is used to increase the temperature. So, . Equate the relations to obtain .
244. The element of 1 kW electric fire has a surface area of 0.006 m2. Estimate the working temperature. Stephen constant is 5.7 x 10-8 W/m2K4 [ 1300 K]
Hints: Use relation .
245. A roof measures 20m x 50m and is blackened. If temperature of the sun’s surface is 6000 K and its radius is 7.8 x 108 m, and the distance from the earth is 1.5 x 1011 m calculate how much solar energy is incident on the roof per minute. Assume that the half of the energy is lost in passing through the earth’s atmosphere and the roof is normal to the sun’s rays. [ 5.6 x 107 J]
Hints: Power radiated by sun and solar constant S = . The additional term ½ is taken according to question that half of the energy is lost in passing through the earth’s atmosphere. Finally multiply area of the roof with solar constant.
246. Calculate the apparent temperature of the sun from the following data. Sun’s radius = 7.04 x 105 Km. Distance from earth = 14.72 x 107 Km Solar constant = 1400 w m-2 Stephen constant = 5.7 x 10-8 W m-2 k-4 [ 5450 0C ]
Hints: Total power released by sun is P = S × 4 r2, where ‘r’ is the distance of sun from earth. Finally use the Stephen’s law , where A be the surface area of sun, given by A = 4 R2.
Hints: Use relation , where A is the curved surface area of cylindrical electric element, equal to 2 r h
243. A solid copper sphere of diameter 10 mm is cooled to a temperature of 150 K and is then placed in an enclosure maintained at 290 K. Assuming that all interchange of heat is by radiation, calculate the initial rate of rise of temperature of the sphere. The sphere may be treated as a black body. Density of copper 893 kg /m3, specific heat capacity of copper 3.7 x 102 J / kgK and Stephen constant 5.7 x 10-8 W/m2K4. [ 0.068 K/s]
Hints: Use relation . This energy is used to increase the temperature. So, . Equate the relations to obtain .
244. The element of 1 kW electric fire has a surface area of 0.006 m2. Estimate the working temperature. Stephen constant is 5.7 x 10-8 W/m2K4 [ 1300 K]
Hints: Use relation .
245. A roof measures 20m x 50m and is blackened. If temperature of the sun’s surface is 6000 K and its radius is 7.8 x 108 m, and the distance from the earth is 1.5 x 1011 m calculate how much solar energy is incident on the roof per minute. Assume that the half of the energy is lost in passing through the earth’s atmosphere and the roof is normal to the sun’s rays. [ 5.6 x 107 J]
Hints: Power radiated by sun and solar constant S = . The additional term ½ is taken according to question that half of the energy is lost in passing through the earth’s atmosphere. Finally multiply area of the roof with solar constant.
246. Calculate the apparent temperature of the sun from the following data. Sun’s radius = 7.04 x 105 Km. Distance from earth = 14.72 x 107 Km Solar constant = 1400 w m-2 Stephen constant = 5.7 x 10-8 W m-2 k-4 [ 5450 0C ]
Hints: Total power released by sun is P = S × 4 r2, where ‘r’ is the distance of sun from earth. Finally use the Stephen’s law , where A be the surface area of sun, given by A = 4 R2.
Thermodynamics
247. For hydrogen, the molar heat capacity at constant volume and pressure are respectively 20.5 J/molK and 28.8 J/molK. (a) Which heat capacity is related to internal energy ? (b) Calculate the molar gas constant. (c) Calculate the heat needed to rise the temperature of 8g hydrogen from
100 C to 150 C at constant pressure? (d) Increase in the internal energy of the gas. (e) External work done.
[ 8.3 J/molK, 576 J, 410 J, 166 J] Hints: (a) Specific heat capacity at constant volume Cv is related to the internal energy. (b) Molar gas constant R = Cp – Cv (c) dQ = N Cp dT, where N is the number of mole = 4/2 = 4. (d) Increase in internal energy dU = N Cv dT (e) External work done dW = dQ – dU
248. A gas has a volume of 0.02 m3 at pressure of 2 x 105 Pa and temperature of 27 0C. It is heated at constant pressure until its volume be 0.03 m3. Calculate the (a) External work done, (b) The new temperature of the gas, (c) Increase in internal energy of the gas. Mass of the gas is 16 g, molar heat capacity at constant volume is 0.8 J/molK and its molar mass is 32 g. [ 2000J, 450K, 60J] Hints: (a) External work done dW = P V (b) Ideal gas equation (c) Increase in internal energy dU = N Cv dT
249. If ratio of the principal specific heat capacities of a gas is 1.4 and its density at stp is 0.090 kg/m3. Calculate the specific heat capacities at constant pressure and at constant volume. [ 144 x104, 1.03 x 104 J/kgK] Hints: Pressure exerted by gas P = r T. So, gas constant r = Now, cp – cv = r …………(i) and cp = 1.4 cv ………..(ii) Solving equation (i) and (ii), cp and cv can be determined.
