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Chapter 15: Problem 32

(I) If an ideal refrigerator keeps its contents at 2.5\(^\circ\)C when the house temperature is 22\(^\circ\)C, what is its COP?

Short Answer

Expert verified
The COP of the refrigerator is 14.13.

Step by step solution

01

Understanding the Problem

The Coefficient of Performance (COP) of a refrigerator is a measure of its efficiency. It is given by the formula: \( \text{COP} = \frac{T_c}{T_h - T_c} \), where \( T_c \) is the cold temperature in Kelvin and \( T_h \) is the hot temperature in Kelvin.
02

Convert Temperatures to Kelvin

To convert the given temperatures from Celsius to Kelvin, add 273.15 to each. The refrigerator's inner temperature is 2.5掳C, which is 275.65 K, and the house temperature is 22掳C, which is 295.15 K.
03

Substitute Temperatures into COP Formula

Substitute the Kelvin temperatures into the COP formula: \( \text{COP} = \frac{275.65}{295.15 - 275.65} \).
04

Calculate the COP

Calculate the value of the COP: \( \text{COP} = \frac{275.65}{19.5} = 14.13 \).

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Key Concepts

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

Refrigerator Efficiency
The efficiency of a refrigerator is measured by its Coefficient of Performance (COP). This key indicator tells us how effectively a refrigerator uses energy to transfer heat from the inside (keeping it cool) to the outside. A higher COP means the refrigerator is more energy-efficient, meaning it uses less energy to achieve its cooling effect.
Here鈥檚 a simple way to understand it:
  • The COP of a refrigerator is much like the miles per gallon for a car, indicating how well it performs its job under specific conditions.
  • The formula for COP is: \( \text{COP} = \frac{T_c}{T_h - T_c} \) where \( T_c \) is the temperature inside the refrigerator in Kelvin and \( T_h \) is the ambient room temperature (also in Kelvin).
  • This formula helps us understand how the energy moves from inside the fridge to the outside.

By calculating the COP, consumers can make informed choices when selecting energy-efficient appliances. Understanding these concepts better can help in reducing energy consumption, in turn saving money and aiding in environmental conservation.
Temperature Conversion Kelvin
The Kelvin scale is a thermodynamic temperature scale that is key to scientific measurements, including the study of thermodynamics and refrigeration. Unlike Celsius, Kelvin doesn鈥檛 have negative numbers because it starts at absolute zero, the theoretical point where particle movement ceases.
Why Kelvin?
  • This makes Kelvin perfect for scientific equations, as it ensures that temperature values are always positive which simplifies calculations.
  • To convert from Celsius to Kelvin, simply add 273.15. For example, to convert 22掳C (room temperature), you compute it as 22 + 273.15 = 295.15 K.
  • This conversion is crucial for calculations involving thermodynamic principles since these require absolute temperatures.

Knowing how to convert temperatures from Celsius to Kelvin helps accurately plug them into formulas like calculating the Coefficient of Performance for refrigerators. It might seem like a small step, but it's crucial for precision in scientific and engineering applications.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat, work, and energy. It explains how energy changes form and the laws governing these transformations. This area of science is fundamental to understanding how machines like refrigerators operate.
Here's a clearer look:
  • Thermodynamics explains why refrigerators can transfer heat from cooler surroundings (inside the fridge) to the warmer kitchen environment.
  • It involves key concepts such as the conservation of energy, where energy cannot be created or destroyed, only transformed.
  • The laws of thermodynamics define how refrigerators must move energy to keep things cold.
    For a refrigerator, the primary concept is the second law of thermodynamics which states that heat naturally flows from hot to cold areas but requires work to flow from cold to hot places.

By understanding thermodynamics, we gain insights into the efficiency of various appliances, helping us design better, more efficient machines. This knowledge is essential for both scientists and engineers working on sustainable technologies.

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Most popular questions from this chapter

Metabolizing 1.0 kg of fat results in about \(3.7 \times 10^7\)J of internal energy in the body. (\(a\)) In one day, how much fat does the body burn to maintain the body temperature of a person staying in bed and metabolizing at an average rate of 95 W? (\(b\)) How long would it take to burn 1.0 kg of fat this way assuming there is no food intake?

(II) Energy may be stored by pumping water to a high reservoir when demand is low and then releasing it to drive turbines during peak demand. Suppose water is pumped to a lake 115 m above the turbines at a rate of \(1.00 \times 10^5\) kg/s for 10.0 h at night. (\(a\)) How much energy (kWh) is needed to do this each night? (\(b\)) If all this energy is released during a 14-h day, at 75\(\%\) efficiency, what is the average power output?

A car engine whose output power is 135 hp operates at about 15\(\%\) efficiency. Assume the engine's water temperature of 85\(^\circ\)C is its cold-temperature (exhaust) reservoir and 495\(^\circ\)C is its thermal "intake" temperature (the temperature of the exploding gas\(-\)air mixture). (\(a\)) What is the ratio of its efficiency relative to its maximum possible (Carnot) efficiency? (\(b\)) Estimate how much power (in watts) goes into moving the car, and how much heat, in joules and in kcal, is exhausted to the air in 1.0 h.

(I) The exhaust temperature of a heat engine is 230\(^\circ\)C.What is the high temperature if the Carnot efficiency is 34\(\%\)?

(I) An ideal gas expands isothermally, performing \(4.30 \times 10^3\) J of work in the process. Calculate (\(a\)) the change in internal energy of the gas, and (\(b\)) the heat absorbed during this expansion.
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