Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Thermodynamics

Question:

An ideal gas heat engine operates in Carnot cycle between 227oC and 127oC. It absorbs 6 × 104 cals of heat at higher temperature. Amount of heat converted to work is :-

Options:

4.8 × 10cals

1.2 × 104 cals

2.4 × 104 cals

3.6 × 10cals

Correct Answer:

1.2 × 104 cals

Explanation:

\(\text{Efficiency : }\eta = \frac{W}{Q}\)

\(\text{Also, } \eta = 1 - \frac{T_2}{T_1}\)

\(\Rightarrow W = Q[1-\frac{T_2}{T_1}] \)

$\Rightarrow W = 6\times 10^4 [1 - \frac{127+273}{227+273}] = 6\times 10^4 \times (1- \frac{400}{500}) = 6\times 10^4 \times \frac{1}{5} = 1.2\times 10^4 cal$