Question

A refrigerator with a coefficient of performance of 4 transfers heat from a cold region at \(-20^{\circ} \mathrm{C}\) to a hot region at \(30^{\circ} \mathrm{C}\). Calculate the total entropy change of the regions when \(1 \mathrm{kJ}\) of heat is transferred from the cold region. Is the second law satisfied? Will this refrigerator still satisfy the second law if its coefficient of performance is \(6 ?\)

Short answer

Based on the given information, calculate the total entropy change when 1 kJ of heat is transferred from a cold region at -20°C to a hot region at 30°C, given a refrigerator with a coefficient of performance (COP) of 4. Additionally, determine if the second law of thermodynamics is satisfied and if it would still be satisfied with a COP of 6. Solution: 1. Convert temperatures to Kelvin: Cold region: 253.15 K; Hot region: 303.15 K 2. Calculate the work done by the refrigerator: W = 250 J 3. Calculate the heat transfer to the hot region: Q_hot = 1250 J 4. Calculate entropy change: ΔS_cold = -3.95 J/K; ΔS_hot = 4.12 J/K; ΔS_total = 0.17 J/K 5. Assess if the second law is satisfied: yes, the second law is satisfied. With a COP of 6, the second law would not be satisfied (ΔS_total = -0.10 J/K).

Step-by-step solution

Convert temperatures to Kelvin

To convert temperatures in Celsius to Kelvin, we simply add 273.15 to the Celsius value. In our case: - Cold region: \(-20^{\circ}\mathrm{C} + 273.15 = 253.15\,\mathrm{K}\) - Hot region: \(30^{\circ}\mathrm{C} + 273.15 = 303.15\,\mathrm{K}\)

Calculate the work done by the refrigerator

The coefficient of performance (COP) of a refrigerator can be calculated by the formula: COP = \(\frac{Q_\mathrm{cold}}{W}\), where \(Q_\mathrm{cold}\) is the heat transfer from the cold region and \(W\) is the work done by the refrigerator. In this case, COP = \(4\) and \(Q_\mathrm{cold} = 1\,\mathrm{kJ} = 1000\,\mathrm{J}\). We need to find \(W\). Rearranging the formula, we get: \(W = \frac{Q_\mathrm{cold}}{\mathrm{COP}} = \frac{1000\,\mathrm{J}}{4} = 250\, \mathrm{J}\)

Calculate the heat transfer to the hot region

The heat transfer to the hot region, \(Q_\mathrm{hot}\) can be found using the energy conservation principle: Energy in (Work) = Energy out (Heat transfers) \(W = Q_\mathrm{hot} - Q_\mathrm{cold}\) \(Q_\mathrm{hot} = W + Q_\mathrm{cold} = 250\,\mathrm{J} + 1000\,\mathrm{J} = 1250\,\mathrm{J}\)

Calculate entropy change

Now we can calculate the entropy change of each region using the formula: \(\Delta S = \frac{Q}{T}\) Entropy change of cold region (\(\Delta S_\mathrm{cold}\)): \(\frac{-1000\,\mathrm{J}}{253.15\,\mathrm{K}} = -3.95\,\mathrm{J/K}\) Entropy change of hot region (\(\Delta S_\mathrm{hot}\)): \(\frac{1250\,\mathrm{J}}{303.15\,\mathrm{K}} = 4.12\,\mathrm{J/K}\) Total entropy change (\(\Delta S_\mathrm{total}\)) = \(\Delta S_\mathrm{cold} + \Delta S_\mathrm{hot} = -3.95\,\mathrm{J/K} + 4.12\,\mathrm{J/K} = 0.17\,\mathrm{J/K}\)

