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Chapter 19: Problem 1

Could you cool the kitchen by leaving the refrigerator open? Explain.

Short Answer

Expert verified
No, a kitchen cannot be cooled effectively by leaving the refrigerator open. Instead, it might cause the room to warm up due to waste heat released by the fridge's motor working harder to cool the larger space.

Step by step solution

01

Understanding How a Refrigerator Works

A refrigerator cools food by removing the heat from inside the refrigerator and releasing it to the outside environment and this cycle keeps repeating.
02

Applying the Thermodynamics Law

According to the second law of thermodynamics, heat naturally flows from hot to cold. When the fridge door is open, the inside of the fridge (cold) is exposed to the warmer room temperature. Thus, the fridge will try to lower the temperature of the room by taking in that heat.
03

Energy Transfer between Fridge and Room

This continuous energy transfer is not 100% efficient. Some of the energy is lost in the form of waste heat, which is released back into the room. This waste heat causes the room to warm up rather than cool down.
04

Effect on the Fridge

As the fridge tries to cool the warmer room, the fridge's motor will work harder, produce more heat, and eventually could burn out faster.

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Most popular questions from this chapter

Refrigerators remain among the greatest consumers of electrical energy in most homes, although mandated efficiency standards have decreased their energy consumption by some \(80 \%\) in the past four decades. In the course of a day, one kitchen refrigerator removes \(30 \mathrm{MJ}\) of energy from its contents, in the process consuming \(10 \mathrm{MJ}\) of electrical energy. The electricity comes from a \(40 \%\) efficient coal-fired power plant. The electrical energy a. is used to run the light bulb inside the refrigerator. b. wouldn't be necessary if the refrigerator had enough insulation. c. retains its high-quality status after the refrigerator has used it. d. ends up as waste heat rejected to the kitchen environment.

A Carnot engine extracts heat from a block of mass \(m\) and specific heat \(c\) initially at temperature \(T_{\mathrm{b} 0}\) but without a heat source to maintain that temperature. The engine rejects heat to a reservoir at constant temperature \(T_{c} .\) The engine is operated so its mechanical power output is proportional to the temperature difference \(T_{\mathrm{h}}-T_{\mathrm{c}}\) $$P=P_{0} \frac{T_{\mathrm{h}}-T_{\mathrm{c}}}{T_{\mathrm{h} 0}-T_{\mathrm{c}}}$$ where \(T_{\mathrm{h}}\) is the instantancous temperature of the hot block and \(P_{0}\) is the initial power. (a) Find an expression for \(T_{\mathrm{h}}\) as a function of time, and (b) determine how long it takes for the engine's power output to reach zero.

A cosmic heat engine might operate between the Sun's \(5800 \mathrm{K}\) surface and the \(2.7 \mathrm{K}\) temperature of intergalactic space. What would be its maximum efficiency?

In an alternative universe, you've got the impossible: an infinite heat reservoir, containing infinite energy at temperature \(T_{\mathrm{b}} .\) But you've only got a finite cool reservoir, with initial temperature \(T_{c 0}\) and heat capacity \(C .\) Find an expression for the maximum work you can extract if you operate an engine between these two reservoirs.

How much energy becomes unavailable for work in an isothermal process at \(440 \mathrm{K},\) if the entropy increase is \(25 \mathrm{J} / \mathrm{K} ?\)
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