Question

A Carnot refrigerator is operating between thermal reservoirs with temperatures of \(27.0^{\circ} \mathrm{C}\) and \(0.00^{\circ} \mathrm{C}\) a) How much work will need to be input to extract \(10.0 \mathrm{~J}\) of heat from the colder reservoir? b) How much work will be needed if the colder reservoir is at \(-20.0^{\circ} \mathrm{C}^{2}\)

Short answer

Answer: (a) The required work input to extract 10 J of heat from the colder reservoir at 0.00°C is approximately 0.988 J. (b) The required work input to extract 10 J of heat from the colder reservoir at -20.0°C is approximately 1.86 J.

Step-by-step solution

Understand the Carnot refrigerator efficiency formula

The efficiency of a Carnot refrigerator is given by the coefficient of performance (COP), which is defined as the ratio of the extracted heat from the colder reservoir (Q_c) to the work input (W). The formula for the COP of a Carnot refrigerator is: COP = \(\frac{Q_c}{W} = \frac{T_c}{T_h - T_c}\), where \(T_c\) is the absolute temperature of the colder reservoir, and \(T_h\) is the absolute temperature of the hotter reservoir.

Convert the given temperatures to Kelvin

To utilize the formula, we need to convert the given temperatures in Celsius to Kelvin by adding 273.15. For the first case: \(T_c = 0.00^\circ\mathrm{C} + 273.15 = 273.15\mathrm{K}\), \(T_h = 27.0^\circ\mathrm{C} + 273.15 = 300.15\mathrm{K}\). For the second case: \(T_c = -20.0^\circ\mathrm{C} + 273.15 = 253.15\mathrm{K}\).

Calculate the COP for both cases

Using the COP formula and the temperatures in Kelvin, we can calculate the COP for both cases. For the first case: COP = \(\frac{273.15}{300.15 - 273.15} = \frac{273.15}{27} = 10.12\). For the second case: COP = \(\frac{253.15}{300.15 - 253.15} = \frac{253.15}{47} = 5.39\).

Calculate the work input for both cases

Now that we have the COP for both cases, we can use the definition of COP to find the required work input. a) To extract \(10.0\mathrm{J}\) of heat from the colder reservoir at \(0.00^\circ \mathrm{C}\): \(W = \frac{Q_c}{\text{COP}} = \frac{10.0}{10.12} \approx 0.988\mathrm{J}\). b) To extract \(10.0\mathrm{J}\) of heat from the colder reservoir at \(-20.0^\circ \mathrm{C}\): \(W = \frac{Q_c}{\text{COP}} = \frac{10.0}{5.39} \approx 1.86\mathrm{J}\).

State the final results

We have calculated the work input required for both cases: a) To extract \(10.0 \mathrm{J}\) of heat from the colder reservoir at \(0.00^\circ\mathrm{C}\), the required work input is approximately \(0.988\mathrm{J}\). b) To extract \(10.0 \mathrm{J}\) of heat from the colder reservoir at \(-20.0^\circ\mathrm{C}\), the required work input is approximately \(1.86\mathrm{J}\).

Key concepts

Coefficient of Performance (COP)

The efficiency of refrigeration systems, like the Carnot refrigerator, is measured using the coefficient of performance, or COP for short. This key concept refers to the ratio of how much heat energy is moved from a colder area compared to the work energy put into the system to move that heat. The higher the COP, the more efficient the refrigerator is.

Mathematically, COP is defined as \( \frac{Q_c}{W} \), where \( Q_c \) is the quantity of heat extracted from the cold reservoir and \( W \) is the work input to the system. For Carnot refrigerators, which represent an idealized version of a real refrigerator, the COP is also equal to \( \frac{T_c}{T_h - T_c} \), where \(T_c\) is the absolute temperature of the cold reservoir and \(T_h\) is the absolute temperature of the hot reservoir. Understanding COP is crucial in assessing the performance and energy efficiency of refrigeration cycles.

Absolute Temperature Conversion

Temperature conversion from Celsius to Kelvin is a foundational aspect of thermodynamics, as equations like the one for COP require absolute temperatures. The Kelvin scale is an absolute temperature scale, meaning it starts at absolute zero, where particles have minimal thermal motion.

To convert temperatures from Celsius to Kelvin, you simply add 273.15. For example, \(27.0^{\textdegree} \mathrm{C}\) becomes \(300.15 \mathrm{K}\). This step is not just about changing numbers; it shifts the framework into one where temperature values represent total energy above absolute zero, which is essential for thermodynamic calculations.

Work Input Calculation

The concept of work input is tied to the first law of thermodynamics, which concerns the conservation of energy. In the context of a Carnot refrigerator, work input is the energy required by the system to transfer heat from a colder to a hotter reservoir.

Once the COP has been determined, we can calculate the work input using the relationship \( W = \frac{Q_c}{\text{COP}} \). This equation helps us understand the energy efficiency of the refrigerator: less work input for a given amount of heat transfer signifies greater efficiency. Calculating the precise amount of work informs us about the system's practicality and sustainability.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy. The laws of thermodynamics govern the principles of energy conservation and entropy, providing vital guidelines for the operation of machines such as the Carnot refrigerator.

In the context of the Carnot cycle, which is a model for understanding the most efficient heat engine cycle, thermodynamics tells us that only a fraction of heat energy can be converted into work, illustrating the inherent limitations due to the entropy increase. It's a fascinating area that helps us optimize energy usage and technology design for thermal systems.