A heat engine operates between a cold reservoir at temperature T2 = 400 K and a hot reservoir at tem

Q: Why do we assume it's a Carnot engine when the question doesn't explicitly say so?

The key is the word 'minimum'. A real, irreversible engine is less efficient than a Carnot engine. To get the same work output for a given heat input, a real engine would require a larger temperature difference (a hotter source). Therefore, the minimum possible source temperature corresponds to the maximum possible efficiency, which is the Carnot efficiency.

Q: What is the work done by this engine?

According to the First Law of Thermodynamics (conservation of energy), the work done is the difference between the heat absorbed and the heat rejected. So, $W = Q_1 - Q_2 = 300\text{ J} - 240\text{ J} = 60\text{ J}$.

Q: Can $T_1$ be less than 500 K?

No. If $T_1$ were less than $500\text{ K}$, the Carnot efficiency would be lower than the actual efficiency calculated from the heat values ($1 - 240/300 = 20\%$). Having an engine more efficient than a Carnot engine violates the Second Law of Thermodynamics.

Q: Do we need to convert Joules to Calories for this formula?

No, you don't need to convert units as long as both $Q_1$ and $Q_2$ are in the same units. The ratio $Q_2/Q_1$ is dimensionless, so the specific energy unit cancels out.

Q: What if the temperatures were given in Celsius instead of Kelvin?

You must ALWAYS convert temperatures to Kelvin (absolute scale) before using thermodynamic ratio formulas like $T_2/T_1$. Using Celsius will give you completely incorrect results because the zero point is arbitrary.

Question

A heat engine operates between a cold reservoir at temperature $T_2 = 400	ext{ K}$ and a hot reservoir at temperature $T_1$. It takes $300	ext{ J}$ of heat from the hot reservoir and delivers $240	ext{ J}$ of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be ............ K.

Sigma Percentile · Accepted Answer

### Decoding the Heat Engine

Imagine a classic heat engine operating between two thermal reservoirs. It acts as a bridge, extracting thermal energy from a hot source, converting a portion of it into useful mechanical work, and dumping the leftover energy into a cold sink.

In our specific problem, the engine absorbs $Q_1 = 300	ext{ J}$ of heat from a hot reservoir at an unknown temperature $T_1$. After doing its job, it rejects $Q_2 = 240	ext{ J}$ of heat into a cold reservoir maintained at $T_2 = 400	ext{ K}$. The question asks us to find the **minimum** possible temperature for the hot reservoir, $T_1$.

### The "Minimum Temperature" Catch

Why does the problem specifically ask for the *minimum* temperature? This is a subtle nod to the Second Law of Thermodynamics.

The efficiency of any heat engine is defined as the ratio of work done to heat absorbed: $\eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1}$.

According to Carnot's theorem, no engine operating between two given temperatures can be more efficient than a reversible Carnot engine. For a Carnot engine, the efficiency is solely dependent on the absolute temperatures of the reservoirs: $\eta_{	ext{Carnot}} = 1 - \frac{T_2}{T_1}$.

If we want to achieve a specific efficiency (dictated by the $300	ext{ J}$ in and $240	ext{ J}$ out) using the lowest possible source temperature $T_1$, our engine must be operating at the absolute maximum theoretical efficiency. In other words, we must assume it is a **Carnot engine**.

### The Master Equation

By equating the general efficiency formula with the Carnot efficiency formula, we get a beautiful, simple relationship:

$$1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}$$

Which simplifies directly to:

$$\frac{Q_2}{Q_1} = \frac{T_2}{T_1}$$

This equation tells us that for a perfectly reversible engine, the ratio of heat exchanged is exactly proportional to the ratio of the absolute temperatures of the reservoirs.

### Final Calculation

Now, we simply substitute our known values into the master equation:

$$\frac{240}{300} = \frac{400}{T_1}$$

Let's simplify the fraction on the left side. Dividing both the numerator and the denominator by $60$ gives us $\frac{4}{5}$.

$$\frac{4}{5} = \frac{400}{T_1}$$

Cross-multiplying to solve for $T_1$:

$$4 \cdot T_1 = 5 \cdot 400$$

$$4 \cdot T_1 = 2000$$

$$T_1 = 500	ext{ K}$$

Thus, the absolute minimum temperature the hot reservoir can have is **$500	ext{ K}$**. If the engine were real (irreversible) and had friction or heat leaks, it would be less efficient, meaning it would require a source hotter than $500	ext{ K}$ to perform the exact same energy transfer!

Enter Numerical Value:

Visualized Solution

Visualizing the Heat Engine

The Principle of Maximum Efficiency

Substituting the Values

Simplifying the Ratio

Final Calculation

The Way Forward

The Sigma Insight: Heat Engines and Refrigerators

Decoding the Heat Engine

The "Minimum Temperature" Catch

The Master Equation

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Heat Engine

The Principle of Maximum Efficiency

Substituting the Values

Simplifying the Ratio

Final Calculation

The Way Forward

The Sigma Insight: Heat Engines and Refrigerators

Decoding the Heat Engine

The "Minimum Temperature" Catch

The Master Equation

Final Calculation

Similar Questions

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