Heat and Thermodynamics — Page 8 | NEET Physics

Q71

An engine has an efficiency of 1/6. When the temperature of sink is reduced by 62 o C, its efficiency is doubled. Temperatures of the source is

A 37 o C

B 62 o C

C 99 o C

D 124 o C.

Correct Answer

Option C

Solution

Since efficiency of engine is

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

According to problem,

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

...(i) When the temperature of the sink is reduced by 62°C, its efficiency is doubled

2\left( {{1 \over 6}} \right) = 1 - {{{T_2} - 62} \over {{T_1}}}

...(ii) Solving (i) and (ii) T 2 = 372 K T 1 = 99°C = Temperature of source.

Q72

A Carnot engine whose sink is at 300 K 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 ?

A 380 K

B 275 K

C 325 K

D 250 K

Correct Answer

Option D

Solution

We know that efficiency of Carnot Engine

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

where, T 1 is temp. of source & T 2 is temp. of sink

\therefore 0.40 = {{{T_1} - 300} \over {{T_1}}} \Rightarrow {T_1} - 300 = 0.40{T_1}

0.6{T_1} = 300 \Rightarrow {T_1} = {{300} \over {0.6}} = {{3000} \over 6} = 500K

Now efficiency to be increased by 50% $\therefore$

0.60 = {{{T_1} - 300} \over {{T_1}}} \Rightarrow {T_1} - 300 = 0.6{T_1}

0.4{T_1} = 300 \Rightarrow {T_1} = {{300} \over {0.4}} = {{300 \times 10} \over 4} = 750

Increase in temp = 750 – 500 = 250 K

Q73

The molar specific heat at constant pressure of an ideal gas is (7/2) R. The ratio of specific heat at constant pressure to that at constant volume is

A 9/7

B 7/5

C 8/7

D 5/7

Correct Answer

Option B

Solution

{C_P} = {7 \over 2}R;{C_V} = {C_P} - R

= {7 \over 2}R - R = {5 \over 2}R

{{{C_P}} \over {{C_V}}} = {{7/2R} \over {5/2R}} = {7 \over 5}

Q74

An ideal gas heat engine operates in Carnot cycle between 227 o C and 127 o C. It absorbs 6

\times

10 4 cal of heat at higher temperature. Amount of heat converted to work is

A 4.8

\times

10 4 cal

B 6

\times

10 4 cal

C 2.4

\times

10 4 cal

D 1.2

\times

10 4 cal.

Correct Answer

Option D

Solution

1 - {{{T_2}} \over {{T_1}}} = 1 - {{{Q_2}} \over {{Q_1}}}

\Rightarrow 1 - {{400} \over {500}} = 1 - {{{Q_2}} \over {6 \times {{10}^4}}}

\Rightarrow {4 \over 5} = {{{Q_2}} \over {6 \times {{10}^4}}} \Rightarrow {Q_2} = 4.8 \times {10^4}cal

Net heat converted into work

= 6.0 \times {10^4} - 4.8 \times {10^4} = 1.2 \times {10^4}cal

Q75

Which of the following processes is reversible?

A Transfer of heat by conduction

B Transfer of heat by radiation

C Isothermal compression

D Electrical heating of a nichrome wire.

Correct Answer

Option C

Solution

For a process to be reversible, it must be quasi-static.

For quasi static process, all changes take place infinitely slowly.

Isothermal process occur very slowly so it is quasi-static and hence it is reversible.

Q76

One mole of an ideal gas at an initial temperature of T K does 6R joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be

A (T + 2.4) K

B (T

-

2.4) K

C (T + 4) K

D (T

-

4) K

Correct Answer

Option D

Solution

{T_1} = T,W = 6R\,joules,\gamma = {5 \over 3}

W = {{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}} = {{nR{T_1} - nR{T_2}} \over {\gamma - 1}}

= {{nR\left( {{T_1} - {T_2}} \right)} \over {\gamma - 1}}

n = 1,{T_1} = T \Rightarrow {{R\left( {T - {T_2}} \right)} \over {5/3 - 1}} = 6R

\Rightarrow {T_2} = \left( {T - 4} \right)K

Q77

The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be (where R is the gas constant)

A PV = (5/32)RT

B PV = 5RT

C PV = (5/2)RT

D PV = (5/16) RT

Correct Answer

Option A

Solution

As PV = nRT

n = {m \over {molecular\,mass}} = {5 \over {32}}

\Rightarrow PV = \left( {{5 \over {32}}} \right)RT

Q78

An ideal gas heat engine operates in a Carnot cycle between 227 o C and 127 o C. It absorbs 6 kcal at the height temperature. The amount of heat (in kcal) converted into work is equal to

A 4.8

B 3.5

C 1.6

D 1.2

Correct Answer

Option D

Solution

Efficiency of Carnot engine

= {W \over {{Q_1}}} = 1 - {{{T_2}} \over {{T_1}}}

{W \over 6} = 1 - {{400} \over {500}} = {1 \over 5} \Rightarrow W = {6 \over 5} = 1.2\,kcal

Q79

The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be

A 100 K

B 600 K

C 400 K

D 500 K

Correct Answer

Option C

Solution

Efficiency ( $\eta$ ) of a carnot engine is given by

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

, where T 1 is the temperature of the source and T 2 is the temperature of the sink. Here, T 2 = 500 K $\therefore$

0.5 = 1 - {{500} \over {{T_1}}} \Rightarrow {T_1} = 1000K

Now,

\eta = 0.6 = 1 - {{{T_2}'} \over {1000}}

(T 2 ' is the new sink temperature)

\Rightarrow {T_2}' = 400\,K

Q80

A scientist says that the efficiency of his heat engine which work at source temperature 127 o C and sink temperature 27 o C is 26%, then

A it is impossible

B it is possible but less probable

C it is quite probable

D data are incomplete.

Correct Answer

Option A

Solution

Efficiency is maximum in Carnot engine which is an ideal engine.

\because \eta = {{400 - 300} \over {400}} \times 100\% = 25\%

$\therefore$ efficiency 26% is impossible for his heat engine.