Heat and Thermodynamics — Page 10 | JEE Physics

Q91

When M1 gram of ice at –10oC (specific heat = 0.5 cal g–1 oC–1 ) is added to M2 gram of water at 50C, finally no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g–1 is :

A

{{50{M_2}} \over {{M_1}}} - 5

B

{{50{M_2}} \over {{M_1}}}

C

{{5{M_2}} \over {{M_1}}} - 5

D

{{5{M_1}} \over {{M_2}}} - 50

Correct Answer

Option A

Solution

Heat lost = Heat gain

\Rightarrow {M_2} \times 1 \times 50 = {M_1} \times 0.5 \times 10 + {M_1}.{L_f}

\Rightarrow {L_f} = {{50{M_2} - 5{M_1}} \over {{M_1}}}

= {{50{M_2}} \over {{M_1}}} - 5

Q92

A Carnot engine has an efficiency of

{1 \over 6}

. When the temperature of the sink is reduced by 62ºC, its efficiency is doubled. The temperatures of the source and the sink are, respectively

A 99oC, 37oC

B 124oC, 62oC

C 37oC, 99oC

D 62oC, 124oC

Correct Answer

Option A

Solution

\eta = 1 - {{{T_2}} \over {{T_1}}}

$\Rightarrow$

{1 \over 6} = 1 - {{{T_2}} \over {{T_1}}}

$\Rightarrow$

{{{T_2}} \over {{T_1}}} = {5 \over 6}

When the temperature of the sink is reduced by 62ºC, then

\eta ' = 1 - {{{T_2} - 62} \over {{T_1}}}

$\Rightarrow$

2 \times {1 \over 6} = 1 - {{{T_2}} \over {{T_1}}} + {{62} \over {{T_1}}}

$\Rightarrow$

{1 \over 3} = {1 \over 6} + {{62} \over {{T_1}}}

$\Rightarrow$ T1 = 372 K = 99o C T2 =

{5 \over 3} \times 372

= 310 K = 37o C

Q93

A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process ?

A 35 J

B 30 J

C 25 J

D 40 J

Correct Answer

Option A

Solution

At constant pressure, W = P

\Delta

V = nR

\Delta

T At constant pressure, heat supplied Q = nCp

\Delta

T

{W \over Q} = {R \over {{C_p}}}

For diatomic gas, Cp =

{7 \over 2}R

$\therefore$

{{10} \over Q} = {R \over {{7 \over 2}R}}

$\Rightarrow$ Q = 10 $\times$

{7 \over 2}

= 35 J

Q94

One kg of water, at 20oC, heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20

\Omega

. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to : [Specific heat of water = 4200 J/(kg oC), Latent heat of water = 2260 kJ/kg]

A 10 minutes

B 22 minutes

C 3 minutes

D 16 minutes

Correct Answer

Option B

Solution

P $\times$ t = mS

\Delta

t + mLv $\Rightarrow$

{{{V^2}} \over R}t

= mS

\Delta

t + mLv $\Rightarrow$

{{{{\left( {200} \right)}^2}} \over {20}}t

=

1 \times 4200 \times \left( {100 - 20} \right)

+

1 \times 2260 \times {10^3}

$\Rightarrow$ 2000t = 336000 + 2260000 $\Rightarrow$ t = 1298 s = 21.6 min

\simeq

22 min

Q95

Two kg of a monoatomic gas is at a pressure of 4

\times

104 N/m2. The density of the gas is 8 kg/m3. What is the order of energy of the gas due to its thermal motion ?

A 104 J

B 103 J

C 105 J

D 106 J

Correct Answer

Option A

Solution

Thermal energy of N molecule = N

\left( {{3 \over 2}kT} \right)

=

{N \over {{N_A}}}{3 \over 2}

RT =

{3 \over 2}

(nRT) =

{3 \over 2}

PV =

{3 \over 2}

P

\left( {{m \over 8}} \right)

=

{3 \over 2}

$\times$ 4 $\times$ 104 $\times$

{2 \over 8}

= 1.5 $\times$ 104 order will 104

Q96

A balloon filled with helium (32oC and 1.7 atm.) bursts. Immediately afterwards the expansion of helium can be considered as

A Irreversible adiabatic

B Reversible adiabatic

C Irreversible isothermal

D Reversible isothermal

Correct Answer

Option A

Solution

Bursting of helium ballon is irreversible adiabatic because Energy can not be restored.

Q97

A metal ball of mass 0.1 kg is heated upto 500oC and dropped into a vessel of heat capacity 800 JK–1 and containing 0.5 kg water. The initial temperature of water and vessel is 30oC. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200 Jkg–1 and 400 Jkg–1 K–1

A 20%

B 25%

C 15%

D 30%

Correct Answer

Option A

Solution

0.1 $\times$ 400 $\times$ (500 $-$ T) = 0.5 $\times$ 4200 $\times$ (T $-$ 30) + 800 (T $-$ 30) $\Rightarrow$ 40(500 $-$ T) = (T $-$ 30) (2100 + 800) $\Rightarrow$ 20000 $-$ 40T = 2900 T $-$ 30 $\times$ 2900 $\Rightarrow$ 20000 + 30 $\times$ 2900 = T(2940) T = 30.4oC

{{\Delta T} \over T} \times 100

=

{{6.4} \over {30}} \times 100

\simeq

20%

Q98

A thermometer graduated according to a linear scale reads a value x0 when in contact with boiling water, and x0/3 when in contact with ice. What is the temperature of an object in oC, if this thermometer in the contact with the object reads x0/2 ?

A 60

B 35

C 25

D 40

Correct Answer

Option C

Solution

$\Rightarrow$ ToC =

{{{x_0}} \over 6}

&

\left( {{x_0} - {{{x_0}} \over 3}} \right)

= (100 $-$ 0oC) x0 =

{{300} \over 2}

$\Rightarrow$ ToC =

{{150} \over 6}

= 25oC

Q99

A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. considering only translational and rotational modes, the total internal energy of the system is :

A 12 RT

B 20 RT

C 4 RT

D 15 RT

Correct Answer

Option D

Solution

U

= {{{f_1}} \over 2}{n_1}RT + {{{f_2}} \over 2}{n_2}RT

= {5 \over 2}\left( {3RT} \right) + {3 \over 2} \times 5RT

U

= 15RT

Q100

A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is

\ell

1, and that below the piston is

\ell

2, such that

\ell

1 >

\ell

2. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by : (R is universal gas constant and g is the acceleration due to gravity)

A

{{nRT} \over g}\left[ {{{{\ell _1} - {\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]

B

{{RT} \over g}\left[ {{{2{\ell _1} + {\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]

C

{{nRT} \over g}\left[ {{1 \over {{\ell _2}}} + {1 \over {{\ell _1}}}} \right]

D

{{RT} \over {ng}}\left[ {{{{\ell _1} - 3{\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]

Correct Answer

Option A

Solution

P2A = P1A + mg

{{nRT.A} \over {A{\ell _2}}}

=

{{nRT.A} \over {A{\ell _1}}}

+ mg nRT

\left( {{1 \over {{\ell _2}}} - {1 \over {{\ell _1}}}} \right)

= mg m =

{{nRT} \over g}\left( {{{{\ell _1} - {\ell _2}} \over {{\ell _1}.{\ell _2}}}} \right)