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Heat And Thermodynamics Mcq
Quiz 1
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Q.1
If a wire of resistance 20Ω is covered with ice and voltage of 210V is applied across the wire then rate of melting of ice will be ..[ AFMC 1997]
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a) 6.56 g/s
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b) 0.85 g/s
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c)3.56 g/s
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d)none of these
Explanation
Heat produced per second=heat absorbed by ice Heat produced per second=Latent heat of fusion × mass; Heat produced per second in calories =V2/ R(4.2) Heat produced=(210)2/20(4.2)=525 calories525=Latent heat of fusion × mass;525=80× (m) m=525 /80=6.56gm/sec Answer: (a)
Q.2
The ratio of speed of sound in Helium and Hydrogen gases at the same temperature is ..[ AFMC 1998]
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a) √42:√25
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b) √25:√42
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c)42:45
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d)25:42
Explanation
formula for velocity of sound in gas isHelium is mono atomic gas degree of freedom is 5 thus γ=5/3 and density=4ρ thus velocity of sound in Helium v1hydrogen is diatomic gas hence γ=7/5 and density -2ρ thus velocity of sound in Hydrogen v2 taking the ratio of v1 / v2=√ (25/42) Answer:(b)
Q.3
When the amount of work done is 300 J and change in internal energy is 100J, then the heat supplied is..[ AFMC 1998]
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a) 400 J
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b) 350 J
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c) 200 J
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d) 150 J
Explanation
According to first law of thermodynamics ΔQ=ΔE + ΔW ΔQ=100 + 300=400 J Answer: (a)
Q.4
Water falls from a height 500m. The rise in temperature of water at bottom if whole of the energy remains in water, will be ( specific heat of water is 4.2 kJ/kg) [AFMC 1997]
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a)0.23°C
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b) 1.16°C
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c)096°C
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d)1.02°C
Explanation
Loss of potential energy converted in to kinetic energy of water molecule or heat energy ∴ Loss of Potential energy=Heat energymgh=mcΔθ m ×9.8 × 500=m×4.2×103× θ By substituting the values and solving above equation we getθ=1.16°CAnswer: (b)
Q.5
16g of oxygen at 37°C is mixed with 14 g of nitrogen at 27°C, the temperature of the mixture will be...[ AFMC 1997]
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a) 30.5°C
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b) 37°C
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c)27°C
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d)32°C
Explanation
Both oxygen and nitrogen will have same specific heat as both are diatomic gasesLet θ be the final temperature of mixtureHeat loss by Oxygen=Heat gain by NitrogenNumber of moles of oxygen× molar specific heat × ( Δθ)=Number of moles of nitrogen × molar specific heat × ( Δθ) 1× Cp×(37-θ)=1× Cp(θ - 27)2θ=37+27=64θ=32°CAnswer: (d)
Q.6
When there is no heat exchange from surrounding to a system, then the process is related with. [ AFMC 1997]
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a) isobaric
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b) isochoric
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c)isothermal
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d)adiabatic
Explanation
The process is known as adiabatic Answer:(d)
Q.7
If the ratio of specific heat of gas at constant pressure to that at constant volume is γ, the change in internal energy of one mole of gas when the volume changes from V to 2V at constant pressure P is ...[ AFMC 1997]
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a) γPV/(γ -1)
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b) PV/(γ -1)
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c) R/(γ -1)
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d) PV
Explanation
According to first law of thermodynamics ΔQ=ΔE + ΔW When Pressure is constant Then ΔQ=ΔW=Cp=and ΔE=0 At constant volume ΔQ=ΔE=Cv Cp / Cv=γ Cp=γ Cv Work done by gas=(2V-V)P=PV Modified first law for one mole of gas is Cp=γ Cv + PVFrom above γ Cv=Cv + PVThus Cv=PV/(γ -1) ΔE=PV/(γ -1) Answer: (b)
Q.8
The original temperature of a black body is 727°C. The temperature to which that black body must be raised so as to double the total radiant energy, is .. [ AFMC 1997]
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a)917°C
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b) 1190°C
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c)1454°C
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d)2000K
Explanation
According to Stefan-Boltzmann law : Total emission power W=eσT4For black body e=1 E1 / E2=T14 / T24 Given E2 / E1 2=(T2 / T1) 4 Given T1=7273+273=1000°K T2=(2)1/4×1000°K T2=1190Kapprox. 1190 - 273 = 917°C Answer: (a)
Q.9
If the heat of 110J is added to a gaseous system, whose internal energy is 40J, then the amount of external work done is ...[ AFMC 1999]
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a)80 J
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b) 70 J
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c)114 J
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d)140 J
Explanation
Heat added hence positiveAccording to first law of thermodynamics ΔQ=ΔE + ΔW 110=40 + ΔW ΔW=70 J Answer: (b)
Q.10
If the coefficient of cubical expansion is 'x' times of the coefficient of superficial expansion, then value of 'x' is [ AFMC 2000]
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a) 1/2
