In a cyclic process, change in internal energy is zero. ⇒∆U=0 Hence, by first law of thermodynamics ‌ ∆ Q=∆ U+∆ W $ ∆Q=∆W ( ∵ ∆ U=0) $ Now, work done by a gas = area under pV graph. Also work done is negative if cycle is anticlockwise. In given cycle, area =
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2
×BC×AC =
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(20−5)×(30−10) =
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×15×20=150N−m So, work done in 10 such complete cycles =−10×150 =−1500J=−1.5‌kJ Heat absorbed = Work done =−1.5‌kJ