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:

For a heat engine we use the First Law of Thermodynamics which says that over one complete cycle, the net work done $$W$$ equals the algebraic sum of all heats exchanged:

$$W=\sum Q_i$$

The heats exchanged in the given cycle are

$$+1915\ \text{J},\; -40\ \text{J},\; +125\ \text{J},\; -Q\ \text{J}$$

(A positive sign means heat is absorbed by the engine, a negative sign means heat is rejected.)

So the net heat, and therefore the work done in one cycle, is

$$W = 1915 + (-40) + 125 + (-Q)$$

Adding the known numbers step by step:

$$1915 + 125 = 2040$$

$$2040 - 40 = 2000$$

Hence

$$W = 2000 - Q$$

Next, efficiency $$\eta$$ of a heat engine is defined as

$$\eta = \frac{W}{Q_{\text{in}}}$$

where $$Q_{\text{in}}$$ is the total heat absorbed (all positive terms only). The positive heats here are $$1915\ \text{J}$$ and $$125\ \text{J}$$, so

$$Q_{\text{in}} = 1915 + 125 = 2040\ \text{J}$$

The given efficiency is $$\eta = 0.500$$, therefore

$$0.500 = \frac{W}{2040}$$

Substituting $$W = 2000 - Q$$ into this relation gives

$$0.500 = \frac{2000 - Q}{2040}$$

Multiplying both sides by $$2040$$:

$$0.500 \times 2040 = 2000 - Q$$

$$1020 = 2000 - Q$$

Rearranging to isolate $$Q$$:

$$-Q = 1020 - 2000$$

$$-Q = -980$$

$$Q = 980\ \text{J}$$

Hence, the correct answer is Option C.