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Chapter 1: Problem 9

On a hot summer day, a student turns his fan on when he leaves his room in the morning. When he returns in the evening, will his room be warmer or cooler than the neighboring rooms? Why? Assume all the doors and windows are kept closed.

Short Answer

Expert verified
Answer: The fan running all day in a closed room makes the room warmer compared to neighboring rooms with no fans due to the heat generated by the fan's inefficiencies. However, it can make the room feel cooler if someone is present because of the increased air circulation.

Step by step solution

01

Understand the heat exchange between the room and the surroundings

Since the doors and windows are closed, the exchange of heat with the surroundings will be minimal. In the closed room, the only significant heat exchange will happen due to the fan, which takes in electrical energy and then converts it into mechanical work and heat.
02

Analyze the work done by the fan

The fan's job is to circulate air within the room, which can make the room feel cooler because of increased air movement across the skin, allowing better heat transfer from the body to air. However, the fan itself doesn't lower the room's temperature.
03

Analyze the heat produced by the fan

Any electrical appliance, including a fan, generates heat as it works. This is because it converts electrical energy into the mechanical work of circulating air, and some of that energy unavoidably gets transformed into heat due to electrical and mechanical inefficiencies.
04

Compare the room temperature with neighboring rooms

The fan stays on throughout the day, continually generating and releasing heat into the room while making some air circulation. In contrast, assuming the neighboring rooms have no electrical appliances generating heat, they lack both additional heat from appliances and air circulation.
05

Determine whether the room will be warmer or cooler

Given that the fan generates heat throughout the day due to its inefficiencies, it will make the room warmer compared to the neighboring rooms, despite making it feel cooler if someone is in the room due to air circulation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Exchange
In thermodynamics, heat exchange refers to the transfer of thermal energy between different areas. In the context of a closed room, heat exchange occurs mainly due to energy transformations and inefficiencies in any operating appliances, like a fan. Since doors and windows are closed, heat exchange with the external environment is minimal. The fan uses electrical energy to perform work, and a portion of this energy is inevitably converted to heat. This additional heat contributes to increasing the room's temperature, as there's limited space for the excess thermal energy to dissipate.
Mechanical Work
Mechanical work involves the movement of objects or substances. In the case of a fan, the mechanical work translates to the circulation of air within the room. The fan generates movement by converting electrical energy into kinetic energy, driving its blades. This air movement can improve comfort by promoting better air circulation, which enhances the body's ability to release heat. However, it's important to note:
  • The actual room temperature is not lowered by the fan.
  • Mechanical work doesn't equate to reducing thermal energy; it simply redistributes air.
Thus, while creating a perception of a cooler environment through enhanced air movement, the fan introduces added heat into the room.
Electrical Energy Conversion
Electrical energy conversion is the process of transforming electrical power into another form of energy. In the case of the fan in the exercise, this conversion process includes turning electrical energy into the mechanical energy needed to rotate the fan blades.
  • During this conversion, not all energy is perfectly transferred.
  • Some is inevitably converted into heat due to imperfections in efficiency.
This latent heat output increases the room鈥檚 overall temperature, despite the air circulation effect. Understanding electrical energy conversion helps explain why appliances like fans don't actually cool the air; they redistribute it while warming the space slightly.
Room Temperature
Room temperature is the measure of thermal energy in the air of a room. When saying a room's temperature rises or falls, we're talking about changes in the average kinetic energy of the air molecules there. In this exercise, the fan continuously operates, converting energy and adding heat. Although the fan might make the room feel cooler due to moving air against the skin, the closed environment means the extra heat from energy conversion raises the actual temperature.
  • The room with the fan will ultimately be warmer than neighboring ones.
  • This contrast occurs if neighboring rooms lack similar processes adding heat.
Hence, the perception of a cooler room from air movement coexists with a higher room temperature due to ongoing energy conversions.

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Most popular questions from this chapter

Consider two walls of a house that are identical except that one is made of 10 -cm-thick wood, while the other is made of 25 -cm-thick brick. Through which wall will the house lose more heat in winter?

In a manufacturing plant, AISI 1010 carbon steel strips \(\left(\rho=7832 \mathrm{~kg} / \mathrm{m}^{3}\right)\) of \(2 \mathrm{~mm}\) thick and \(3 \mathrm{~cm}\) wide are conveyed into a chamber at a constant speed to be cooled from \(527^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\). Determine the speed of a steel strip being conveyed inside the chamber, if the rate of heat being removed from a steel strip inside the chamber is \(100 \mathrm{~kW}\).

A room is heated by a baseboard resistance heater. When the heat losses from the room on a winter day amount to \(9000 \mathrm{~kJ} / \mathrm{h}\), it is observed that the air temperature in the room remains constant even though the heater operates continuously. Determine the power rating of the heater, in \(\mathrm{kW}\).

Determine a positive real root of this equation using \(E E S\) : $$ 3.5 x^{3}-10 x^{0.5}-3 x=-4 $$

Heat is lost steadily through a \(0.5-\mathrm{cm}\) thick \(2 \mathrm{~m} \times 3 \mathrm{~m}\) window glass whose thermal conductivity is \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner and outer surface temperatures of the glass are measured to be \(12^{\circ} \mathrm{C}\) to \(9^{\circ} \mathrm{C}\). The rate of heat loss by conduction through the glass is (a) \(420 \mathrm{~W}\) (b) \(5040 \mathrm{~W}\) (c) \(17,600 \mathrm{~W}\) (d) \(1256 \mathrm{~W}\) (e) \(2520 \mathrm{~W}\)
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