A cannot engine operates between two reservoirs of temperature `T_(1)` and `T_(2)` .The efficiency of engine is 25 %. The engine performs 100 J of work percycle. The heat energy (in J) delivered by the engine to the low temperature reservoir in a cycle is _______

Explanation

To solve the problem, we need to find the heat energy delivered by the engine to the low-temperature reservoir (Q2) in a cycle, given the efficiency of the engine and the work done per cycle.

### Step-by-Step Solution:

1. **Understand the Efficiency Formula:**
The efficiency (η) of a heat engine is defined as the ratio of the work done (W) by the engine to the heat absorbed from the hot reservoir (Q1):
\[ \eta = \frac{W}{Q_1} \]
Given that the efficiency is 25%, we can express this as:
\[ \eta = 0.25 \]

2. **Express Q1 in terms of W:**
Rearranging the efficiency formula gives:
\[ Q_1 = \frac{W}{\eta} \]
Substituting the values we have:
\[ Q_1 = \frac{100 \, \text{J}}{0.25} = 400 \, \text{J} \]

3. **Use the First Law of Thermodynamics:**
According to the first law of thermodynamics, the heat absorbed from the hot reservoir (Q1) is equal to the work done (W) plus the heat delivered to the cold reservoir (Q2):
\[ Q_1 = W + Q_2 \]
Rearranging this gives:
\[ Q_2 = Q_1 - W \]

4. **Substitute the values to find Q2:**
Now we can substitute the values of Q1 and W into the equation:
\[ Q_2 = 400 \, \text{J} - 100 \, \text{J} = 300 \, \text{J} \]

5. **Final Answer:**
The heat energy delivered by the engine to the low-temperature reservoir in a cycle is:
\[ \boxed{300 \, \text{J}} \]