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NEET 2006
Exam
NEET
Year
2006
Subject
Physics
Chapter
Thermodynamics

NEET 2006 · Physics · Thermodynamics

A Carnot engine whose sink is at 300 ~K has an efficiency of 40 \%...

A Carnot engine whose sink is at 300 K300 \mathrm{~K} has an efficiency of 40%40 \%. By how much should the temperature of source be increased so as to increase its efficiency by 50%50 \% of original efficiency?

Options

  1. A380 K380 \mathrm{~K}
  2. B275 K275 \mathrm{~K}
  3. C325 K325 \mathrm{~K}
  4. D250 K250 \mathrm{~K}

Correct answer

D. 250 K250 \mathrm{~K}

Step-by-step solution

Efficiency of carnot engine = η=1−T2T1 or T2T1=1−η=1−40100=35∴T1=50100n=32n=500 K New efficiency η′=η+50100η=32η=60%∴T2T1=1−60100=25∴T1′=52\timesT2=52×300=750. \begin{aligned}& \eta=1-\frac{T_2}{T_1}\\ & \text{ or } \quad \frac{T_2}{T_1}=1-\eta \\ & =1-\frac{40}{100}=\frac{3}{5}\\ & \therefore \quad T_1=\frac{50}{100}n=\frac{3}{2}n \\ & =500 \mathrm{~K}\\ & \begin{array}{l}\text{ New efficiency }\\ \eta^{\prime}=\eta+\frac{50}{100} \eta=\frac{3}{2} \eta=60 \% \end{array} \\ & \therefore \quad \frac{T_2}{T_1}=1-\frac{60}{100}=\frac{2}{5}\\ & \therefore T_1^{\prime}=\frac{5}{2} \timesT_2=\frac{5}{2} \times300=750 \text{. }\\ & \end{aligned}Increaseintemperatureofsourcein temperature of source=T1′−T1=750−300=250 K.\begin{aligned}& =T_1^{\prime}-T_1 \\ & =750-300=250 \mathrm{~K} . \end{aligned}$
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