Second Law of Thermodynamics

148. Which one of the following statements is correct?

(a) A machine which violates Clausius statement will violate the first law of thermodynamics

(b) A machine which violates Kelvin-Plank statement will violate the first law of thermodynamics

(d) A machine which violates Kelvin-Plank statement will violate the Clausius statement

UPSC JWM 2017

Ans. (d) :Kelvin Plank Statement- It is impossible for a heat engine to produced net work in a complete cycle if it exchange heat only with bodies at a single fixed temperature.

Clausius statement- It is impossible to construct an engine which operates in a cycle, transfer heat from cooler body to hotter body without any work input. A machine which violated Kelvin plank statement will violate the clausius statement.

149. Which of the followings is correct statement?

(a) Entropy of isolated system always decreases

(b) Energy always degrades during the real process

(d) Heat transfer through a finite temperature difference is reversible process

UPRVUNL AE 2016

Ans. (b) :

• Entropy of isolated system always increase or remain constant.

• Energy always degrades during the real process.

• Energy losses occur during the real process not destroyed.

• Heat transfer through finite temperature difference is irreversible process.

150. Which of the following devices complies with the Clausius statement of the second law of thermodynamics?

(a) Closed-cycle gas turbine

(b) Internal combustion engine

(d) Domestic refrigerator

ESE 2018

Ans. (d) : Clausius statement is related to refrigerator and heat pump not with heat engine.

151. If a closed system is undergoing an irreversible process, the entropy of the system:

(a) Must decrease

(b) Zero

(d) Remain constant

OPSC AEE 2019 PAPER – II

Ans : (c) : The entropy may increase, decrease or remains constant depending upon heat involved and internal irreversibility.

153. Consider the following statements:

1. Heat pumps and air conditioners have the same mechanical components.

2. The same system can be used as heat pump in winter and as air conditioner in summer.

3. The capacity and efficiency of a heat pump fall significantly at high temperatures.

Which of the above statements are correct?

(a) 1 and 2 only

(b) 1 and 3 only

(d) 1, 2 and 3

ESE 2018

Ans. (d) : Heat pumps and air conditioners have the same mechanical components.

Therefore it is not economical to have two separate systems to meet the heating and cooling requirements of a building. One system can be used as a heat pump in winter and an air conditioner in summer.

Heat pump is a device used to maintain the temperature of system higher than that of surroundings by supplying heat to system To evaluate the performance of a heat pump, COP is calculated. It is defined as the ratio of desired effect to external work supplied.

Note : The answer given by UPSC is option (b).

154. The area of a p-v diagram for a Carnot cycle represents

(a) Heat supplied

(b) Heat rejected

(d) Temperature drop

UPPSC AE 12.04.2016 Paper-II

TPSC AE 2015

Ans : (c)

155. A reversible Carnot engine operates between 27ºC and 1527ºC, and produces 400 kW of net power. The change of entropy of the working fluid during the heat addition process is

(a) 0.222 kW/K (b) 0.266 kW/K

ESE 2018

Ans. (b) :

156. An engine works on the basis of Carnot cycle operating between temperatures of 800 K and 400 K. If the heat supplied is 100 kW, the output is

(a) 50 kW (b) 60 kW

ESE 2018

Ans. (a) :

157. A heat engine receives 1 kW of heat transfer at 1200 K and gives out 600 W as work, with the rest as heat transfer to the ambient at 300 K. The second law efficiency of the engine is :

(a) 70% (b) 90%

BHEL ET 2019

Ans. (c) :

158. A Carnot cycle engine receives and rejects heat with a 200C temperature differential between itself and the thermal energy reservoirs. The expansion and compression processes have a pressure ratio of 50. For 1 kg of air as the working substance, cycle temperature limits of 1000 K and 300 K and T0= 280 K, determine the second law efficiency.