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b) 1
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c)1.5
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d)4
Explanation
If α is coefficient of linear expansion then cubical expansion=3α Superficial expansion=2α Thus 3α=(x)2α x=3/2=1.5Answer: (c)
Q.11
A body radiates 5W energy at a temperature of 400K. If the temperature is increased to 1200K then it radiates energy at the rate of ...[ AFMC 2000]
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a) 419 W
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b) 405 W
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c)210 W
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d)80 W
Explanation
According to Stefan-Boltzmann law : Total emissive power W=eσT4 E1 / E2=T14 / T24 5 / E2=(400 /1200)4 E2=5×(81)=405W Answer:(b)
Q.12
The maximum energy in the thermal radiation from a heat source occurs at a wave length of 11×10-5cm. according to wein's displacement Law the temperature of this source will be 'n' times. The temperature of another source for which the wavelength at maximum energy is 5.5×10-5cm. Then the value of n is [ AFMC 2000]
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a) 1/2
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b) 1
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c) 2
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d) 4
Explanation
According to Wein's displacement law λ1T1=λ2T2 11×10-5nT=5.5×10-5T n=1/2 Answer: (a)
Q.13
A ideal gas A and real gas B have their volumes increased from V to 2V under isothermal conditions. The increase in internal energy. [ CBSE-PMT 1993]
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a)will be same in both A and B
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b) will be zero in both the gas
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c)of B will be more than that of A
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d)of A will be more than that of B
Explanation
Under isothermal condition s, there is no change in internal energyAnswer: (b)
Q.14
The temperature of source and sink of a heat engine are 127° and 27°C respectively. An inventor claims its efficiency to be 26%, then...[ CBSE-PMT 2001]
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a) it is impossible
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b) it is possible with high probability
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c)it is possible with low probability
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d)data are insufficient
Explanation
η=1 - T1 / T2 η=1 - (300/400)=1/4 Therefore it is note possible to have efficiency more than 25%Answer: (a)
Q.15
A diatomic gas initially at 18°C is compressed adiabatically to one eight of its original volume. The temperature after compression will be ... [ CBSE-PMT 1996]
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a) 18°C
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b) 668.4°K
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c)395.4°C
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d)144°C
Explanation
for adiabatic process T1V1γ-1=T2V2γ-1for diatomic gas γ=7/5 Now T1=18°C=291KV1=V V2=V/8 γ -1=2/5V2/5 ×(291)=(V/8)2/5 TT=668.4K Answer:(b)
Q.16
A mass of diatomic gas (γ=1.4) at a pressure of 2atmosphere is compressed adiabatically so that its temperature rises from 27°C to 927°C. The pressure of the gas is final state is .. [ CBSE-PMT 2011]
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a) 28 atm
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b) 68.7 atm
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c) 256 atm
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d) 8 atm
Explanation
T1=273+27=300K T2=273+927=1200K For adiabatic process P11-γ T1γ=P21-γ T2γ Answer: (c)
Q.17
When 1kg of ice at 0°C metals to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80cal/°C, is .. [ CBSE-PMT 2011]
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a)273 cal/K
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b) 8×104 cal/K
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c)80 cal/K
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d)293 cal/K
Explanation
Change in entropy is given by ΔS=dQ/T dQ=mLf m is mass and Lf is latent heat of fusion ΔS=1000×80 / 273=293 cal/KAnswer: (d)
Q.18
Which of the following is not thermo dynamical function? [ CBSE-PMT 1993]
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a) Enthalpy
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b) Work done
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c)Gibbs's energy
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d)Internal energy
Explanation
Work done is not thermo dynamical functionAnswer: (b)
Q.19
The internal energy change in a system that has absorbed 2 kcal of heat and done 500J of work is [ CBSE-PMT 2009]
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a) 6400 J
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b) 5400 J
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c)7900 J
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d)8900 J
Explanation
According to first law of thermodynamics Q=ΔU + W ΔU=Q-WΔU=2×103×4.2 -500 ΔU=7900 J Answer:(c)
Q.20
A sample of gas expands from volume V1 to VThe amount of work done by the gas is greatest hen the expansion is ... [ CBSE-PMT 1997]
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a) adiabatic
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b) isobaric
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c) isothermal
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d) equal in all cases
Explanation
In thermodynamic for same change in volume, the work done is maximum in isobaric process. As area enclosed by curve in P-V graph is maximum Answer: (b)
Q.21
A Carnot engine whose sink is at 300K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency? [ CBSE-PMT 2006]