250. Given that the volume of a gas at stp is 2.24 x 10-2 m3 /mol , calculate the molar gas constant R and use it to find the difference between the quantities of heat required to raise the temperature of 0.01 kg of oxygen from 0 0C to 10 0C when (a) pressure (b) volume of the gas is kept constant. Given relative molecular mass of the oxygen = 32. [ 8.3 J/molK, 25.9 J] Hints: Molar gas equation PV = NRT gives the molar gas constant ‘R’. Energy required to heat at constant pressure dQ = N Cp T …………..(i) Energy required to heat at constant volume dU = N Cv T ……………..(ii) Subtracting, the obtained result is dQ – dU = N T ( Cp – Cv) = N T R.
251. An ideal gas at a temperature of 290 K and a pressure of 1.0 x 105 N/m2, occupies the volume of 1.0 x 10-3 m3. Its density under these conditions is 0.30 kg/m3. It expands at a constant pressure to a volume of 1.5 x 10-3 m3.Calculate the energy added. It is now compressed isothermally to its original volume. Calculate its final pressure and temperature and the difference between its initial and final internal energy. Given Cv = 7.1 x 102 J / Kg K . [ 81J, 1.5 x 105 N/m2, 435K, 30.9 J] Hints: Ideal gas equation gives the final temperature T2. Pressure exerted by gas P = r T. So, gas constant r = . Also, specific heat capacity at constant pressure cp = cv + r and Heat energy added dQ = m cp T Again, Ideal gas equation P2V2 = P3V3 gives the final pressure P3. Difference between initial and final internal energy is dU = m x cv x dT
252. The density of ideal gas is 1.6 kg/m3 at 27 0C and 1.00 x 105 N/m2 pressure and specific heat capacity at constant volume is 312 J/ kgK. Find the ratio of specific heat capacities at constant pressure and constant volume. [1.67] Hints: Pressure exerted by gas P = r T. So, gas constant r = . Now, specific heat capacity at constant pressure cp = cv + r. And, ratio of specific heat capacities = cp / cv
253. A gas in a cylinder of temperature 17 0C and a pressure of 1.01 x 105 N/m2 is to be compressed to one-eighth of its volume. What would be the difference between the final pressures if the compression is done (a) isothermally, (b) adiabatically ? (c) What would be the final pressure in the later case? Given = 1.40 [10 .48 x 105 N/m2, 666 K] Hints: When gas is compressed isothermally, P1 V1 = P2 V2 When gas is compressed adiabatically, P1 V1 = P2 V2
254. A liter of air initially at 20 0C and 760 mm of Hg is heated at constant pressure until its volume is doubled. Find (a) its final temperature, (b) external work done, (c) the quantity of heat supplied. Given density of air at stp is 1.293 kg/m3 and Cv = 714 J/kgK. [ 586 K, 101.2 J, 355J] Hints: (a) Ideal gas equation gives final temperature (b) External work done, W = P V, where V = 2 lit – 1 lit = 1 lit = 10-3 m3 (c) Volume of gas at 00C is given by Or, Or, V1 = 9.32 x 10-4 m3 Therefore , mass of the gas m = x V = 1.293 x 9.32 x 10-4 = 1.205 x 10-3 kg Quantity of heat supplied dQ = dU + dW = m x cv x dT + P V
255. A mass of air occupying initially a volume 2 x 10-3 m3 at a pressure 760 mm of Hg and temperature 20 0C is expanded adiabatically reversibly to twice its volume, and then compressed isothermally and reversibly to a volume of 3 x 10-3 m3. Find the final temperature and pressure. Given = 1.40
[ 222K, 384 mm of Hg] Hints: Expression for adiabatic expansion P1 V1 = P2 V2 Expression for isothermal compression P2 V2 = P3 V3. Where, P3 is the final pressure. Similarly, the expression for adiabatic expansion T1V1-1 = T2V2-1. In isothermal process, temperature remains constant. So, T2 is the final temperature after isothermal compression.
