A heat engine is involved with exchange of heat of 1915 J, – 40J, + 125 J and –Q J, during one cycle achieving an efficiency of 50.0%. The value of Q is

Solution

Physics sir · 4 steps

Sort the heats into absorbed and rejected, then make work exactly half of the heat taken in.

1. Heat absorbed. Positive heats flow into the engine: Qᵢₙ = 1915 + 125 = 2040 J.
2. Net work. Over one full cycle the internal energy returns to its starting value. So the work done equals the net heat: W = 1915 - 40 + 125 - Q = 2000 - Q.

   η = W/Qᵢₙ ⇒ 0.5 = (2000 - Q)/2040

3. Solve. 2000 - Q = 1020, so Q = 980 J. As a check, total rejected heat is 40 + 980 = 1020 J, which is half of 2040 J.
4. So. Q = 980 J, option (a).

The idea to keep. Efficiency divides work by the heat absorbed only, not by the net heat. In a cycle, work equals the sum of all heat exchanges with their signs.

Answer from the NTA final answer key. Solution by PrepWiser.