Assess if the second law is satisfied

The second law of thermodynamics will be satisfied if the total entropy change (\(\Delta S_\mathrm{total}\)) is greater than or equal to zero. Since our \(\Delta S_\mathrm{total} = 0.17\,\mathrm{J/K} > 0\), the second law is satisfied. Now, let's check if the second law would still be satisfied if the COP was increased to 6. Using the same calculations as before, with COP = 6, we would obtain: \(W = \frac{1000\,\mathrm{J}}{6} = 166.67\,\mathrm{J}\) \(Q_\mathrm{hot} = 166.67\,\mathrm{J} + 1000\,\mathrm{J} = 1166.67\,\mathrm{J}\) \(\Delta S_\mathrm{cold} = \frac{-1000\,\mathrm{J}}{253.15\,\mathrm{K}} = -3.95\,\mathrm{J/K}\) \(\Delta S_\mathrm{hot} = \frac{1166.67\,\mathrm{J}}{303.15\,\mathrm{K}} = 3.85\,\mathrm{J/K}\) \(\Delta S_\mathrm{total} = -3.95\,\mathrm{J/K} + 3.85\,\mathrm{J/K} = -0.10\,\mathrm{J/K}\) With a COP of 6, \(\Delta S_\mathrm{total} < 0\), therefore, the second law of thermodynamics would not be satisfied.

Key concepts

Coefficient of Performance

The Coefficient of Performance (COP) is a crucial concept in thermodynamics, particularly when discussing refrigerators and heat pumps. It is a measure of efficiency, representing the ratio of heat removal to work input in a cooling system.

In mathematical terms, the COP for a refrigerator is given by the equation:
\[\begin{equation}COP = \frac{Q_{cold}}{W}\end{equation}\]
where \(Q_{cold}\) is the heat extracted from the cold region (in joules) and \(W\) is the work done by the refrigerator (also in joules). In simpler terms, it tells us how much heat is moved per unit of work. A higher COP indicates a more efficient refrigerator, requiring less work to transfer a certain amount of heat.

To further clarify, in our example where the refrigerator has a COP of 4, it means that for every joule of work the refrigerator uses, it moves 4 joules of heat from the cold region. Efficiency is a primal concern in thermodynamics, and understanding the COP helps gauge this efficiency in real-world applications.

Second Law of Thermodynamics

The Second Law of Thermodynamics is one of the most important principles in the field and dictates that the total entropy of an isolated system can never decrease over time. Entropy can be considered a measure of disorder or randomness in a system.

In practical terms, this law has profound implications for energy transformations, such as in refrigerators. When a refrigerator transfers heat from a colder to a hotter area, there must be accompanying increases in the entropy of the surroundings to ensure that the total entropy does not decrease, thus complying with the second law.

When assessing whether a process satisfies the Second Law, one must calculate the total entropy change (\[\begin{equation}\text{\Delta} S_{total}\end{equation}\]). If this value is positive or zero, the law is upheld. In our exercise, we calculated that with an initial COP of 4, the increase in entropy was sufficient, satisfying the Second Law. However, changing the COP to 6 resulted in the total entropy decreasing, which violates this law.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or medium to another, and it happens when there is a temperature difference. It's fundamental to refrigeration, where heat is transferred from a cooler to a warmer area. There are three main modes of heat transfer: conduction (through materials), convection (through liquids or gases), and radiation (through electromagnetic waves).

Understanding how to calculate heat transfer is essential for various applications. In the given exercise, we saw that heat transfer calculations are vital for determining the heat removed from a cold region and the work required for a refrigerator to function. By considering the heat transferred and the temperatures involved, we can also calculate changes in entropy, which serves as a performance and compliance check with thermodynamic laws.

In our specific example, we calculated heat transfer from the cold region (\[\begin{equation}Q_{cold}\end{equation}\]) and used the principle of energy conservation to find the heat transferred to the hot region (\[\begin{equation}Q_{hot}\end{equation}\]). This exercise also demonstrates how heat transfer is interrelated with work, energy conservation, and entropy in thermodynamics.

Temperature Conversion

Temperature conversion between different scales is a fundamental skill required to solve thermodynamic problems, as temperatures are often measured in Celsius (\[\begin{equation}^{\text{\circ}C}\end{equation}\]) but need to be converted to Kelvin (K) for calculations. The Kelvin scale is the thermodynamic temperature scale, and it starts at absolute zero, the point where particles have minimum thermal motion.

The conversion is straightforward:

\[\begin{equation}T(K) = T(^{\text{\circ}}C) + 273.15\end{equation}\]
As seen in the exercise solution, understanding how to properly convert temperatures allows for accurate thermodynamic calculations. In practical problems, using the Kelvin scale is essential because it ensures that the values used in formulas, such as those for entropy change, are thermodynamically valid, since temperatures in Kelvin directly relate to the absolute thermal energy of the system.