Refrigerators: Videos & Practice Problems

Refrigerators and heat pumps operate on similar principles, utilizing work to transfer heat from colder to hotter areas, in accordance with the second law of thermodynamics. A refrigerator extracts heat from its interior (cold reservoir) and expels it outside (hot reservoir), while a heat pump reverses this process, drawing heat from the cold outside air to warm the interior. The coefficient of performance (COP) for refrigerators is defined as K = Q_CW, and for heat pumps as K = Q_HW.

Refrigerators

In thermodynamics, the principles governing heat engines and refrigerators share significant similarities, particularly in their operation and the laws of thermodynamics that apply to them. A fundamental concept to understand is that heat naturally flows from hotter to colder bodies, a principle rooted in the second law of thermodynamics. This law, often referred to as the Clausius statement, asserts that heat cannot spontaneously flow from a colder object to a hotter one without the input of work.

A refrigerator operates on this principle by utilizing work to transfer heat from a cold reservoir (such as the interior of the refrigerator) to a hot reservoir (the surrounding environment). This process is essentially the reverse of a heat engine, which converts heat from a hot reservoir into work while transferring some heat to a cold reservoir. In a refrigerator, work is supplied, typically in the form of electrical energy, to facilitate the movement of heat against its natural flow.

The relationship between the heat absorbed from the cold reservoir (\(Q_C\)) and the heat expelled to the hot reservoir (\(Q_H\)) can be expressed through the equation for work (\(W\)) done by the refrigerator:

\[ W = Q_H - Q_C \]

Both refrigerators and heat engines are cyclic processes, meaning that their internal energy change over one complete cycle is zero. This leads to the understanding that the work done is simply the difference between the heat transferred to the hot reservoir and the heat extracted from the cold reservoir.

To evaluate the performance of a refrigerator, we use the coefficient of performance (\(K\)), which measures how effectively the refrigerator transfers heat from the cold reservoir. The coefficient of performance is defined as:

\[ K = \frac{Q_C}{W} \]

Alternatively, it can also be expressed as:

\[ K = \frac{Q_C}{Q_H - Q_C} \]

In practical terms, if a refrigerator extracts 600 kilojoules of heat from its interior and releases 720 kilojoules to the surrounding environment, the work required to operate the refrigerator can be calculated as follows:

\[ W = 720 \, \text{kJ} - 600 \, \text{kJ} = 120 \, \text{kJ} \]

Subsequently, the coefficient of performance can be determined by substituting the values into the equation:

\[ K = \frac{600 \, \text{kJ}}{120 \, \text{kJ}} = 5 \]

This coefficient indicates that the refrigerator is quite efficient, as typical values range from 3 to 10. Understanding these concepts is crucial for analyzing the efficiency and functionality of refrigeration systems in various applications.

A refrigerator has a coefficient of performance of 2.4. Each cycle, it takes in 3×10⁴ J of heat from the cold reservoir. How much is expelled to the hot reservoir?

• A. 2.5×10⁴ J
• B. 4.25×10⁴ J
• C. 2.1×10⁴ J
• D. 1.25×10⁴ J

Answer: (B)

How long to freeze water?

In this example, we explore the process of freezing water using a household mini refrigerator, focusing on the relationship between the coefficient of performance (COP), power input, and the time required to freeze the water. The coefficient of performance, denoted as \( k \), is defined by the equation:

\( k = \frac{Q_C}{W} \)

Here, \( Q_C \) represents the heat extracted from the cold reservoir, and \( W \) is the work input, which can be expressed in terms of power and time. Since power (\( P \)) is related to work by the equation:

\( P = \frac{W}{\Delta t} \)

we can rearrange this to find work as:

\( W = P \cdot \Delta t \)

Substituting this into the COP equation gives us:

\( k = \frac{Q_C}{P \cdot \Delta t} \)

From this, we can derive the time required to freeze the water:

\( \Delta t = \frac{Q_C}{P \cdot k} \)

To find \( Q_C \), we need to consider both the temperature change and the phase change of the water. The total heat extracted can be calculated using the formula:

\( Q_C = m \cdot c \cdot \Delta T + m \cdot L \)

where:

• \( m \) is the mass of the water (0.5 kg in this case),
• \( c \) is the specific heat capacity of water (approximately 4186 J/kg·°C),
• \( \Delta T \) is the change in temperature (from 20°C to 0°C), and
• \( L \) is the latent heat of fusion for water (approximately \( 3.34 \times 10^5 \) J/kg).

