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

Short Answer

Expert verified
No, leaving the refrigerator open wouldn't cool the kitchen, but would rather increase the temperature due to the heat expelled by the refrigerator's increased efforts to cool its interior.

Step by step solution

01

Understanding the function of a refrigerator

A refrigerator works by utilizing the principle of thermodynamics. It removes heat from the inside, cooling the contents, and expels that heat to the outside, effectively warming the surroundings.
02

Consider the effect of leaving the refrigerator open

When the refrigerator door is open, the refrigerator detects the rise in interior temperature and starts working harder to reach the desired coolness, thereby expelling more heat into the kitchen.
03

Conclusion

Even though the inside of the fridge is cooler and seems like it might cool the kitchen, this isn't the case. The refrigerator would have to work harder to try to cool down, and the end result would be more heat being expelled, warming up the kitchen instead of cooling it.

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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 refrigerator's COP is a. \(\frac{1}{3}\). b. 2. c. 3. d. 4.

An ideal gas undergoes a process that takes it from pressure \(p_{1}\) and volume \(V_{1}\) to \(p_{2}\) and \(V_{2},\) such that \(p_{1} V_{1}^{\gamma}=p_{2} V_{2}^{\gamma},\) where \(\gamma\) is the specific heat ratio. Find the entropy change if the process consists of constant-pressure and constant-volume segments. Why does your result make sense?

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