256. A petrol engines consumes 5 kg petrol per hour. If the power of the engine is 20 KW and the calorific value of petrol is 11 x 103 K Cal per kg, calculate the efficiency of the engine. [ 31.15%] Hints: Efficiency of an engine Where, output = 20 KW = 20000 W Input = 5 Kg petrol per hour = 5 / 60 x 60 kg per sec. = Watt
257. An ideal heat engine operates in a Carnot Cycle between 227 0C and 127 0C. It absorbs 6 x 104 J at the higher temperature . How much work per cycle is this engine capable of performing? [ 5 x 104 J ] Hints: Efficiency of an engine = So, = Where, T2 = temperature of sink in absolute scale T1 = temperature of source in absolute scale Input = 6 x 104 J
258. A Carnot engine operates between a hot reservoir at 320 K and a cold reservoir 260 K. If it absorbs 500 J of heat from hot reservoir, how much work does it deliver? If the same engine working in reverses as a refrigerator, how much work must be supplied to remove 1000 J of heat from the cold reservoir? [ 4333 J] Hints: Efficiency of a heat engine is , = where, T2 = temperature of sink, 260 K T1 = temperature of source, 320 K Input = 500 J Output = ? If the same engine works as a refrigerator, its efficiency Where, = efficiency obtained from above , 18.75 % Q2 = heat taken from cold body, 1000 J W = work supplied, to determine.
259. A motor in a refrigerator has a power of output 200 W. If the freezing compartment is at 270 K and room temperature is 300 K, assuming ideal efficiency, what is the maximum amount of heat that can be extracted from the freezing compartment in 10 min? [ 1.1 x 106 ]
100 C to 150 C at constant pressure? (d) Increase in the internal energy of the gas. (e) External work done.
[ 8.3 J/molK, 576 J, 410 J, 166 J] Hints: (a) Specific heat capacity at constant volume Cv is related to the internal energy. (b) Molar gas constant R = Cp – Cv (c) dQ = N Cp dT, where N is the number of mole = 4/2 = 4. (d) Increase in internal energy dU = N Cv dT (e) External work done dW = dQ – dU
248. A gas has a volume of 0.02 m3 at pressure of 2 x 105 Pa and temperature of 27 0C. It is heated at constant pressure until its volume be 0.03 m3. Calculate the (a) External work done, (b) The new temperature of the gas, (c) Increase in internal energy of the gas. Mass of the gas is 16 g, molar heat capacity at constant volume is 0.8 J/molK and its molar mass is 32 g. [ 2000J, 450K, 60J] Hints: (a) External work done dW = P V (b) Ideal gas equation (c) Increase in internal energy dU = N Cv dT
249. If ratio of the principal specific heat capacities of a gas is 1.4 and its density at stp is 0.090 kg/m3. Calculate the specific heat capacities at constant pressure and at constant volume. [ 144 x104, 1.03 x 104 J/kgK] Hints: Pressure exerted by gas P = r T. So, gas constant r = Now, cp – cv = r …………(i) and cp = 1.4 cv ………..(ii) Solving equation (i) and (ii), cp and cv can be determined.
250. Given that the volume of a gas at stp is 2.24 x 10-2 m3 /mol , calculate the molar gas constant R and use it to find the difference between the quantities of heat required to raise the temperature of 0.01 kg of oxygen from 0 0C to 10 0C when (a) pressure (b) volume of the gas is kept constant. Given relative molecular mass of the oxygen = 32. [ 8.3 J/molK, 25.9 J] Hints: Molar gas equation PV = NRT gives the molar gas constant ‘R’. Energy required to heat at constant pressure dQ = N Cp T …………..(i) Energy required to heat at constant volume dU = N Cv T ……………..(ii) Subtracting, the obtained result is dQ – dU = N T ( Cp – Cv) = N T R.
