Thermodynamics: Physics

 

• Answers are in BOLD Option name.

Q.1.  2 kg of ice at - 20C is mixed with 5 kg of water at 20C in an insulating vessel having a negligible heat capacity . Calculate the final mass of water in the vessel . It is given that the specific heat of water and ice are 1 kcal / kg / C and 0.5 kcal / kg /C respectively and the latent heat of fusion of ice is 80 kcal / kg 

A) 7 kg

B) 6 kg

C) 4 kg

D) 2 kg

Explanation -

Let m kg be the mass f ice melted into water. Heat lost by 5 kg of water=5kg1 kcal/kg/ 20=100 kcal. Heat gained=m kg80kcal/kg +2 kg+2 kg 0.5 kcal/kg/20=80m kcal+20 kcal. Now, heat gained=heat lost. Therefore

80m+20=100

or m=1 kg. Therefore , final mass of water=5kig+1kg=6kg. 

Q.2.  An ideal gas ( y = 1.4 ) expands from 5  to 25  at a constant pressure of 1   Pa . The heat energy supplied to the gas in this process is 

A) 7 J

B) 70 J

C) 700 J

D) 7000 J

Explanation-

Work done on the gas is 

From the first law of thermodynamics , the heat energy supplied to the gas is 

Q.3.  How much mechanical work must be done to completely melt 1 gram of ice 0 C ?

 A) 4.2 J

 B) 80 J

 C) 336 J 

 D) 2268 J

Explanation- 

Work done=latent heat of function of ice = 80 calories=804.2=336 J. 

Q.4.  A copper block of mass 2 kg is heated to a temperature of 500C and then placed in a large block of ice at 0C . What is the maximum amount of ice that can melt ? The specific heat of copper is 400 J kg C and latent heat of fusion of water is 3.5 J kg .

A)  kg

B)  kg

C)  kg

D)  kg

Explanation-

Heat energy in copper block=2 The amount of ice that melts will be maximum if the entire heat energy of the copper block is used up in melting ice. Now, 3.5J of heat energy is needed to melt 1 kg of ice into water. Therefore, the amount of ice melted by    of heat energy is 

 

Q.5. Two monoatomic ideal gases 1 and 2 of molecular masses M and M  respectively are enclosed in separate containers kept at the same temperature . The ratio of the speed of sound in gas 1 to that in gas 2 is 

A) 

B) 

C) 

D)  

Explanation-

The speed of sound in a gas of bulk modulus B and density  is given by

 

Q.6. One mole of a monoatomic ideal gas is contained in a insulated and rigid container . It is heated by passing a current of 2 A for 10 minutes through a filament of resistance 100  . The change in the internal energy of the gas is 

A) 30 kJ

B) 60 kJ

C) 120 kJ

D) 240 kJ

Explanation-

The heat energy supplied is 

Since V = 0 ; work done W = 0 . From first law of thermodynamics . U = Q = 240 kJ .

Q.7.  A metal sphere of radius r and specific heat S is rotated about an axis passing through its center at a speed of n rotation per second . It is suddenly stopped and 50% o its energy is used in increasing its temperature . Then the rise in temperature of the sphere is : 

A) 

B) 

C) 

D)  

Explanation-

 

Q.8.  5 moles  of Hydrogen   initially at S. T. P.  are compressed adiabatically so that its temperature becomes 400 C . The increase in the internal energy of the gas in kilo - joules is ( R = 8.30 J molK)

A) 21.55

B) 41.50

C) 65.55

D) 80.55

Explanation-

 By convention , the work done on the gas is taken to be negative , i.e. W = - 41.5 kJ . From the first laws of thermodynamics dQ = dU + dW . For an adiabatic process , dQ = 0 . Hence dU = - dW = - ( - 41.5 ) = 41.5 kJ . The positive sign of dU implies that the internal energy increases .

Q.9.  During an adiabatic process , the pressure of a gas is proportional to the cube of its absolute temperature . The value of   for that gas is : 

A) 

B) 

C) 

D)  

Explanation-

 

Q.10.  A uniform metallic circular disc , mounted on frictionless bearings , is rotating at an angular frequency   about ab axis passing through its center and perpendicular to its plane . The coefficient of linear expansion of the metal is a . If the temperature of the disc is increased by  t , the angular frequency of rotation of the disc will 

A) remain unchanged 

B) increased by 

C) increased by 2 

D) decrease by 2 

Explanation-

The moment of inertia of the disc about the given axis of rotation is 

where M is the mass of the disc and R its radius . If the disc is heated , it expands . Hence R increases . The resulting increase in I is obtained by partially differentiating (1).

Now , the angular momentum of the disc is given by 

Since no external torque acts , J remains constant . Partially differentiating (3), we have 

The negative sign indicates that the angular frequency decreases due to increase in temperature .