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Heat and Thermodynamics question

2000 · Q160
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Heat and Thermodynamics question

2000 · Q160

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
The (W/Q) of a Carnot engine is 1/6, now the temperature of sink is reduced by 62oC, then this ratio becomes twice, therefore the initial temperature of the sink and source are respectively
  1. A
    33oC, 67oC
  2. B
    37oC, 99oC
  3. C
    67oC, 33oC
  4. D
    97 K, 37 K
View written solutionFree

Correct answer: B

Initially the efficiency of the engine was 16{1 \over 6}61​ which increases to 13{1 \over 3}31​ when the sink temperature reduces by 62º C.

η=16=1−T2T1\eta = {1 \over 6} = 1 - {{{T_2}} \over {{T_1}}}η=61​=1−T1​T2​​ when T2 = sink temperature , T1 = source temperature

⇒T2=56T1 \Rightarrow {T_2} = {5 \over 6}{T_1}⇒T2​=65​T1​

Secondly,
13=1−T2−62T1=1−T2T1+62T1{1 \over 3} = 1 - {{{T_2} - 62} \over {{T_1}}} = 1 - {{{T_2}} \over {{T_1}}} + {{62} \over {{T_1}}}31​=1−T1​T2​−62​=1−T1​T2​​+T1​62​

=1−56+62T1 = 1 - {5 \over 6} + {{62} \over {{T_1}}}=1−65​+T1​62​

⇒T1=62×6=372K=372−273=99oC \Rightarrow {T_1} = 62 \times 6 = 372K = 372 - 273 = {99^o}C⇒T1​=62×6=372K=372−273=99oC

& T2=56×372=310K=310−273=37oC{T_2} = {5 \over 6} \times 372 = 310K = 310 - 273 = {37^o}CT2​=65​×372=310K=310−273=37oC

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