Question

A refrigerator removes heat from a refrigerated space at \(0^{\circ} \mathrm{C}\) at a rate of \(2.2 \mathrm{kJ} / \mathrm{s}\) and rejects it to an environment at \(20^{\circ} \mathrm{C}\). The minimum required power input is \((a) 89 \mathrm{W}\) \((b) 150 \mathrm{W}\) \((c) 161 \mathrm{W}\) \((d) 557 \mathrm{W}\) \((e) 2200 \mathrm{W}\)

Short answer

Answer: The minimum required power input for the refrigerator is 161 W.

Step-by-step solution

Find the Coefficient of Performance (COP) for the refrigerator

The Coefficient of Performance (COP) for a refrigerator is defined as the ratio of the heat removed from the refrigerated space (Q_c) to the work input (W). In mathematical terms, it can be represented as: COP = \(\frac{Q_c}{W}\) However, in the case of a refrigerator, the COP can also be expressed in terms of the temperatures of the refrigerated space (T_c) and the environment (T_h): COP = \(\frac{T_c}{T_h - T_c}\) Remember that we need to convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature.

Convert the temperatures to Kelvin

To convert the temperatures from Celsius to Kelvin, simply add 273.15 to each temperature: \(T_c = 0^{\circ}C + 273.15 = 273.15 K\) \(T_h = 20^{\circ}C + 273.15 = 293.15 K\)

Calculate the COP using the Kelvin temperatures

Now that we have the temperatures in Kelvin, we can calculate the Coefficient of Performance (COP) using the previously mentioned formula: COP = \(\frac{T_c}{T_h - T_c}\) COP = \(\frac{273.15}{293.15 - 273.15}\) COP = \(\frac{273.15}{20}\) COP ≈ 13.66

Calculate the minimum required work input

Now that we have the COP, we can find the minimum required work input (W) by rearranging the original COP formula: \(W = \frac{Q_c}{COP}\) The heat removed from the refrigerated space at a rate of 2.2 kJ/s can be converted to Watts (1 kJ/s = 1000 W): \(Q_c = 2.2 \mathrm{kJ/s} \times 1000 = 2200 \mathrm{W}\) Substitute the values for Q_c and COP into the equation: \(W = \frac{2200}{13.66}\) \(W \approx 161 \mathrm{W}\) Therefore, the minimum required power input for the refrigerator is \((c) 161 \mathrm{W}\).

Key concepts

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a key concept when assessing the efficiency of a refrigerator or any other heat pump system. In simple terms, it measures how effectively a refrigerator transfers heat versus the energy it consumes. Higher COP values indicate a more efficient refrigerator, as it means more cooling is achieved per unit of work input.

COP is calculated using the formula \[\begin{equation}COP = \frac{Q_c}{W}\end{equation}\]

where \begin{itemize} \item \(Q_c\) represents the amount of heat removed from the refrigerated space (in watts or kilojoules per second), and \item \(W\) is the work input (also in watts or kilojoules per second).\begin{itemize}

To improve understanding of this concept, one might consider this analogy: A vehicle's fuel efficiency can be thought of as its 'COP.' A car that can travel further on less fuel has a higher 'COP' than one that needs more fuel for the same distance. Similarly, a refrigerator with a higher COP needs less electrical energy to transfer a given amount of heat from inside to outside the refrigerator.

Kelvin Temperature Scale

The Kelvin temperature scale is vital in the field of thermodynamics and for problems involving thermal efficiency, like refrigerator performance. Unlike the Celsius and Fahrenheit scales, Kelvin is an absolute temperature scale which starts at absolute zero - the point at which particles have minimum thermal motion.

To convert Celsius to Kelvin, one adds 273.15 to the Celsius temperature. So, room temperature (approximately 20°C) on the Kelvin scale is \[T_K = 20 \degree C + 273.15 = 293.15 K\].

The Kelvin scale is important because it allows for absolute measurements of thermal energy, which is crucial when calculating heat transfer and work in thermodynamic systems. Moreover, the COP formula for refrigerators is applied using the Kelvin scale to eliminate negative numbers, thereby easing calculations and interpretations of the system's behavior.

Heat Transfer Rate

Heat transfer rate is a measure of the thermal energy transferred per unit of time. In the context of refrigerators, it signifies the rate at which heat is removed from the interior to keep it cool. This rate can be measured in Watts (W), where 1 Watt is equivalent to 1 Joule per second (1 W = 1 J/s).

In the given exercise, the refrigerated space has a heat removal rate of \[2.2 \text{ kJ/s}\],

which means that to maintain its internal temperature, the refrigerator effectively extracts 2.2 kilojoules of heat every second. A higher heat transfer rate often requires a refrigerator to perform more work, that is, use more electrical power to maintain the low temperature.

Work Input Calculation

Work input calculation for a refrigeration system is directly linked to its efficiency and performance. It represents the amount of energy the refrigerator needs to draw from the electrical outlet in order to remove heat from the inside. This calculation is crucial when designing and evaluating refrigeration systems as it determines the operational cost.

From the exercise, we can see that the minimum required work input can be found by rearranging the COP formula to \[W = \frac{Q_c}{COP}\].

By substituting the known values for the heat transfer rate and the COP, we can calculate the minimum electrical power the refrigerator needs to operate effectively. This value is the minimal because any real-world system would also include losses such as friction and non-ideal insulation, which means that the actual power input would likely be slightly higher than what is calculated.