A Carnot engine absorbs 6 × 10^5 cal at 227^ C . The work done per cycle by the engine, if its sink is maintained at 127^ C is

Correct answer: C. 5 × 10^5 ~J Here, Q_1=6 × 10^5 cal aligned & T_1=227^ C=227+273=500 ~K & ~T_2=127^ C=127+273=400 ~K aligned Work done/cycle, W= ? As Q_2 Q_1 = T_2 ~T_1 or Q_2= T_2 ~T_1 × Q_1=(400)/(500) × 6 × 10^5 =4.8 × 10^5 cal aligned As, & W=Q_1-Q_2=6 × 10^5-4.8 × 10^5 ~W & =1.2 × 10^5 cal=1.2 × 10^5 × 4.2 ~J ~W & =5.04 × 10^5 ~J aligned

JIPMER 2018

Exam: JIPMER
Year: 2018
Subject: Physics
Chapter: Thermodynamics

JIPMER 2018 · Physics · Thermodynamics

A Carnot engine absorbs 6 × 10^5 cal at 227^ C . The work done per...

A Carnot engine absorbs

6 \times 10^5 \mathrm{cal}

227^{\circ} \mathrm{C}

. The work done per cycle by the engine, if its sink is maintained at

127^{\circ} \mathrm{C}

Options

A $15 \times 10^8 \mathrm{~J}$
B $15 \times 10^4 \mathrm{~J}$
C $5 \times 10^5 \mathrm{~J}$
D $2 \times 10^4 \mathrm{~J}$

Correct answer

C. $5 \times 10^5 \mathrm{~J}$

Step-by-step solution

Here,

\mathrm{Q}_1=6 \times 10^5 \mathrm{cal}

\begin{aligned} & \mathrm{T}_1=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K} \\ & \mathrm{~T}_2=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}\end{aligned}

Work done/cycle, $\mathrm{W}=$ ? As $\frac{\mathrm{Q}_2}{\mathrm{Q}_1}=\frac{\mathrm{T}_2}{\mathrm{~T}_1}$ or $\mathrm{Q}_2=\frac{\mathrm{T}_2}{\mathrm{~T}_1} \times \mathrm{Q}_1=\frac{400}{500} \times 6 \times 10^5 =4.8 \times 10^5 \mathrm{cal}

\begin{aligned} \text{ As, } & \mathrm{W}=\mathrm{Q}_1-\mathrm{Q}_2=6 \times 10^5-4.8 \times 10^5 \\ \mathrm{~W} & =1.2 \times 10^5 \mathrm{cal}=1.2 \times 10^5 \times 4.2 \mathrm{~J} \\ \mathrm{~W} & =5.04 \times 10^5 \mathrm{~J}\end{aligned}$$$

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