A Carnot's heat engine works between the temperatures 427^ ^ C and 27^ ^ C . What amount of heat should it consume per second to deliver mechanical work at the rate of 1.0 kW ?

Correct answer: A. 0.417 kcal / s The efficiency of the heat engine is =1- T_ 2 T_ 1 =1-((273+27 K)/(273+427 K))=(4)/(7) But = W Q_ 1 Q_ 1 = W = 1.0 kW 4 / 7 =1.75 kW=0.417 kcal / s Thus, the engine would require 417 cal of heat per second, to deliver the requisite amount of work.

BITSAT 2013

Exam: BITSAT
Year: 2013
Subject: Physics
Chapter: Thermodynamics

BITSAT 2013 · Physics · Thermodynamics

A Carnot's heat engine works between the temperatures 427^ ^ C and...

A Carnot's heat engine works between the temperatures

427^{{}^{\circ}} \mathrm{C}

and

27^{{}^{\circ}} \mathrm{C}

. What amount of heat should it consume per second to deliver mechanical work at the rate of

1.0 \mathrm{ kW} ?

Options

A $0.417 \mathrm{kcal} / \mathrm{s}$
B $4.17 \mathrm{kcal} / \mathrm{s}$
C $41.7 \mathrm{kcal} / \mathrm{s}$
D $0.212 \mathrm{kcal} / \mathrm{s}$

Correct answer

A. $0.417 \mathrm{kcal} / \mathrm{s}$

Step-by-step solution

The efficiency of the heat engine is \eta=1-

\frac{T_{2}}{T_{1}}=1-\left(\frac{273+27 K}{273+427 K}\right)=\frac{4}{7}

But

\quad \eta=\frac{\mathrm{W}}{\mathrm{Q}_{1}}

\therefore

\mathrm{Q}_{1}=\frac{\mathrm{W}}{\eta}=\frac{1.0 \mathrm{ kW}}{4 / 7}=1.75 \mathrm{ kW}=0.417 \mathrm{kcal}

\mathrm{s}

Thus, the engine would require 417 cal of heat per second, to deliver the requisite amount of work.

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