Question

In regions with dry climates, evaporative coolers are used to cool air. A typical electric air conditioner is rated at \(1.00 \times\) \(10^{4} \mathrm{Btu} / \mathrm{h}(1 \mathrm{Btu}, \text { or British thermal unit }=\) amount of energy to raise the temperature of 1 lb water by \(1^{\circ} \mathrm{F}\) ). What quantity of water must be evaporated each hour to dissipate as much heat as a typical electric air conditioner?

Short answer

To dissipate the same amount of heat as a typical electric air conditioner, approximately 4.67 kilograms of water must be evaporated each hour.

Step-by-step solution

Understand the given information

We are given the power rating of a typical electric air conditioner as \(1.00 \times 10^{4} \mathrm{Btu/h}\). We need to calculate the amount of water evaporated each hour to dissipate the same amount of heat.

Convert Btu to calories

We need to convert the given power rating from Btu to calories. To do this, use the conversion factor: 1 Btu = 252 calories. So, \[1.00 \times 10^{4} \mathrm{Btu/h} \times \frac{252 \mathrm{cal}}{1 \mathrm{Btu}} = 2.52 \times 10^{6} \mathrm{cal/h}\]

Compute the heat needed to evaporate water

We will now compute how much heat is needed to evaporate a certain quantity of water. We'll use the heat of vaporization of water, which is defined as the amount of heat required to turn 1 gram of water into vapor. The heat of vaporization for water is 540 calories per gram at 100°C. Given that \(Q = mc\Delta T\), where Q is the heat needed to evaporate water, m is the mass of the water in grams, and c is the specific heat of vaporization (540 cal/g), we can write the equation as: \[Q = m \times 540\]

Calculate the mass of water needed to be evaporated

Now we need to find the mass of water (m) that needs to be evaporated to dissipate the same amount of heat as the air conditioner. Using the results from step 2 and step 3, we can write the equation as follows: \[2.52 \times 10^{6} \mathrm{cal/h} = m \times 540\] To find the mass of water (m), divide both sides by 540: \[m = \frac{2.52 \times 10^{6} \mathrm{cal/h}}{540}\]

Solve for m

Now, solve for m: \[m = \frac{2.52 \times 10^{6} \mathrm{cal/h}}{540} = 4666.67 \mathrm{g/h}\] So, to dissipate the same amount of heat as a typical electric air conditioner, 4666.67 grams or approximately 4.67 kilograms of water must be evaporated each hour.

Key concepts

Heat of Vaporization

The heat of vaporization is the energy required to transform a given amount of a substance from a liquid into a gas at a constant temperature and pressure. It's a crucial concept in evaporative cooling, the process behind things like sweat helping to cool our bodies or evaporative air coolers making hot air bearable in arid climates.

Understanding heat of vaporization is essential for solving problems where energy transfer due to phase change is involved. For instance, in the textbook exercise, using the heat of vaporization of water allows us to calculate the amount of energy necessary to vaporize water which directly relates to how much cooling can be achieved. The higher the heat of vaporization, the more energy it takes to vaporize a substance and consequently, the more heat that substance can remove from its surroundings as it vaporizes.

Practical Implication in Evaporative Cooling

When water evaporates, it absorbs a significant amount of energy from the environment, which results in a cooling effect. This process is used in so-called 'swamp coolers' or evaporative coolers. The exact quantity of water that needs to evaporate to match the cooling power of a traditional air conditioner can be deduced by applying the heat of vaporization.

Energy Conversion

The process of changing one form of energy into another is known as energy conversion. In thermodynamics, this is a core principle, often observed when heat energy is transformed into mechanical energy or vice versa.

In the context of our textbook problem, we're looking at the conversion of thermal energy into the latent energy of vaporization. The energy used by a conventional air conditioner is initially electrical, which then gets converted into thermal energy as it removes heat from the air, cooling it down. By contrast, an evaporative cooler doesn't convert electrical energy into cool air as directly. Instead, it uses a portion of the heat in the air to evaporate water, thus converting the thermal energy of the air into the latent heat (energy) required for the vaporization process.

Understanding the Cooling Mechanism

Understanding the mechanics of energy conversion helps in grasping how different air cooling methods work, and why evaporative coolers are preferable in certain climates. These coolers are more energy-efficient under the right conditions, as the main energy input is the heat from the air itself rather than electricity.

Thermodynamics

Thermodynamics is the branch of physics concerned with heat and temperature and their relation to energy and work. It encompasses all aspects of energy transformation, including how energy is used and exchanged, how it impacts matter, and how it can lead to work being done, which is essential in understanding heat engines, refrigerators, and even living cells.

In the exercise we've discussed, thermodynamics principles help explain how the energy from the ambient heat is used to evaporate water. By following thermodynamic laws, specifically the first law of thermodynamics (conservation of energy), we know that the energy must come from somewhere—in this case, the surrounding air. Evaporative cooling is an example of thermodynamics in action: as the water evaporates, it absorbs heat from the surrounding air, thereby reducing the air temperature.

The Role of Thermodynamics in Cooling Systems

Any efficient cooling system design, including evaporative coolers, must adhere to the laws of thermodynamics. These systems manage the flow and transformation of energy, ensuring our environments remain comfortably cool. Educating students on thermodynamics can thus not only help them solve textbook problems but also understand and apply this knowledge to real-world systems.