(a) 0.935 (b) 0.945

(e) 0.975

CGPSC AE 2014 –II

Ans. (c)

159. Gas A at 1 MPa, 100°C and Gas B at 5 MPa, 100°C are mixed such that final temperature after mixing remains 100°C. The process is adiabatic. The entropy of the gases after mixing:-

(a) Will increase

(b) Will remain same

(d) Cannot be calculated

UKPSC AE-2013, Paper-II

Ans. (b) :

160. If the COP of a Carnot refrigerator is 6, then the ratio of higher temperature to lower is :

(a) 6 : 1 (b) 3 : 2

BHEL ET 2019

Ans. (d) :

161. An engine operates between temperature limitsof 900 K and T2 and the other engine operates between T2 and 400 K. For both engines to be equally efficient, T2 should be equal to

(a) 600 K (b) 625 K

BPSC AE Mains 2017 Paper - V

UKPSC AE-2013, Paper-II

Ans : (a) : Given as,

T1 = 900 K

T3 = 400 K

If the efficiency are equal then

we know that

162. A frictionless heat engine can be 100% efficient only if its exhaust temperature is

(a) equal to its input temperature

(b) less than its input temperature

(d) 0 K

JPSC AE 2013 Ist Paper

Ans. (d) : A frictionless heat engine can be 100% efficiency only if its exhaust temperature is 0 K.

163. An inventor claims a thermal engine operates between ocean layers at 27°C and 10°C. It produces 10 kW and discharges 9900 kJ/min. Such an engine is

(a) Impossible b) Reversible

RPSC Vice Principal ITI 2018

Ans. (a)

164. The efficiency of a reversible cycle depends upon the-

a) nature of the working substance

(b) amount of the working substance

(d) type of cycle followed

RPSC INSP. OF FACTORIES AND BOILER 2016

Ans : (c)

165. Carnot cycle consists of

(a) Two constant volume & two isentropic processes

(b) Two isothermal and two isentropic processes

(d) One constant volume, one constant pressure and two isentropic processes

Vizag Steel (MT) 2017

TNPSC AE 2018

KPSC AE. 2015

TSPSC AEE 2015

Ans. (b) :

166. Carnot cycle has maximum efficiency for–

(a) Petrol engine

(b) Diesel engine

(d) Irreversible engine

Vizag Steel (MT) 2017

Ans. (c) : Carnot cycle has maximum efficiency for Reversible engine. According to the Carnot theorem,the reversible engine will always have a greater efficiency than the irreversible.

167. One reversible heat engine operates between 1600 K and T2 K and another reversible heat engine operates between T2 K and 400 K. If both the engines have the same heat input and output, then temperature T2 is equal to

(a) 800 K (b) 1000 K

Vizag Steel (MT) 2017

OPSC AEE 2015 PAPER - II

Ans. (a)

168.The more effective way of increasing efficiency of Carnot Engine is

(a) Increase of Source temperature

(b) Decrease of Source temperature

(d) Decrease of Sink temperature

APPSC AEE 2016

Ans. (d) : The more effective way of increasing efficiency of Carnot Engine is decrease of Sink temperature.

169.If carnot engine rejects heat at temperature of 400 K and accepts at 750K. What shall be heat absorbed, if heat rejected is 1000 kJ.

(a) 946 kJ (b) 800 kJ

TNPSC AE 2017

Ans. (c) :For Carnot engine, ηideal = ηactual11LLHHTQTQ−= −1000 750400HQ×=QH = 1875 kJ

170. 300 kJ/s of heat is supplied at a constant fixed temperature of 290oC to a heat engine. 150 kJ/s of heat are rejected at 8.5 oC. Then the cycle is reported as

(a) Reversible

(b) Irreversible

(d) Random

TNPSC AE 2018

Ans. (a) : TH = 290oC = 563 K TL = 8.5 oC = 281.5 K SRHLQQSTTδδ∆ =+300150563281.5−=+S 0∆ =It means the cycle is reversible

171. Carnot cycle efficiency is maximum when

(a) initial temperature is 0oK

(b) final temperature is 0oK

(d) final temperature is 0oC

TNPSC AE 2018

JPSC AE PRE 2019

Ans.(b): We known that Carnot cycle efficiency depends on source and sink temperature limits, LcHT1Tη = −If TL→ 0o K then ()cmax100%η=

172.If the temperature of the source is increased, the efficiency of the Carnot engine-

(a) decreases

(b) increases

(d) does not change

RPSC AE 2018

Ans. (b) : We know that the efficiency of Carnot engine. sinksource1TcTη= −If temperature of source Tsource increase than cη↑

173.The device that produces network in a complete cycle by exchanging heat only with single thermal energy reservoir is known as:

(a)PMM3

(b)Heat pump

(c)PMM2

(d)PMM1

UPRVUNL AE 2016

Ans. (c) :PMM-1—Perpetual motion machine of first kind does not exist because such machines violate the first law of thermodynamics such machines will produce the energy by itself and as we know that according to the law of energy conservation, energycould not be created or destroyed but could be converted from one form of energy to other form of energy. PMM-2—Perpetual motion machines of second kind are those machines that violate the second law of thermodynamics because such machines will absorb continuously heat energy from a single thermal reservoir and will convert the absorbed heat energycompletely into work energy. Such machine will have 100% efficiency. PMM-3—Perpetual motion machine of third kind of imaginary machine that has zero friction.

