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A refrigerator maintaining a \(5^{\circ} \mathrm{C}\) inside temperature is located in a \(30^{\circ} \mathrm{C}\) room. It must have a high temperature \(\Delta T\) above room temperature and a low temperature \(\Delta T\) below the refrigerated space in the cycle to actually transfer the heat. For a \(\Delta T\) of \(0^{\circ}, 5^{\circ}\), and \(10^{\circ} \mathrm{C}\), respectively, calculate the \(\mathrm{COP}\), assuming a Carnot cycle.

Short Answer

Expert verified
The COP values are 11.12 for \(\Delta T = 0^{\circ}\), 7.8 for \(\Delta T = 5^{\circ}\), and 5.96 for \(\Delta T = 10^{\circ}\).

Step by step solution

01

Understand the Carnot Cycle and COP

The Coefficient of Performance (COP) for a refrigerator operating on a Carnot cycle is given by the formula \( \text{COP}_{\text{ref}} = \frac{T_L}{T_H - T_L} \), where \( T_L \) is the absolute temperature of the refrigerated space and \( T_H \) is the absolute temperature of the room. This formula assumes the temperatures are in Kelvin.
02

Convert Given Temperatures to Kelvin

The refrigerated space is initially at \( 5^{\circ} \text{C} \), and the room is at \( 30^{\circ} \text{C} \). These temperatures need to be in Kelvin: \( T_L = 5 + 273 = 278 \text{ K} \)\( T_H = 30 + 273 = 303 \text{ K} \)
03

Calculate Effective Temperatures

With \( \Delta T = 0^{\circ}, 5^{\circ}, \text{ and } 10^{\circ} \), calculate the effective temperatures:For \( \Delta T = 0^{\circ} \) :- \( T_L' = 278 \text{ K} \), \( T_H' = 303 \text{ K} \)For \( \Delta T = 5^{\circ} \) :- \( T_L' = 278 - 5 = 273 \text{ K} \)- \( T_H' = 303 + 5 = 308 \text{ K} \)For \( \Delta T = 10^{\circ} \) :- \( T_L' = 278 - 10 = 268 \text{ K} \)- \( T_H' = 303 + 10 = 313 \text{ K} \)
04

Calculate the COP for Each \( \Delta T \)

Using the COP formula \( \text{COP}_{\text{ref}} = \frac{T_L'}{T_H' - T_L'} \):For \( \Delta T = 0^{\circ} \) :\( \text{COP}_{0} = \frac{278}{303 - 278} = \frac{278}{25} = 11.12 \)For \( \Delta T = 5^{\circ} \) :\( \text{COP}_{5} = \frac{273}{308 - 273} = \frac{273}{35} = 7.8 \)For \( \Delta T = 10^{\circ} \) :\( \text{COP}_{10} = \frac{268}{313 - 268} = \frac{268}{45} = 5.96 \)

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Key Concepts

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

Coefficient of Performance (COP)
In the world of thermodynamics, the coefficient of performance (COP) is a measure of the efficiency of a refrigeration cycle. While other machines measure efficiency as the amount of work they can do, a refrigerator is different. It measures how well it can move heat from a cold area (where you want things cool) to a warm area (the room, typically).
This movement or process doesn't come free; it requires some work. The COP is the ratio of the effective heat removed from the cold region to the work input required to effectuate this process. For refrigerators operating under ideal conditions, like a Carnot cycle, the COP is denoted by the formula:
  • \[ \text{COP}_{\text{ref}} = \frac{T_L}{T_H - T_L} \]
Here, \( T_L \) is the low or refrigerated space's temperature and \( T_H \) is the temperature of the surrounding room. What makes COP fascinating is that higher values mean a more efficient cycle at cooling, since the same amount of input work moves more heat.
Kelvin Scale in Thermodynamics
In thermodynamics, precision and uniformity are critical. The Kelvin scale provides this standard measure for temperature, offering consistency across different calculations. Unlike Celsius or Fahrenheit, Kelvin doesn’t include negative values, simplifying the math, especially for formulas like those of the Carnot cycle.
  • Kelvin is the absolute temperature scale, starting at absolute zero, the theoretical point where molecular energy is at a minimum.
  • Conversions are straightforward: \( 0^{\circ} \text{C} \) converts to \( 273 \text{K} \). Similarly, simply add 273.15 to a Celsius reading to turn it into Kelvin.
When dealing with refrigeration calculations, such as determining the COP of a Carnot cycle, using Kelvin ensures no awkward negative numbers or discrepancies from differing zero-start points.
Temperature Conversion for COP Calculations
When calculating the COP of a refrigeration cycle, properly converting temperatures to the Kelvin scale is crucial. A typical temperature setup might show you a refrigerated space and the ambient room temperature in Celsius. For the computations to be correct, they first need to switch to Kelvin.
To convert the temperatures provided:
  • Add 273 to the Celsius measurement of the refrigerated space to find \( T_L \) in Kelvin.
  • Similarly, add 273 to the ambient room temperature for \( T_H \).
The challenge can intensify when given a temperature difference, \( \Delta T \). This \( \Delta T \) makes the room seem slightly warmer, or the refrigerated space slightly cooler. You incorporate these changes by adjusting \( T_H \) by adding \( \Delta T \) and adjusting \( T_L \) by subtracting \( \Delta T \). Without making these conversions and adjustments, any calculation of the COP will be inaccurate.

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Most popular questions from this chapter

In a few places where the air is very cold in the winter, such as \(-30^{\circ} \mathrm{C}\), it is possible to find a temperature of \(13^{\circ} \mathrm{C}\) below ground. What efficiency will a heat engine have when operating between these two thermal reservoirs?

A car engine takes atmospheric air in at \(20^{\circ} \mathrm{C}\), no fuel, and exhausts the air at \(-20^{\circ} \mathrm{C}\), producing work in the process. What do the first and second laws say about that?

Air in a rigid \(1-\mathrm{m}^{3}\) box is at \(300 \mathrm{~K}, 200 \mathrm{kPa}\). It is heated to \(600 \mathrm{~K}\) by heat transfer from a reversible heat pump that receives energy from the ambient at \(300 \mathrm{~K}\) besides the work input. Use constant specific heat at \(300 \mathrm{~K}\). Since the COP changes, write \(d Q=m_{\text {air }} C_{v} d T\) and find \(d W\). Integrate \(d W\) with the temperature to find the required heat pump work.

An air conditioner on a hot summer day removes \(8 \mathrm{Btu} / \mathrm{s}\) of energy from a house at \(70 \mathrm{~F}\) and pushes energy to the outside, which is at \(88 \mathrm{~F}\). The house has \(50000 \mathrm{lbm}\) mass with an average specific heat of \(0.23 \mathrm{Btu} / \mathrm{lbm}-\mathrm{R}\). In order to do this, the cold side of the air conditioner is at \(40 \mathrm{~F}\) and the hot side is at \(100 \mathrm{~F}\). The air conditioner (refrigerator) has a COP that is \(50 \%\) that of a corresponding Carnot refrigerator. Find the actual COP of the air conditioner and the power required to run it.

Two heat engines operate between the same two energy reservoirs, and both receive the same \(Q_{H}\). One engine is reversible and the other is not. What can you say about the two \(Q_{L}\) 's?

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