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

Correct answer: C. 400 ~K aligned & \% =(1-(T_2)/(T_1)) × 100 & For 50 \%, (50)/(100)=1-(50)/(11) & T_1=1000 K & aim for 60 \%, (60)/(100)=1-(T_2)/(1000) & T_2=400 K aligned

NEET 2002

Exam: NEET
Year: 2002
Subject: Physics
Chapter: Thermodynamics

NEET 2002 · Physics · Thermodynamics

The efficiency of the Carnot engine is 50 \% and the temperature of...

The efficiency of the Carnot engine is

50 \%

and the temperature of the sink is

500 \mathrm{~K}

. If the temperature of the source is kept constant and its efficiency raised to

60 \%

, then the required temperature of the sink will be:

Options

A $100 \mathrm{~K}$
B $600 \mathrm{~K}$
C $400 \mathrm{~K}$
D $500 \mathrm{~K}$

Correct answer

C. $400 \mathrm{~K}$

Step-by-step solution

\begin{aligned} & \% \eta=\left(1-\frac{T_2}{T_1}\right) \times 100 \\ & \therefore \text{\text{ For } } 50 \%, \frac{50}{100}=1-\frac{50}{11} \\ & \Rightarrow T_1=1000 \mathrm{ K} \\ & \text{ aim for } 60 \%, \frac{60}{100}=1-\frac{T_2}{1000} \\ & \Rightarrow T_2=400 \mathrm{ K} \end{aligned}

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