174. Second law of thermodynamics is known as the law of

(a) Energy

(b) Entropy

(d) Internal energy

HPPSC AE 2018

HPPSC LECT. 2016

UKPSC AE 2012 Paper–II

Ans. (b) : Second law of thermodynamics is known as the law of Entropy or qualitative law of thermodynamics whereas first law of thermodynamics is known as the law of internal energy or quantitative law of thermodynamics.

175. Perpetual motion machine of second kind (PMM-II) violates the

(a) Zeroth law of thermodynamics

(b) First law of thermodynamics

(d) Third law of thermodynamics

RPSC LECTURER 16.01.2016

Ans. (c) : Perpetual motion machine of second kind (PMM-II) violates the second law of thermodynamics.ηHE = 100%

176. A heat pump operates between two heat reservoirs, one at 800 K and other at 400 K. What will be the coefficient of performance (COP) of the heat pump?

(a) 0.5 (b) 1.0

(CGPCS Polytechnic Lecturer 2017)

Ans. (d) : [COP]H.P. =112TTT−=800800 400−[COP]H.P. =800400= 2

177. The enthalpy of the fluid before throttling is _______ the enthalpy of the fluid after throttling.

(a) equal to

(b) greater than

(d) greater than or equal to

(e) less than or equal to

(CGPCS Polytechnic Lecturer 2017)

Ans. (a) :Throttling Process–A throttling process is defined as a process in which there is no change inenthalpy from state 1 to state 2. [h1 = h2] It is highly irreversible process There is no work done (W = 0) Process is adiabatic (Q = 0) Enthalpy will remain constant.

178. The efficiency (η) of a reversible heat engine receiving heat solely at T1 and rejecting heat solely at T2 is given by:

(a) 122TTT−η =

(b) 212TTT−η =

(d) 211TTT−η =

(e) 2121TTT−η =

(CGPCS Polytechnic Lecturer 2017)

Ans. (c) :ηrev. =1WQ =121QQQ−ηrev. =21Q1Q−

34For reversible engine 12QQ=12TT, So 2rev.1T1Tη = −12rev.1TTT−η =

179. A thermal reservoir is a body of

(a) Small heat capacity

(b) Large heat capacity

(d) Large work capacity

JPSC AE PRE 2019

Ans. (c) :Thermal Reservoir—It is a large body with infinite heat capacity which can supply or absorbs unlimited quantity of heat without appreciable change in its temperature. Example—Atmosphere, ocean, river, industrial furnace, two phase mixture of liquid and vapour. • It is supplies heat then it is called a source. • It is absorbs heat then it is called a sink.

180. An engine operates between temperature of 900oK and T2 and another engine between T2 and 400 oK for both to do equal work, value of T2will be

(a) 650 oK (b) 600 oK

SJVN ET 2013

Ans. (a) : For work to be equal 132TTT2+=900 4002+=2T650º K

181. The statement "The efficiency of all reversible heat engines operating between the same temperature levels is the same" is known as:

(a) Corollary of Carnot theorem

(b) Zeroth law of thermodynamics

(d) Third law of thermodynamics

SJVN ET 2019

Ans. (a) : The efficiency of all reversible heat engines operating between the same temperature level is thesame is known as corollary of carnot theorem.

182. Dry saturated steam enters a frictionless adiabatic nozzle with negligible velocity at a temperature of 300oC [h1 = 2751 kJ/kg]. It is expanded to a pressure of 5 MPa isentropically [h2 = 2651 kJ/kg]. What will be the exit velocity of steam?

