A heat engine is involved with exchange of heat of 1915 J , – 40 J , + 125 J and – Q J , during one cycle achieving and efficiency of 50 . 0 % . The value of Q is:

Step-by-step solution

I st case when Q 4 is negative η = W Σ Q + = Q 1 + Q 2 + Q 3 + Q 4 Q 1 + Q 3 = 0 . 5 ⇒    1915 - 40 + 125 + Q 4 1915 + 125 = 0 . 5 ⇒    1915 - 40 + 125 + Q 4 = 1020 ⇒    Q 4 = 1020 - 2000 ⇒ Q 4 = - Q = - 980 J Q = 980 J II nd case when Q 4 W Σ Q + = Q 1 + Q 2 + Q 3 + Q 4 Q 1 + Q 3 + Q 4 = 0 . 5 is positive Q 4 = - 1960