A carnot engine operates between two reservoirs of temperature 900K and 300K. The engine performs 1200 J of work per cycle. The heat energy delivered by the engine to the low temperature reservoir in a cycle is:

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Carnot Engine Efficiency Formula
The efficiency (η) of a Carnot engine is given by the formula:
\[ η = 1 - \frac{T_2}{T_1} \]
where \(T_1\) is the temperature of the hot reservoir and \(T_2\) is the temperature of the cold reservoir.

### Step 2: Substitute the Given Temperatures
In this case, \(T_1 = 900 \, K\) and \(T_2 = 300 \, K\). Plugging these values into the efficiency formula:
\[ η = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 3: Relate Efficiency to Heat Transfer
The efficiency can also be expressed in terms of heat absorbed from the hot reservoir (\(Q_1\)) and heat rejected to the cold reservoir (\(Q_2\)):
\[ η = \frac{W}{Q_1} \]
where \(W\) is the work done by the engine. Rearranging gives:
\[ Q_1 = \frac{W}{η} \]

### Step 4: Calculate \(Q_1\)
Given that the work done \(W = 1200 \, J\), we can now calculate \(Q_1\):
\[ Q_1 = \frac{1200 \, J}{\frac{2}{3}} = 1200 \, J \times \frac{3}{2} = 1800 \, J \]

### Step 5: Use the First Law of Thermodynamics
According to the first law of thermodynamics for the Carnot engine:
\[ W = Q_1 - Q_2 \]
We can rearrange this to find \(Q_2\):
\[ Q_2 = Q_1 - W \]

### Step 6: Substitute the Values to Find \(Q_2\)
Now substituting the values we have:
\[ Q_2 = 1800 \, J - 1200 \, J = 600 \, J \]

### Final Answer
The heat energy delivered by the engine to the low-temperature reservoir in a cycle is:
\[ \boxed{600 \, J} \]

---