Refrigerators: Videos & Practice Problems

Refrigerators

In this scenario, we are analyzing a Carnot engine, which is known for operating at maximum efficiency. The efficiency of this particular engine is given as 70%. The engine is cooled by water at a temperature of 10 degrees Celsius, which serves as the cold reservoir. To convert this temperature to Kelvin, we add 273, resulting in a cold reservoir temperature of 283 K.

To determine the temperature of the hot reservoir, we utilize the Carnot efficiency equation, expressed as:

\[E_{\text{Carnot}} = 1 - \frac{T_c}{T_h}\]

Here, \(E_{\text{Carnot}}\) represents the efficiency, \(T_c\) is the temperature of the cold reservoir, and \(T_h\) is the temperature of the hot reservoir. Given that the efficiency is 70%, we convert this to decimal form, yielding \(0.7\). Substituting the known values into the equation gives:

\[0.7 = 1 - \frac{283}{T_h}\]

Rearranging the equation allows us to isolate \(T_h\). First, we rewrite it as:

\[\frac{283}{T_h} = 0.3\]

Next, we cross-multiply to solve for \(T_h\):

\[T_h = \frac{283}{0.3}\]

Calculating this results in:

\[T_h \approx 943.3 \, \text{K}\]

This value confirms that the temperature of the hot reservoir is indeed higher than that of the cold reservoir, as expected in a heat engine. To verify the accuracy of our calculations, substituting \(T_c = 283 \, \text{K}\) and \(T_h = 943.3 \, \text{K}\) back into the efficiency equation should yield an efficiency of 0.7, confirming our solution is correct.

Thus, the temperature of the hot reservoir is approximately 943.3 K, which aligns with the principles governing Carnot engines and their efficiencies.

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.