Question

A certain refrigerator is rated as being \(32.0 \%\) as efficient as a Carnot refrigerator. To remove \(100 .\) J of heat from the interior at \(0.00^{\circ} \mathrm{C}\) and eject it to the outside at \(22.0^{\circ} \mathrm{C}\), how much work must the refrigerator motor do?

Short answer

Answer: The refrigerator motor must do about 349.79 Joules of work.

Step-by-step solution

Finding the efficiency of the Carnot refrigerator

We know that the efficiency of a Carnot refrigerator is given by the formula: \(eff_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}}\) To use this equation, we will need to convert the given temperatures from Celsius to Kelvin.

Converting temperatures to Kelvin

To convert the temperature in Celsius to Kelvin, we simply add 273.15. So we have, \(T_{cold} = 0^{\circ} \mathrm{C} + 273.15 = 273.15\,K\) \(T_{hot} = 22.0^{\circ} \mathrm{C} + 273.15 = 295.15\,K\)

Plugging temperatures into the Carnot efficiency equation

Now that we have the temperatures in Kelvin, we can plug them into the efficiency equation: \(eff_{Carnot} = 1- \frac{273.15}{295.15} = 1 - 0.9253 = 0.0747\)

Finding the efficiency of the given refrigerator

We are given that the refrigerator is 32.0% as efficient as a Carnot refrigerator. So we have, \(eff_{given} = 0.32 \times eff_{Carnot}\) Plug the value of \(eff_{Carnot}\) to find \(eff_{given}\): \(eff_{given} = 0.32 \times 0.0747 = 0.0239\)

Using heat and efficiency to find the work

The relationship between the heat removed from the interior (Q_removed), the work done to remove the heat (W), and the efficiency is given by: \(eff_{given} = \frac{Q_{removed}}{Q_{removed} + W}\) Rearranging the equation to find the work: \(W = \frac{Q_{removed} - Q_{removed} \times eff_{given}}{eff_{given}}\) Now plug in the given values for the heat removed and the efficiency: \(W = \frac{100\,\text{J} - 100\,\text{J} \times 0.0239}{0.0239} = 349.79\,\text{J}\)

Answer

The refrigerator motor must do about 349.79 Joules of work to remove 100 Joules of heat from the interior at \(0.00^{\circ}\mathrm{C}\) and eject it to the outside at \(22.0^{\circ}\mathrm{C}\).

Key concepts

Thermodynamic Temperature Conversion

To accurately analyze and predict the behavior of a thermodynamic system such as a refrigerator, temperatures must be measured on an absolute scale. The Kelvin scale is the SI unit for thermodynamic temperature, important because it starts at absolute zero, unlike Celsius or Fahrenheit scales.

For practical calculations, often you will find temperatures given in Celsius which necessitates a conversion to Kelvin. The conversion is straightforward: add 273.15 to the Celsius temperature to get the Kelvin equivalent. This is crucial because the efficiency equations for refrigerators, such as the one for a Carnot refrigerator, require temperatures to be in Kelvin. If we mistakenly use a relative temperature scale like Celsius or Fahrenheit, our calculations for efficiency will be erroneous.

Carnot Cycle

The Carnot cycle is a theoretical model that defines the maximum possible efficiency of a heat engine or refrigerator operating between two temperatures. It's composed of four reversible processes: two isothermal (constant temperature) and two adiabatic (no heat exchange).

In a Carnot refrigerator, heat is absorbed isothermally from a cold reservoir and expelled to a warm reservoir. The efficiency of such a system is determined by the temperatures of the hot and cold reservoirs. The closer these temperatures, the less work is needed for the transfer, and thus the system is more efficient. This high efficiency, however, is only theoretically achievable because real refrigerators are not perfectly reversible and thus cannot achieve Carnot cycle efficiency.

Heat Transfer in Refrigeration

Understanding Heat Flow

Refrigerators operate by manipulating the flow of heat. In essence, they transfer heat from a cooler interior space to the warmer environment outside. This is contrary to the natural direction of heat flow, which requires work.

Heat transfer in refrigeration is grounded in the second law of thermodynamics, where the refrigeration cycle works to reverse the spontaneous direction of heat flow through the application of energy, usually in the form of mechanical work by a motor or compressor. This forced transfer of heat is what keeps the fridge's interior cool.

• Evaporation: Absorbing heat from the freezer's interior.
• Compression: Compressor raises the refrigerant’s pressure and temperature.
• Condensation: Refrigerant releases heat to the outside.
• Expansion: Pressure drops, refrigerant cools down, and cycle repeats.

Refrigerator Work Calculation

The work required by a refrigerator to transfer heat is directly related to its efficiency. A refrigerator's efficiency is defined as the ratio of heat removed from the inside to the total energy consumed to move that heat.

As seen in the Carnot refrigerator model, efficiency depends on the temperature difference between the heat source (inside the refrigerator) and the heat sink (outside environment). From the efficiency, we can calculate the work needed using the formula given in the solution steps above. Remember, the less efficient a refrigerator is, the more work it requires to remove a certain amount of heat. By using the proportionality of the given refrigerator's efficiency to the Carnot efficiency, we find the actual work needed to maintain the cold environment inside the refrigerator.