Heat and Thermodynamics — Page 7 | JEE Physics

Q61

100g

of water is heated from

{30^ \circ }C

to

{50^ \circ }C

. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is

4184

J/kg/K

):

A

8.4

kJ

B

84

kJ

C

2.1

kJ

D

4.2

kJ

Correct Answer

Option A

Solution

\Delta U = \Delta Q = mc\Delta T

= 100 \times {10^{ - 3}} \times 4184\left( {50 - 30} \right) \approx 8.4\,kJ

Q62

Three perfect gases at absolute temperatures

{T_1},\,{T_2}

and

{T_3}

are mixed. The masses of molecules are

{m_1},{m_2}

and

{m_3}

and the number of molecules are

{n_1},

{n_2}

and

{n_3}

respectively. Assuming no loss of energy, the final temperature of the mixture is:

A

{{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}} \over {{n_1} + {n_2} + {n_3}}}

B

{{{n_1}T_1^2 + {n_2}T_2^2 + {n_3}T_3^2} \over {{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}}}

C

{{n_1^2T_1^2 + n_2^2T_2^2 + n_3^2T_3^2} \over {{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}}}

D

{{\left( {{T_1} + {T_2} + {T_3}} \right)} \over 3}

Correct Answer

Option A

Solution

Number of moles of first gas

= {{{n_1}} \over {{N_A}}}

Number of moles of second gas

= {{{n_2}} \over {{N_A}}}

Number of moles of third gas

= {{{n_3}} \over {{N_A}}}

If there is no loss of energy then

{P_1}{V_1} + {P_2}{V_2} + {P_3}{V_3} = PV

{{{n_1}} \over {{N_A}}}R{T_1} + {{{n_2}} \over {{N_A}}}R{T_2} + {{{n_3}} \over {{N_A}}}R{T_3}

= {{{n_1} + {n_2} + {n_3}} \over {{N_A}}}R{T_{mix}}

\Rightarrow {T_{mix}} = {{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}} \over {{n_1} + {n_2} + {n_3}}}

Q63

A thermally insulated vessel contains an ideal gas of molecular mass

M

and ratio of specific heats

\gamma .

It is moving with speed

v

and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, Its temperature increases by:

A

{{\left( {\gamma - 1} \right)} \over {2\gamma R}}M{v^2}K

B

{{\gamma {M^2}v} \over {2R}}K

C

{{\left( {\gamma - 1} \right)} \over {2R}}M{v^2}K

D

{{\left( {\gamma - 1} \right)} \over {2\left( {\gamma + 1} \right)R}}M{v^2}K

Correct Answer

Option C

Solution

Here, work done is zero. So, loss in kinetic energy $=$ change in internal energy of gas

{1 \over 2}m{v^2} = n{C_v}\Delta T = n{R \over {\gamma - 1}}\Delta T

{1 \over 2}m{v^2} = {m \over M}{R \over {\gamma - 1}}\Delta T

$\therefore$

\Delta T = {{M{v^2}\left( {\gamma - 1} \right)} \over {2R}}K

Q64

A Carnot engine operating between temperatures

{{T_1}}

and

{{T_2}}

has efficiency

{1 \over 6}

. When

{T_2}

is lowered by

62

K

its efficiency increases to

{1 \over 3}

. Then

{T_1}

and

{T_2}

are, respectively:

A

372

K

and

330

K

B

330

K

and

268

K

C

310

K

and

248

K

D

372

K

and

310

K

Correct Answer

Option D

Solution

Efficiency of engine

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

and

{\eta _2} = 1 - {{{T_2} - 62} \over {{T_1}}} = {1 \over 3}

$\therefore$

{T_1} = 372\,K

and

{T_2} = {5 \over 6} \times 372 = 310\,K

Q65

Three rods of Copper, Brass and Steel are welded together to form a

Y

shaped structure. Area of cross - section of each rod

= 4c{m^2}.

End of copper rod is maintained at

{100^ \circ }C

where as ends of brass and steel are kept at

{0^ \circ }C

. Lengths of the copper, brass and steel rods are

46,

13

and

12

cms

respectively. The rods are thermally insulated from surroundings excepts at ends. Thermal conductivities of copper, brass and steel are

0.92, 0.26

and

0.12

CGS

units respectively. Rate of heat flow through copper rod is:

A

1.2

cal/s

B

2.4

cal/s

C

4.8

cal/s

D

6.0

cal/s

Correct Answer

Option C

Solution

Rate of heat flow is given by,

Q = {{KA\left( {{\theta _1} - {\theta _2}} \right)} \over l}

Where,

K=

coefficient of thermal conductivity

l=

length of rod and

A=

Area of cross-section of rod If the junction temperature is

T,

then

{Q_{Copper}}\,\, = \,\,{Q_{Brass}}\,\, + \,\,{Q_{Steel}}

{{0.92 \times 4\left( {100 - T} \right)} \over {46}} = {{0.26 \times 4 \times \left( {T - 0} \right)} \over {13}} + {{0.12 \times 4 \times \left( {T - 0} \right)} \over {12}}

