Question

A Carnot heat engine works between the temperatures ${427}^{\circ }\text{C}$ and ${27}^{\circ }\text{C}$. What amount of heat should it consume per second to deliver mechanical work at the rate of $1.0\text{kW}$?

• A. $418\text{cal/s}$
• B. $41.8\text{cal/s}$
• C. $4.18\text{cal/s}$
• D. $0.418\text{cal/s}$

Answer 1

The correct option is A $418\text{cal/s}$

The efficiency of the carnot heat engine is
$\eta =1-\frac{T_2}{T_1}...\left(i\right)$
$T_2=(273+27)=300\text{K}$
$T_1=(273+427)=700\text{K}$
$\Rightarrow \eta =1-\left(\frac{300}{700}\right)=\frac{4}{7}$
Also the efficiency is defined as,
$\eta =\frac{W}{Q_1}...\left(ii\right)$
$\Rightarrow Q_1=\frac{W}{\eta }=\frac{1.0\text{kW}}{4/7}=1.75\text{kJ/s}$
since $1\text{cal}=4.186\text{J}$
$\therefore Q_1=\frac{1.75}{4.186}=0.418\text{kcal/s}=418\text{cal/s}$
Thus, the engine would require $418\text{cal}$ of heat per second, to deliver the requisite amount of work.