Calculating \( Q_C \) involves two parts: first, cooling the water from 20°C to 0°C, and then freezing it. The temperature change is:

\( \Delta T = 0 - 20 = -20 \, \text{°C} \)

Thus, the heat extracted for cooling is:

\( Q_{cooling} = m \cdot c \cdot \Delta T = 0.5 \cdot 4186 \cdot (-20) \)

Next, for the phase change, we have:

\( Q_{freezing} = m \cdot L = 0.5 \cdot 3.34 \times 10^5 \)

Combining these gives:

\( Q_C = Q_{cooling} + Q_{freezing} \)

After calculating, we find that the total heat extracted is approximately \( 2.1 \times 10^5 \) joules. Now, substituting this value into the time equation:

\( \Delta t = \frac{2.1 \times 10^5}{85 \cdot 3} \)

Solving this results in a time of about 823 seconds, which is roughly 15 minutes. This aligns with our expectations, as freezing a small amount of water typically takes some time in a refrigerator.

Heat Pumps

A heat pump operates similarly to a refrigerator, functioning to transfer heat from a colder area to a hotter one. Understanding the distinctions between a heat engine, a refrigerator, and a heat pump is crucial. A heat engine utilizes the natural flow of heat from hot to cold, extracting energy from a hot reservoir to perform work while expelling waste heat to a cold reservoir. Common examples include car engines and household generators.

In contrast, a refrigerator reverses this process by extracting heat from a cold reservoir and expelling it to a hot reservoir, requiring an input of work, such as electricity. This is evident in household refrigerators and air conditioners, which cool indoor spaces by removing heat.

A heat pump, however, inverts the reservoirs compared to a refrigerator. It extracts heat from the cold outside air and transfers it indoors, making it ideal for heating in colder climates. Like refrigerators, heat pumps require work to operate, distinguishing them from heat engines.

The coefficient of performance (COP) is a key concept in understanding the efficiency of these devices. For a refrigerator, the COP is defined as:

$$COP_{refrigerator} = \frac{Q_C}{W}$$

where \(Q_C\) is the heat extracted from the cold reservoir and \(W\) is the work input. For a heat pump, the equation is modified to reflect the heat delivered to the hot reservoir:

$$COP_{heat\ pump} = \frac{Q_H}{W}$$

Here, \(Q_H\) represents the heat delivered into the home. The coefficient of performance for heat pumps is denoted as \(K_{HP}\).

To illustrate, consider a heat pump with a coefficient of performance of 3.6 and a power supply of \(7 \times 10^3\) watts. To calculate the heat energy delivered into a home over 4 hours, we can use the relationship between work and power:

$$W = P \times \Delta t$$

Substituting this into the heat pump equation gives:

$$K_{HP} = \frac{Q_H}{P \times \Delta t}$$

Rearranging for \(Q_H\) yields:

$$Q_H = K_{HP} \times P \times \Delta t$$

For 4 hours, converting to seconds gives \(4 \times 60 \times 60 = 14400\) seconds. Plugging in the values:

$$Q_H = 3.6 \times (7 \times 10^3) \times 14400$$

Calculating this results in:

$$Q_H \approx 3.63 \times 10^8 \text{ joules}$$

This value represents the total heat energy delivered into the home over the specified time period, demonstrating the effective heating capability of the heat pump.

Here’s what students ask on this topic:

What is the coefficient of performance (COP) of a refrigerator?

The coefficient of performance (COP) of a refrigerator, denoted as K, measures the efficiency of the refrigerator. It is defined as the ratio of the heat extracted from the cold reservoir (Q_C) to the work input (W). Mathematically, it is expressed as:

$K=\frac{Q_C}{W}$

This ratio indicates how effectively the refrigerator uses work to remove heat from its interior. A higher COP means the refrigerator is more efficient.

How does a refrigerator work according to the second law of thermodynamics?

A refrigerator works by transferring heat from a colder area (inside the fridge) to a hotter area (outside the fridge) using work, in accordance with the second law of thermodynamics. This law states that heat cannot spontaneously flow from a colder to a hotter body without an input of work. The refrigerator uses electrical energy to perform this work, thereby pumping heat from the cold interior to the warmer exterior.

What is the difference between a heat engine and a refrigerator?

A heat engine and a refrigerator operate on opposite principles. A heat engine converts heat from a hot reservoir into work, expelling some waste heat to a cold reservoir. In contrast, a refrigerator requires work to transfer heat from a cold reservoir to a hot reservoir. Essentially, a refrigerator is a heat engine running in reverse.

What is the role of work in the operation of a refrigerator?

Work is essential in the operation of a refrigerator as it enables the transfer of heat from the cold interior to the warmer exterior. According to the second law of thermodynamics, heat cannot flow from a colder to a hotter region without an input of work. The refrigerator uses electrical energy to perform this work, thereby maintaining a cold environment inside the fridge.

How is the coefficient of performance (COP) of a heat pump different from that of a refrigerator?

The coefficient of performance (COP) of a heat pump, denoted as K_HP, is similar to that of a refrigerator but focuses on the heat delivered to the hot reservoir. It is defined as the ratio of the heat delivered to the hot reservoir (Q_H) to the work input (W). Mathematically, it is expressed as:

$K_{\mathrm{HP}}=\frac{Q_H}{W}$

In contrast, the COP of a refrigerator is the ratio of the heat extracted from the cold reservoir (Q_C) to the work input (W).