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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

A 2-kW electric resistance heater in a room is turned on and kept on for 50 minutes. The amount of energy transferred to the room by the heater is (a) \(2 \mathrm{~kJ}\) (b) \(100 \mathrm{~kJ}\) (c) \(6000 \mathrm{~kJ}\) (d) \(7200 \mathrm{~kJ}\) (e) \(12,000 \mathrm{~kJ}\)

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor is flowing at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) inside a pipe with \(5 \mathrm{~cm}\) in diameter and \(10 \mathrm{~m}\) in length. The pipe is located in a power plant at \(20^{\circ} \mathrm{C}\), and has a uniform pipe surface temperature of \(100^{\circ} \mathrm{C}\). If the temperature drop between the inlet and exit of the pipe is \(30^{\circ} \mathrm{C}\), and the specific heat of the vapor is \(2190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine the heat transfer coefficient as a result of convection between the pipe surface and the surrounding.

A 4-m \(\times 5-\mathrm{m} \times 6-\mathrm{m}\) room is to be heated by one ton ( \(1000 \mathrm{~kg}\) ) of liquid water contained in a tank placed in the room. The room is losing heat to the outside at an average rate of \(10,000 \mathrm{~kJ} / \mathrm{h}\). The room is initially at \(20^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\), and is maintained at an average temperature of \(20^{\circ} \mathrm{C}\) at all times. If the hot water is to meet the heating requirements of this room for a 24-h period, determine the minimum temperature of the water when it is first brought into the room. Assume constant specific heats for both air and water at room temperature. Answer: \(77.4^{\circ} \mathrm{C}\)

Consider a person whose exposed surface area is \(1.7 \mathrm{~m}^{2}\), emissivity is \(0.5\), and surface temperature is \(32^{\circ} \mathrm{C}\). Determine the rate of heat loss from that person by radiation in a large room having walls at a temperature of \((a) 300 \mathrm{~K}\) and (b) \(280 \mathrm{~K}\). Answers: (a) \(26.7 \mathrm{~W}\), (b) \(121 \mathrm{~W}\)

A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by radiation. Both the air in the room and the surrounding surfaces are at \(20^{\circ} \mathrm{C}\). The exposed surface of the person is \(1.5 \mathrm{~m}^{2}\) and has an average temperature of \(32^{\circ} \mathrm{C}\), and an emissivity of \(0.90\). If the rates of heat transfer from the person by convection and by radiation are equal, the combined heat transfer coefficient is (a) \(0.008 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(5.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(8.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(10.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)
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