251. An ideal gas at a temperature of 290 K and a pressure of 1.0 x 105 N/m2, occupies the volume of 1.0 x 10-3 m3. Its density under these conditions is 0.30 kg/m3. It expands at a constant pressure to a volume of 1.5 x 10-3 m3.Calculate the energy added. It is now compressed isothermally to its original volume. Calculate its final pressure and temperature and the difference between its initial and final internal energy. Given Cv = 7.1 x 102 J / Kg K . [ 81J, 1.5 x 105 N/m2, 435K, 30.9 J] Hints: Ideal gas equation gives the final temperature T2. Pressure exerted by gas P = r T. So, gas constant r = . Also, specific heat capacity at constant pressure cp = cv + r and Heat energy added dQ = m cp T Again, Ideal gas equation P2V2 = P3V3 gives the final pressure P3. Difference between initial and final internal energy is dU = m x cv x dT
252. The density of ideal gas is 1.6 kg/m3 at 27 0C and 1.00 x 105 N/m2 pressure and specific heat capacity at constant volume is 312 J/ kgK. Find the ratio of specific heat capacities at constant pressure and constant volume. [1.67] Hints: Pressure exerted by gas P = r T. So, gas constant r = . Now, specific heat capacity at constant pressure cp = cv + r. And, ratio of specific heat capacities = cp / cv
253. A gas in a cylinder of temperature 17 0C and a pressure of 1.01 x 105 N/m2 is to be compressed to one-eighth of its volume. What would be the difference between the final pressures if the compression is done (a) isothermally, (b) adiabatically ? (c) What would be the final pressure in the later case? Given = 1.40 [10 .48 x 105 N/m2, 666 K] Hints: When gas is compressed isothermally, P1 V1 = P2 V2 When gas is compressed adiabatically, P1 V1 = P2 V2
254. A liter of air initially at 20 0C and 760 mm of Hg is heated at constant pressure until its volume is doubled. Find (a) its final temperature, (b) external work done, (c) the quantity of heat supplied. Given density of air at stp is 1.293 kg/m3 and Cv = 714 J/kgK. [ 586 K, 101.2 J, 355J] Hints: (a) Ideal gas equation gives final temperature (b) External work done, W = P V, where V = 2 lit – 1 lit = 1 lit = 10-3 m3 (c) Volume of gas at 00C is given by Or, Or, V1 = 9.32 x 10-4 m3 Therefore , mass of the gas m = x V = 1.293 x 9.32 x 10-4 = 1.205 x 10-3 kg Quantity of heat supplied dQ = dU + dW = m x cv x dT + P V
255. A mass of air occupying initially a volume 2 x 10-3 m3 at a pressure 760 mm of Hg and temperature 20 0C is expanded adiabatically reversibly to twice its volume, and then compressed isothermally and reversibly to a volume of 3 x 10-3 m3. Find the final temperature and pressure. Given = 1.40
[ 222K, 384 mm of Hg] Hints: Expression for adiabatic expansion P1 V1 = P2 V2 Expression for isothermal compression P2 V2 = P3 V3. Where, P3 is the final pressure. Similarly, the expression for adiabatic expansion T1V1-1 = T2V2-1. In isothermal process, temperature remains constant. So, T2 is the final temperature after isothermal compression.
256. A petrol engines consumes 5 kg petrol per hour. If the power of the engine is 20 KW and the calorific value of petrol is 11 x 103 K Cal per kg, calculate the efficiency of the engine. [ 31.15%] Hints: Efficiency of an engine Where, output = 20 KW = 20000 W Input = 5 Kg petrol per hour = 5 / 60 x 60 kg per sec. = Watt
257. An ideal heat engine operates in a Carnot Cycle between 227 0C and 127 0C. It absorbs 6 x 104 J at the higher temperature . How much work per cycle is this engine capable of performing? [ 5 x 104 J ] Hints: Efficiency of an engine = So, = Where, T2 = temperature of sink in absolute scale T1 = temperature of source in absolute scale Input = 6 x 104 J
258. A Carnot engine operates between a hot reservoir at 320 K and a cold reservoir 260 K. If it absorbs 500 J of heat from hot reservoir, how much work does it deliver? If the same engine working in reverses as a refrigerator, how much work must be supplied to remove 1000 J of heat from the cold reservoir? [ 4333 J] Hints: Efficiency of a heat engine is , = where, T2 = temperature of sink, 260 K T1 = temperature of source, 320 K Input = 500 J Output = ? If the same engine works as a refrigerator, its efficiency Where, = efficiency obtained from above , 18.75 % Q2 = heat taken from cold body, 1000 J W = work supplied, to determine.
259. A motor in a refrigerator has a power of output 200 W. If the freezing compartment is at 270 K and room temperature is 300 K, assuming ideal efficiency, what is the maximum amount of heat that can be extracted from the freezing compartment in 10 min? [ 1.1 x 106 ]
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