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 step-by-step, we will use the principles of thermodynamics related to the Carnot engine.

### Step 1: Understand the Carnot Engine
A Carnot engine operates between two thermal reservoirs at temperatures \( T_H \) (hot reservoir) and \( T_L \) (cold reservoir). The work done by the engine \( W \) is related to the heat absorbed from the hot reservoir \( Q_H \) and the heat rejected to the cold reservoir \( Q_L \) by the equation:
\[ W = Q_H - Q_L \]

### Step 2: Calculate the Efficiency of the Carnot Engine
The efficiency \( \eta \) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_L}{T_H} \]
Substituting the values \( T_H = 900 \, K \) and \( T_L = 300 \, K \):
\[ \eta = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 3: Relate Efficiency to Work and Heat
The efficiency can also be expressed in terms of work and heat absorbed from the hot reservoir:
\[ \eta = \frac{W}{Q_H} \]
Substituting the efficiency we found:
\[ \frac{2}{3} = \frac{1200}{Q_H} \]

### Step 4: Solve for \( Q_H \)
Rearranging the equation to solve for \( Q_H \):
\[ Q_H = \frac{1200 \times 3}{2} = 1800 \, J \]

### Step 5: Calculate \( Q_L \)
Now, we can use the work equation to find \( Q_L \):
\[ W = Q_H - Q_L \implies Q_L = Q_H - W \]
Substituting the values we have:
\[ Q_L = 1800 \, J - 1200 \, J = 600 \, J \]

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

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