Question

Using a timer (or watch) and a thermometer, conduct the following experiment to determine the rate of heat gain of your refrigerator. First make sure that the door of the refrigerator is not opened for at least a few hours so that steady operating conditions are established. Start the timer when the refrigerator stops running and measure the time \(\Delta t_{1}\) it stays off before it kicks in. Then, measure the time \(\Delta t_{2}\) it stays on. Noting that the heat removed during \(\Delta t_{2}\) is equal to the heat gain of the refrigerator during \(\Delta t_{1}+\Delta t_{2}\) and using the power consumed by the refrigerator when it is running, determine the average rate of heat gain for your refrigerator, in \(\mathrm{W}\). Take the COP (coefficient of performance) of your refrigerator to be 1.3 if it is not available.

Short answer

Question: Determine the average rate of heat gain for a refrigerator using the given experimental data and COP value of 1.3. Solution: Given data: Δt1 (off period) = ____ Δt2 (on period) = ____ Power consumed (P_consumed) = ____ 1. Measure the time intervals Δt1 and Δt2. 2. Calculate the total time of one cycle: t_total = Δt1 + Δt2 3. Find the power consumed (P_consumed) during the on time from the refrigerator's user manual or a similar model. 4. Calculate the heat removed (Q_removed) during the on time using COP = 1.3: Q_removed = COP × P_consumed 5. Calculate the average rate of heat gain (Q_gain) in Watts: Q_gain(W) = Q_removed / t_total The average rate of heat gain for the refrigerator is Q_gain(W) Watts.

Step-by-step solution

Measure the time intervals

Perform the experiment by measuring the time intervals \(\Delta t_{1}\) and \(\Delta t_{2}\). Ensure that the refrigerator's door remains closed for a few hours for the steady operating conditions to establish.

Calculate the total time of one cycle

Add \(\Delta t_{1}\) and \(\Delta t_{2}\) to determine the total time of one complete cycle: $$t_{total} = \Delta t_{1} + \Delta t_{2}$$

Determine the power consumed during the on time

Refer to the refrigerator's user manual or backside of the refrigerator to find the power consumed when the refrigerator is running. Let this be \(P_{consumed}\) (in Watts). If this information is not available, one can potentially estimate it by referring to similar models from the same manufacturer or by online searches.

Calculate the heat removed during the on time

Use the given or estimated COP value of 1.3 to calculate the heat removed (\(Q_{removed}\)) by the refrigerator during its on period: $$Q_{removed} = COP \times P_{consumed} = 1.3 \times P_{consumed}$$

Calculate the rate of heat gain

As the heat removed during on period is equal to the heat gain during the full cycle, we can now determine the average rate of heat gain, \(Q_{gain}(W)\): $$Q_{gain}(W) = \frac{Q_{removed}}{t_{total}}$$ The calculated value of \(Q_{gain}(W)\) is the average rate of heat gain for the refrigerator, in Watts.

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationship between heat and other forms of energy. It encompasses principles that govern the conversion of energy from one form to another and the direction that heat flows. In the context of a refrigerator, thermodynamics is at play when the appliance removes heat from its interior and expels it to the environment. This process is driven by the laws of thermodynamics that dictate that heat moves from a warmer area to a cooler one until equilibrium is reached.

Within this field, the first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transferred or transformed. In our case, the refrigerator uses electrical energy to facilitate the transfer of thermal energy out of its cooled space. Understanding these principles is crucial for calculating heat gain, as it ensures the conservation of energy principle is respected.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a term used in thermodynamics to describe the efficiency of a heat pump or a refrigeration system. It's the ratio of useful heating or cooling provided to work required. In simpler terms, COP measures how effective a refrigerator is at using electrical energy to move heat out of its interior. For the given exercise, a COP of 1.3 suggests that for every unit of energy consumed by the refrigerator, 1.3 units of heat energy are removed from its interior.

Higher COP values indicate greater efficiency, as more heat is moved per unit of work done. This parameter is critical when calculating heat gain since it directly influences the amount of heat energy that can be removed for a given amount of electrical energy consumed.

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics concerning the movement of thermal energy from one place to another. In the case of refrigerators, heat is transferred from inside the unit, which is to be kept cold, to the outside environment. This transfer occurs via different mechanisms: conduction, where heat is transferred through materials; convection, where it is carried away by moving fluids (such as the refrigerant); and radiation, which involves heat moving as electromagnetic waves.

In a steady-state, the rate of heat transfer into the refrigerator (heat gain) should equal the rate of heat transfer out of it (heat removal). Therefore, understanding the principles of heat transfer is vital for accurately calculating the average rate of heat gain for the refrigerator, which is the purpose of the given exercise.

Steady Operating Conditions

Steady operating conditions refer to a state where the variables involved in the operation of a device remain constant over time. For a refrigerator, this pertains to maintaining a consistent internal temperature and having regular on and off cycles for its compressor, which is essential for homogeneous cooling. These conditions are necessary for our experiment so that we can measure an average rate of heat gain that accurately reflects the normal operation of the fridge.

It is important to allow the refrigerator to reach these conditions before taking measurements to avoid data skewed by transient adjustments. This means that the appliance should not be opened or have its settings changed for several hours prior to conducting the exercise, ensuring that the heat gain and removal are in equilibrium.

Energy Efficiency

Energy efficiency in appliances like refrigerators refers to the proportion of input energy that is effectively used for its intended purpose as opposed to being wasted. In the context of refrigeration, it's the efficiency with which the electrical energy is used to remove heat from the interior. An energy-efficient refrigerator uses less energy to accomplish the same amount of cooling, resulting in reduced electricity bills and lower environmental impact.

A critical aspect of energy efficiency is identifying and minimizing heat gains, which require the refrigerator to work harder and consume more energy. Understanding the rate of heat gain is therefore not only central to performing our experiment but also to improving the overall energy efficiency of the refrigerator. Evaluating this correctly can inform better design, usage habits, and ultimately lead to substantial cost and energy savings.