\Rightarrow 200 - 2T = 2T + T

\Rightarrow T = {40^ \circ }C

$\therefore$

{Q_{Copper}} = {{0.92 \times 4 \times 60} \over {46}} = 4.8\,cal/s

Q66

Consider a spherical shell of radius

R

at temperature

T

. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume

u = {U \over V}\, \propto \,{T^4}

and pressure

p = {1 \over 3}\left( {{U \over V}} \right)

. If the shell now undergoes an adiabatic expansion the relation between

T

and

R

is:

A

T\, \propto {1 \over R}

B

T\, \propto {1 \over {{R^3}}}

C

T\, \propto \,{e^{ - R}}

D

T\, \propto \,{e^{ - 3R}}

Correct Answer

Option A

Solution

As,

P = {1 \over 3}\left( {{U \over V}} \right)

But

\,\,\,\,

{U \over V} = KT{}^4

So,

\,\,\,\,\,P = {1 \over 3}K{T^4}

or

\,\,\,\,{{uRT} \over V} = {1 \over 3}K{T^4}\,\,\,\,

\left[ \, \right.

As

PV = uRT

\left. \, \right]

{4 \over 3}\pi {R^3}{T^3} =

constant

$\therefore$

\,\,\,\,

T \propto {1 \over R}

Q67

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as

{V^q},

where

V

is the volume of the gas. The value of

q

is:

\left( {\gamma = {{{C_p}} \over {{C_v}}}} \right)

A

{{\gamma + 1} \over 2}

B

{{\gamma - 1} \over 2}

C

{{3\gamma + 5} \over 6}

D

{{3\gamma - 5} \over 6}

Correct Answer

Option A

Solution

\tau = {1 \over {\sqrt 2 \pi {d^2}\left( {{N \over V}} \right)\sqrt {{{3RT} \over M}} }}

\tau \propto {V \over {\sqrt T }}

As,

\,\,\,\,T{V^{\gamma - 1}} = K

So,

\,\,\,\,\tau \propto {V^{\gamma + 1/2}}

Therefore,

q = {{\gamma + 1} \over 2}

Q68

A pendulum clock loses

12

s

a day if the temperature is

{40^ \circ }C

and gains

4

s

a day if the temperature is

{20^ \circ }C.

The temperature at which the clock will show correct time, and the co-efficient of linear expansion

\left( \alpha \right)

of the metal of the pendulum shaft are respectively :

A

{30^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 3}}/{}^ \circ C

B

{55^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 2}}/{}^ \circ C

C

{25^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 5}}/{}^ \circ C

D

{60^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 4}}/{}^ \circ C

Correct Answer

Option C

Solution

Time lost/gained per day

= {1 \over 2} \propto \Delta \theta \times 86400

second

12 = {1 \over 2}\alpha \left( {40 - \theta } \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

4 = {1 \over 2}\alpha \left( {\theta - 20} \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

On dividing we get,

\,\,\,3 = {{40 - \theta } \over {\theta - 20}}

3\theta - 60 = 40 - \theta

4\theta = 100 \Rightarrow \theta = {25^ \circ }C

Q69

A Carnot freezer takes heat from water at 0oC inside it and rejects it to the room at a temperature of 27oC. The latent heat of ice is 336×103 J kg−1. If 5 kg of water at 0oC is converted into ice at 0oC by the freezer, then the energy consumed by the freezer is close to :

A 1.67

\times

105 J

B 1.68

\times

106 J

C 1.51

\times

105 J

D 1.71

\times

107 J

Correct Answer

Option A

Solution

Total heat required to freeze 5 kg water, = 5 $\times$ 336 $\times$ 103 J = 1680 $\times$ 103 Joule = 1680 kJ For carnot cycle,

{{{Q_2}} \over {{Q_1}}}

=

{{{T_2}} \over {{T_1}}}

$\Rightarrow$

{{{Q_2}} \over {1680}}

=

{{300} \over {273}}

$\Rightarrow$ Q2 = 1680 $\times$

{{300} \over {273}}

kJ $\therefore$ W = Q2 $-$ Q1 = 1680

\left( {{{300} \over {273}} - 1} \right)

= 1680 $\times$

{{27} \over {273}}

= 166.15 kJ = 166.15 $\times$ 103 J = 1.66 $\times$ 105 J

Q70

The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is :

A

{3 \over 5}

B

{2 \over 3}

C

{3 \over 2}

D

{2 \over 5}

Correct Answer

Option D

Solution

In an isobaric process, Heat supplied, Q = n Cp

\Delta

T Work done, w = nR

\Delta

T $\therefore$ Ratio =

{w \over Q}

=

{{nR\Delta T} \over {n{C_p}\Delta T}}

=

{R \over {{5 \over 2}R}}

=

{2 \over 5}

[Cp =

{5 \over 2}

R for monoatomic gas]