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a)325K
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b) 250K
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c) 380K
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d)275K
Explanation
We know that efficiency of carnot engineη=(T1 - T2) / T1 Where T1 is temperature of source and T2 is temperature of sink Now efficiency to be increased by 50% thus new efficiency will be 60% Increase in temperature=750-500=250KAnswer: (b)
Q.22
Which of the following process is reversible? [ CBSE-PMT 2005]
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a) Transfer of heat by conduction
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b) Transfer of heat by radiation
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c)Isothermal compression
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d)Electrical heating of a nichrome wire
Explanation
For process to be reversible, it must be quasi-static. For quasi static process, all changes take place infinitely slow. Isothermal process occur very slowly so it is quasi-static and hence it is reversibleAnswer: (c)
Q.23
We consider a thermodynamic system. If represents the increase in its internal energy of W the work done by the system, which of the following statement is true? [ CBSE-PMT 1998]
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a) ΔU=-W in an adiabatic process
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b) ΔU=W in an isothermal process
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c)Δ U=-W in an isothermal process
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d)ΔU=W in an adiabatic process
Explanation
According to first law of thermodynamics ΔQ=ΔU + W for adiabatic process ΔQ=0ΔU=-W Answer:(a)
Q.24
An engine has efficiency of 1/When the temperature of sink is reduced by 62°C, its efficiency is doubled. Temperature of the source is [ CBSE-PMT 2007]
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a) 37°
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b) 62°
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c) 99°
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d) 124°
Explanation
Efficiency of engine η=(T1 - T2) / T1 Where T1 temperature of source and T2 temperature of sink According given When temperature of th sink is decreased by 62° efficiency is doubledSolving above equations we get T1-T2=62substituting value of T1-T2 in equation first we get T1=372K or 99° Answer: (c)
Q.25
An ideal carnot engine, whose efficiency is 40% receives heat at 500K. If its efficiency is 50% then intake temperature for the same exhaust temperature is [ CBSE-PMT 1995]
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a)600 K
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b) 700 K
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c)800 K
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d)900 K
Explanation
η=(T1 - T2) / T1Thus T2 / T1=1-η First case T2 /500=1-0.4=0.6∴ T2=0.6×500=300KSecond case 300/T1=1-0.5=0.5 ∴ T1=300 / 0.5=600KAnswer: (a)
Q.26
110 joules of heat is added to a gaseous system. Whose internal energy is 40J, then the amount of external work done is .. [ CBSE-PMT 1993]
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a) 150J
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b) 70J
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c)110J
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d)40 J
Explanation
ΔW=ΔQ - ΔU ΔW=110-40=70JAnswer: (b)
Q.27
If Q, E and W denote respectively the heat added change in internal energy and the work done in a closed cyclic process then [ CBSE-PMT 2008]
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a) W=0
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b) Q=W=0
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c)E=0
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d)Q=0
Explanation
In a cyclic process, the initial state coincides with the final state. Hence, the change in internal energy is zero, as it depends only on the initial and final states. But Q and W are nonzero during a cycle process. Answer:(c)
Q.28
If the ratio of specific heat of a gas constant pressure to that at constant volume is γ, the change in internal energy of mass of gas, when the volume changes from V and 2V at constant pressure P, is ... [ CBSE-PMT 1998]
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a)
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b) PV
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c)
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d)
Explanation
Change in internal energy is equal to work done in adiabatic system ΔW=-ΔU ( Expansion in the system) here V1=V , V2=2V Thus Answer: (c)
Q.29
The volume of gas reduced adiabatically to 1/4 of its volume at 27°C. If γ=1.4 the new temperature is ..[ MPPMT 1990]
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a) (300)(2)0.4 K
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b) (300)(4)1.4 K
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c)(300)(4)0.4 K
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d)(300)(2)1.4 K
Explanation
For adiabatic expansion Answer:(c)
Q.30
A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by 62°C, the efficiency of the engine is doubled. The temperatures of the source and sink are... [ CBSE-PMT 2000]
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a)99°C, 37°C
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b) 80°C, 37°C
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c)95°C, 37°C
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d)90°C, 37°C
Explanation
Efficiency of engine η=(T1 - T2) / T1 Where T1 temperature of source and T2 temperature of sink According given1/6=1 - (T2 / T1)T2=(5/6) T1 Second conditionon simplifying we getOn solving we getT1=372K=99°CT2=310K=37°C Answer: (a)
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