A heat engine operates between a cold reservoir at temperature $$T_2 = 400$$ K and a hot reservoir at temperature $$T_1$$. 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 _________.

We have a heat engine that absorbs heat $$Q_1 = 300\ \text{J}$$ from a hot reservoir at temperature $$T_1$$ and rejects heat $$Q_2 = 240\ \text{J}$$ to a cold reservoir at temperature $$T_2 = 400\ \text{K}$$.

First, we find the work done in one complete cycle. The First Law of Thermodynamics for a cyclic engine states that the net work output $$W$$ equals the net heat absorbed:

$$W = Q_1 - Q_2.$$

Substituting the given heats,

$$W = 300\ \text{J} - 240\ \text{J} = 60\ \text{J}.$$

Now we determine the actual thermal efficiency $$\eta_{\text{actual}}$$ of this engine. By definition, thermal efficiency is the ratio of work output to heat absorbed from the hot reservoir:

$$\eta_{\text{actual}} = \frac{W}{Q_1}.$$

Substituting $$W = 60\ \text{J}$$ and $$Q_1 = 300\ \text{J},$$ we get

$$\eta_{\text{actual}} = \frac{60}{300} = 0.20.$$

According to the Second Law of Thermodynamics, no real engine can be more efficient than a reversible (Carnot) engine operating between the same two temperature reservoirs. For a Carnot engine the efficiency is

$$\eta_{\text{Carnot}} = 1 - \frac{T_2}{T_1},$$

where $$T_1$$ is the temperature of the hot reservoir and $$T_2 = 400\ \text{K}$$ is the temperature of the cold reservoir.

To allow the given engine to operate without violating the Second Law, its efficiency must not exceed the Carnot efficiency. Hence we require

$$\eta_{\text{actual}} \le \eta_{\text{Carnot}}.$$

Substituting the expressions,

$$0.20 \le 1 - \frac{T_2}{T_1}.$$

Rewriting the right-hand side,

$$0.20 \le 1 - \frac{400}{T_1}.$$

Now we isolate the fraction:

$$\frac{400}{T_1} \le 1 - 0.20 = 0.80.$$

Next, we solve for $$T_1$$ by inverting and rearranging:

$$\frac{1}{T_1} \le \frac{0.80}{400} \quad\Longrightarrow\quad T_1 \ge \frac{400}{0.80}.$$

Carrying out the division,

$$T_1 \ge 500\ \text{K}.$$

This value represents the minimum possible temperature of the hot reservoir that ensures the engine’s efficiency does not exceed the Carnot limit.

So, the answer is $$500\ \text{K}$$.