Question

A heat pump is used to heat a building. The outside temperature is \(-5.0^{\circ} \mathrm{C}\) and the temperature inside the building is to be maintained at \(22^{\circ} \mathrm{C}\). The coefficient of performance is \(3.8\), and the pump delivers \(7.6 \mathrm{MJ}\) of heat to the building each hour. At what rate must work be done to run the pump?

Short answer

The rate at which work must be done to run the pump is 555.5 W.

Step-by-step solution

Understanding the problem

A heat pump is a device that extracts heat from one place and transfers it to another. The rate at which work must be done to run the pump can be calculated using the definition of Coefficient of performance (COP) which is the ratio of heat extracted or delivered (Qh) to the work done (W) by the pump, that is, COP = Qh/W.

Conversion of units

The heat delivered is given as \(7.6 \mathrm{MJ/hour}\), we need to convert this to the SI unit of power (Watt or W), which is equivalent to Joules per second. Therefore, multiply the given value by \(10^6\) (to convert from mega to base unit) and divide by 3600, (to convert hours to seconds), giving \(2111.1 \mathrm{W}\).

Calculate the work

Now that we have Qh in the appropriate units and the value of the COP, we can use the COP equation to find W. Rearranging the equation to W = Qh / COP, and substituting the known values, we find W = \(2111.1 \mathrm{W} / 3.8 = 555.5 \mathrm{W}\). This is the rate at which work must be done to run the pump.

Key concepts

Coefficient of Performance

When discussing the effectiveness of heat pumps, the Coefficient of Performance (COP) is a crucial concept that denotes the ratio of heating or cooling provided to the work required for operation. Think of it in simple terms as a measure of efficiency - the higher the COP, the more heating or cooling you get for each unit of work put in. It's analogous to getting more bang for your buck.
To solidify this, consider our example where the COP is 3.8. This means for every unit of energy used to run the pump, 3.8 units of heat energy are delivered to the building. When calculating the work needed for the heat pump, one simply rearranges the formula for COP. Here, rearranging gives us the work rate (W) as the heat output (Qh) divided by the COP. In relation to the problem we're trying to solve, knowing the COP allows us to determine how much work the pump must perform to maintain the desired temperature.

Thermodynamics

The field of thermodynamics is central to understanding how heat pumps work. At its core, it's the study of energy, heat, work, and the transformations between them. A heat pump operates on the principles of thermodynamics, essentially transferring heat from a cooler to a warmer place.

The Second Law of Thermodynamics

A relevant law here is the second law of thermodynamics, which states that heat flows naturally from a high-temperature area to a lower one; yet, a heat pump defies this natural flow by consuming work to achieve the reverse. In our textbook exercise, thermodynamic principles help us determine the work required for this process - to 'pump' heat against the natural gradient from the outside’s chilly temperature into the warmth of a building.

Energy Conversion

The process of energy conversion is at the heart of a heat pump's functionality. Essentially, energy conversion is about changing energy from one form to another. In the case of our heat pump, electrical (or sometimes mechanical) work energy is converted into thermal energy to heat the building.
One can picture the heat pump as an energy transformation wizard, taking less useful energy (work) and turning it into valuable energy (heat) to keep us cozy. However, no energy conversion process is perfect; some energy will always be lost, typically as waste heat to the environment. The COP gives us an indication of just how good a heat pump is at minimizing these losses and maximizing the conversion of work into useful heat.

SI Units

When it comes to science and engineering, consistency is key, hence the use of the International System of Units (SI units). They provide a standard way to measure and report physical quantities like energy, power, and temperature.
Our heat pump exercise illustrates why SI units are so important - calculations require consistent units for accuracy. The original energy output was provided in megajoules per hour, but to calculate the work rate in terms of power, we needed to convert this to watts (joules per second). This ensures that when we apply the formula to find the required work rate for the pump, we avoid misinterpretation and errors that could arise from unit discrepancies. It's akin to making sure all ingredients in a recipe are measured using the same type of cups for the recipe to work flawlessly.