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 will follow these steps:

### Step 1: Understand the efficiency of the engine
The efficiency (η) of a heat engine is defined as the ratio of the work done (W) by the engine to the heat energy (Q1) supplied to it from the hot reservoir.

\[ \eta = \frac{W}{Q_1} \]

Given that the efficiency is 25%, we can express this as:

\[ \eta = \frac{1}{4} \]

### Step 2: Relate work done to heat supplied
We know the work done by the engine per cycle is 100 J. Using the efficiency formula, we can find the heat energy supplied by the hot reservoir (Q1):

\[ \frac{1}{4} = \frac{100}{Q_1} \]

### Step 3: Solve for Q1
Rearranging the equation to find Q1 gives:

\[ Q_1 = 4 \times W = 4 \times 100 = 400 \text{ J} \]

### Step 4: Use the first law of thermodynamics
According to the first law of thermodynamics, the heat energy delivered to the low-temperature reservoir (Q2) can be calculated using the equation:

\[ Q_2 = Q_1 - W \]

### Step 5: Substitute the known values
Now we can substitute the values we have:

\[ Q_2 = 400 \text{ J} - 100 \text{ J} = 300 \text{ J} \]

### Final Answer
The heat energy delivered by the engine to the low-temperature reservoir in a cycle is **300 J**.

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