A heat engine operates between a cold reservoir at temperature `T_(2)=400 K` and a hot reservoir at temperature T . It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be _____________ K.

Answer

To solve the problem, we will use the principles of thermodynamics, specifically the Carnot efficiency for a heat engine operating between two reservoirs.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Cold reservoir temperature, \( T_2 = 400 \, K \)
- Heat taken from the hot reservoir, \( Q_H = 300 \, J \)
- Heat delivered to the cold reservoir, \( Q_C = 240 \, J \)

2. **Calculate Work Done by the Engine:**
The work done \( W \) by the engine can be calculated using the formula:
\[ W = Q_H - Q_C \]
Substituting the values:
\[ W = 300 \, J - 240 \, J = 60 \, J \]

3. **Calculate Efficiency of the Engine:**
The efficiency \( \eta \) of the heat engine is given by:
\[ \eta = \frac{W}{Q_H} \]
Substituting the values:
\[ \eta = \frac{60 \, J}{300 \, J} = \frac{1}{5} = 0.2 \]

4. **Relate Efficiency to Temperatures:**
For a Carnot engine, the efficiency is also given by:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \( T_1 \) is the temperature of the hot reservoir. Rearranging this gives:
\[ \frac{T_2}{T_1} = 1 - \eta \]
Substituting the known values:
\[ \frac{400 \, K}{T_1} = 1 - 0.2 = 0.8 \]

5. **Solve for \( T_1 \):**
Rearranging the equation to find \( T_1 \):
\[ T_1 = \frac{400 \, K}{0.8} = 500 \, K \]

Thus, the minimum temperature of the hot reservoir \( T_1 \) is **500 K**.