An ideal gas heat engine operates in carnot cycle between 227^ C and 127^ C . It absorbs 6 × 10^4 cal of heat at higher temperature. Amount of heat converted into work, is:

Correct answer: D. 1.2 × 10^4 cals For a carnot engine efficiency aligned e & =1-(T_2)/(T_1) & =1-(127+273)/(227+273) & =1-(400)/(500)=(1)/(5) aligned Now, e= Work Output Heat Output = W 6 × 10^4 aligned W& =e ×6 ×10^4 & =(1)/(5) ×6 ×10^4 & =1.2 ×10^4 cal . aligned

NEET 2005

Exam: NEET
Year: 2005
Subject: Physics
Chapter: Thermodynamics

NEET 2005 · Physics · Thermodynamics

An ideal gas heat engine operates in carnot cycle between 227^ C...

An ideal gas heat engine operates in carnot cycle between

227^{\circ} \mathrm{C}

and

127^{\circ} \mathrm{C}

. It absorbs

6 \times 10^4

cal of heat at higher temperature. Amount of heat converted into work, is:

Options

A $4.8 \times 10^4 \mathrm{cals}$
B $6 \times 10^4 \mathrm{cals}$
C $2.4 \times 10^4$ cals
D $1.2 \times 10^4$ cals

Correct answer

D. $1.2 \times 10^4$ cals

Step-by-step solution

For a carnot engine

\text{ efficiency }

\begin{aligned}e & =1-\frac{T_2}{T_1}\\ & =1-\frac{127+273}{227+273}\\ & =1-\frac{400}{500}=\frac{1}{5} \end{aligned}

Now, $\quad e=\frac{\text{ Work Output }}{\text{ Heat Output }}=\frac{\mathrm{W}}{6 \times 10^4}

\begin{aligned} \mathrm{W}& =e \times6 \times10^4 \\ & =\frac{1}{5} \times6 \times10^4 \\ & =1.2 \times10^4 \mathrm{cal} . \end{aligned}$$$

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