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ISRO IPRC Technical Assistant Mechanical held on 28/08/2016

Option 4 : 400 kJ

ST 1: Engineering Mechanics

1210

20 Questions
20 Marks
16 Mins

**Concept:**

The efficiency of Reversible Heat Engine ** \(η_R =1-\frac {Heat~ output}{Heat~input}=1- \frac {T_L}{T_H} \)**

where T_{H} = Higher temeperature (Source temperature), T_{L} = Lower temperature (Sink temperature)

**Calculation:**

__Given:__

TL = 300 K, TH = 600 K, Heat output = 200 kJ

\(1-\frac {Heat~ output}{Heat~input}=1- \frac {T_L}{T_H} \)

\(1-\frac {200}{Heat~input}=1- \frac {300}{600} \)

\(1-\frac {200}{Heat~input}=\frac 12 \)

\(\frac {200}{Heat~input}=1-\frac 12 \)

\(\frac {200}{Heat~input}=\frac 12 \)

Heat input = 200 × 2

**Heat input = 400 kJ **