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Ch. 20 - Second Law of Thermodynamics
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 20 - Second Law of ThermodynamicsProblem 88
Chapter 20, Problem 88

A dehumidifier removes water vapor from air and has been referred to as a “refrigerator with an open door.” The humid air is pulled in by a fan and passes over a cold coil, whose temperature is less than the dew point, and some of the air’s water condenses. After this water is extracted, the air is warmed back to its original temperature and sent into the room. In a well-designed dehumidifier, the heat that is removed by the cooling coil mostly comes from the condensation of water vapor to liquid, and this heat is used to re-warm the air. Estimate how much water is removed in 1.0 h by an ideal dehumidifier, if the temperature of the room is 25°C, the water condenses at 8°C, and the dehumidifier does work at the rate of 550 W of electrical power.

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Step 1: Understand the problem. The dehumidifier removes water vapor by cooling the air below its dew point, causing condensation. The heat released during condensation is used to re-warm the air. The goal is to estimate the amount of water removed in 1 hour, given the power input of 550 W, the room temperature of 25°C, and the condensation temperature of 8°C.
Step 2: Calculate the total energy input to the dehumidifier in 1 hour. Use the formula for energy: \( E = P \cdot t \), where \( P \) is the power (550 W) and \( t \) is the time (1 hour = 3600 seconds). This gives the total energy available for the dehumidification process.
Step 3: Relate the energy to the mass of water condensed. The energy required to condense water is given by \( Q = m \cdot L_v \), where \( m \) is the mass of water condensed and \( L_v \) is the latent heat of vaporization of water. At 8°C, \( L_v \) is approximately 2.45 × 10^6 J/kg. Rearrange the formula to solve for \( m \): \( m = \frac{Q}{L_v} \).
Step 4: Assume that the dehumidifier operates ideally, meaning all the electrical energy input is used for condensation. Substitute the total energy \( E \) from Step 2 into \( Q \) in the formula \( m = \frac{Q}{L_v} \). This will give the mass of water condensed in kilograms.
Step 5: Convert the mass of water from kilograms to liters if needed, using the fact that 1 kg of water is approximately equal to 1 liter. This will provide the final estimate of the volume of water removed by the dehumidifier in 1 hour.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Dew Point

The dew point is the temperature at which air becomes saturated with moisture and water vapor begins to condense into liquid. In the context of a dehumidifier, the air is cooled below this temperature, causing water vapor to condense on the cold coils. Understanding the dew point is crucial for determining how much moisture can be removed from the air.
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Heat Transfer

Heat transfer refers to the movement of thermal energy from one object or substance to another. In a dehumidifier, heat is removed from the humid air as it passes over the cold coils, leading to condensation. This process involves both sensible heat (temperature change) and latent heat (energy absorbed or released during phase changes), which are essential for calculating the efficiency of moisture removal.
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Power and Energy Calculations

Power is the rate at which energy is used or transferred, measured in watts (W). In this scenario, the dehumidifier operates at 550 W, indicating how much electrical energy it consumes per second. To estimate the amount of water removed, one must convert this power into energy over the time period (1 hour) and relate it to the latent heat of vaporization of water, which quantifies the energy required to condense water vapor.
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