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Heat and Thermodynamics question

2006 · Q165
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Heat and Thermodynamics question

2006 · Q165

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
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 ?
  1. A
    380 K
  2. B
    275 K
  3. C
    325 K
  4. D
    250 K
View written solutionFree

Correct answer: D

We know that efficiency of Carnot Engine

=T1−T2T1 = {{{T_1} - {T_2}} \over {{T_1}}}=T1​T1​−T2​​

where, T1 is temp. of source & T2 is temp. of sink

∴0.40=T1−300T1⇒T1−300=0.40T1 \therefore 0.40 = {{{T_1} - 300} \over {{T_1}}} \Rightarrow {T_1} - 300 = 0.40{T_1}∴0.40=T1​T1​−300​⇒T1​−300=0.40T1​

0.6T1=300⇒T1=3000.6=30006=500K0.6{T_1} = 300 \Rightarrow {T_1} = {{300} \over {0.6}} = {{3000} \over 6} = 500K0.6T1​=300⇒T1​=0.6300​=63000​=500K

Now efficiency to be increased by 50%

∴\therefore∴ 0.60=T1−300T1⇒T1−300=0.6T10.60 = {{{T_1} - 300} \over {{T_1}}} \Rightarrow {T_1} - 300 = 0.6{T_1}0.60=T1​T1​−300​⇒T1​−300=0.6T1​

0.4T1=300⇒T1=3000.4=300×104=7500.4{T_1} = 300 \Rightarrow {T_1} = {{300} \over {0.4}} = {{300 \times 10} \over 4} = 7500.4T1​=300⇒T1​=0.4300​=4300×10​=750

Increase in temp = 750 – 500 = 250 K

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