Question

The efficiency of ideal heat engine working between ${32}^{0}F$ and ${212}^{0}F$ is

• A. $84.9\%$
• B. $100\%$
• C. $27\%$
• D. $20\%$

Answer 1

The correct option is C $27\%$

We know efficiency is given by
$\eta =1-\frac{T_L}{T_H}$
where $T_L$ is the lower temperature and $T_H$ is the higher temperature
${32}^{0}F=0^{\circ }\text{C = 273 K}$
${212}^{0}F={100}^{\circ }\text{C = 373 K}$
$\eta =1-\frac{273}{373}$
$=\frac{100}{373}\simeq 27\%$