(a) 447.21 m/s

(b) 572.33 m/s

(d) 150.32 m/s

SJVN ET 2019

Ans. (a) : As per SFEE 22121vvhw20002000+= + (if h in kJ/kg) w = 0, v1 = 0 ()2v20002751 2651447.21 m / s=×−=

183.A heat engine working on the Carnot cycle receives heat at the rate of 50 kW from a source at 1300K and rejects it to a sink at 400K. The heat rejected is

(a) 20.3 kW (b) 15.4 kW

ESE 2017

Ans. (b) :Given, T1 = 1300 K T2 = 400 K Q1 = 50 kW Efficiency of engine η =WorkdoneHeat Suppliedη =1WQ=121TTT− For carnot engine 121QQQ−=1300 4001300−21Q1Q−=913⇒21QQ= 0.3076 Q2 = 50 × 0.3076 Q2 = 15.38 kW

184. A reversible heat engine rejects 80% of the heat supplied during a cycle of operation. If the engine is reversed and operates as a refrigerator, then its coefficient of performance shall be

(a) 6 (b) 5

ESE 2017

Ans. (c) :Given, Q2 = 0.8 Q1For heat engine For refrigerator Q1 = W + Q2(COP)R =2QWW = Q1− Q2(COP)R =110.8Q0.2QW = 0.2 Q2(COP)R = 4 Second Method ηE =1WQηE =110.2QQηE = 0.2 1 + (COP)R =E1η(COP)R =110.2−(COP)R = 4

185. A reversible polytropic process is given by

(a) 1n2121TT−ρρ=

(b) n2121PPρρ=

(d) n1n2121TT−ρρ=

BPSC Poly. Lect. 2016

Ans : (a) For reversible polytropic process we know that n 1n1122TPTP−   =       ......(i) (For reversible Process) For ideal gas eqn. PV = mRT P1 = ρ1RT1 ρ = Density of gas P2 = ρ2RT2value of P1 & P2 putting in eqn. (i) n 1n111222TRTTRT−ρ  =  ρ  n 1n 1nn1122TT−−ρ =× ρ n 1n1122TTTT−−   ×      n 1n12−ρ=ρn 1n1122TpTp− =  =n 112ρρ−   = n 121VV−   

186.If the thermal efficiency of Carnot heat engine is 50 percent, then coefficient of performance of a refrigerator working within the same temperature limit would be :

(a) 1 (b) 2

(KPSC AE. 2015)

Ans : (a) Efficiency of carnot head engine = 0.50 HLEHTTT−η =LLHHTT0.5 10.5TT= −⇒=()LRRHLT0.5COP(COP)TT1 0.5=⇒=−−R(COP)1=

187. A cycle of pressure- volume diagram is shown in the figure : Same cycle on temperature-entropy diagram will be represented by

MPPSC AE 2016

Ans : (b)

188. A heat pump operating on Carnot cycle pump heat from a reservoir at 300 K to a reservoir at 600 K. The coefficient of performance is

(a) 1.5 (b) 0.5

MPPSC AE 2016

Ans : (c) TH = 600K TL = 300K ()HHPHLHPT(C.O.P.)TT600C.O.P.2600 300=−==−

189. Heat transfer takes place according to:

(a) Zeroth law of thermodynamics

(b) First law of thermodynamics

(d) Third law of thermodynamics

MPPSC AE 2016

Ans : (c) Heat transfer take place according to second law ofthermodynamics.

190. For a heat engine operating on Carnot cycle, the work output is 25% of heat rejected to the sink. The thermal efficiency of the engine would be

(a) 10% (b) 20%

RPSC AE 2016

Ans : (b) QA = 1.25kJ W=0.25kJ QR = 1kJ E0.251.25η =E20%η

191. Three engines A, B and C operating on Carnot cycle respectively use air, steam and helium as the working fluid. If all the engines operate within the same high and low temperature limits. then which engine will have the highest efficiency?

(a) Engine A

(b) Engine B

(d) All engines will have the same efficiency

RPSC AE 2016

Ans :(d) Efficiency of Carnot cycle ( )HLHTTT−η =Efficiency of Carnot engines does not depend on thetype of fuel use so engine A; B and C operating same high and low temperature limit then engine will have the same thermal efficiency.

192. A Carnot engine Working between 600K and 300K produces 200 KJ of work. The heat supplied is

(a) 200 KJ (b) 400KJ

TSPSC AEE 2015

Ans : (b) TH = 600 K TL = 300K W = 200KJ HLcHTTT−η =ηc = 0.5 cWork outputheat suppliedη =0.5 =A200QQA = 400KJ

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