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 to work is

Correct answer: C. 1.2 × 10^4 cal The heat converted to work is the amount of heat that remains after going through sink. From the relation (Q_2)/(Q_1)=(T_2)/(T_1) Given, aligned Q_1 & =6 × 10^4 cal, T_1 & =227+273=500 ~K T_2 & =127+273=400 ~K aligned (Q_2)/(6 × 10^4)=(400)/(500) Q_2=(4)/(5) × 6 × 10^4=4.8 × 10^4 cal Now, heat converted to work aligned & =Q_1-Q_2 \& =6.0 × 10^4-4.8 × 10^4 & =1.2 × 10^4 cal aligned

JIPMER 2007

Exam: JIPMER
Year: 2007
Subject: Physics
Chapter: Thermodynamics

JIPMER 2007 · 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 \mathrm{cal}

of heat at higher temperature. Amount of heat converted to work is

Options

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

Correct answer

C. $1.2 \times 10^4 \mathrm{cal}$

Step-by-step solution

The heat converted to work is the amount of heat that remains after going through sink. From the relation

\frac{Q_2}{Q_1}=\frac{T_2}{T_1}

Given,

\begin{aligned} Q_1 & =6 \times 10^4 \mathrm{cal}, \\ T_1 & =227+273=500 \mathrm{~K} \\ T_2 & =127+273=400 \mathrm{~K}\end{aligned}

\therefore \frac{Q_2}{6 \times 10^4}=\frac{400}{500} \Rightarrow Q_2=\frac{4}{5} \times 6 \times 10^4=4.8 \times 10^4 \mathrm{cal}

Now, heat converted to work

\begin{aligned}& =Q_1-Q_2 \& =6.0 \times 10^4-4.8 \times 10^4 \\ & =1.2 \times 10^4 \mathrm{cal}\end{aligned}

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