/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 88 A Carnot heat pump operates betw... [FREE SOLUTION] | 91Ó°ÊÓ

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A Carnot heat pump operates between an outdoor temperature of 265 \(\mathrm{K}\) and an indoor temperature of 298 \(\mathrm{K}\) . Find its coefficient of performance.

Short Answer

Expert verified
The COP of the Carnot heat pump is 9.03.

Step by step solution

01

Understand the Problem

The problem asks us to find the coefficient of performance (COP) for a Carnot heat pump. A heat pump is a device that transfers heat from a cooler area (outdoor temperature) to a warmer area (indoor temperature). The Carnot heat pump is an ideal heat pump operating on a reversible cycle (Carnot cycle).
02

Recall the Formula for COP of a Heat Pump

The coefficient of performance (COP) for a heat pump is given by the formula: \[ \text{COP} = \frac{T_{hot}}{T_{hot} - T_{cold}}, \] where \( T_{hot} \) is the temperature of the hot reservoir (indoor temperature) and \( T_{cold} \) is the temperature of the cold reservoir (outdoor temperature).
03

Substitute Given Values

Substitute the given temperatures into the COP formula: \( T_{hot} = 298 \ \mathrm{K} \) (indoor temperature) and \( T_{cold} = 265 \ \mathrm{K} \) (outdoor temperature).\[ \text{COP} = \frac{298}{298 - 265}. \]
04

Perform the Subtraction in the Denominator

Calculate the difference in the denominator: \( 298 - 265 = 33 \). This simplifies the formula to \[ \text{COP} = \frac{298}{33}. \]
05

Calculate the COP

Divide 298 by 33 to find the COP: \[ \text{COP} = 9.03 \] (rounded to two decimal places).

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Key Concepts

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

Coefficient of Performance
The coefficient of performance (COP) is a key measure used to evaluate the efficiency of a heat pump. It signifies how effectively the pump moves heat compared to the energy it consumes. Simply put, it's a ratio of the heat output to the energy input.
A higher COP means the heat pump is more efficient. For a Carnot heat pump, the COP is calculated using the formula:
  • \[ \text{COP} = \frac{T_{\text{hot}}}{T_{\text{hot}} - T_{\text{cold}}} \]
Where \( T_{\text{hot}} \) and \( T_{\text{cold}} \) are the temperatures of the hot and cold reservoirs, respectively. In our problem, these are the indoor and outdoor temperatures. Understanding and calculating COP is essential because it helps us compare the performance of different heat pumps.
Carnot Cycle
The Carnot cycle is a theoretical model that represents the most efficient way a heat engine can operate. It consists of four reversible processes: two isothermal (constant temperature) and two adiabatic (no heat transfer).
This cycle provides an ideal benchmark for real-world engines and heat pumps. For a Carnot heat pump, the cycle moves heat from a cooler space to a warmer space, leveraging the temperature difference to do so efficiently.
While actual pumps cannot achieve the perfection of a Carnot cycle, they can be designed to get as close as possible. This concept is crucial for thermodynamics because it defines an upper limit for efficiency. Understanding the Carnot cycle helps in recognizing what makes the Carnot heat pump theoretic in operation.
Thermodynamics
Thermodynamics is the branch of physics that deals with how heat, work, and energy interrelate. When we talk about heat pumps, thermodynamics provides the foundational principles for how heat transfer occurs.
The Carnot heat pump follows these principles, demonstrating how energy conversions can be made more efficient. Concepts like thermal equilibrium, the efficiency of processes, and the laws of thermodynamics guide our understanding of these systems.
  • 1st Law: Energy cannot be created or destroyed, only transformed.
  • 2nd Law: Heat naturally flows from hot to cold unless work is applied to reverse it.
These laws help explain why the Carnot cycle is the ideal, and they guide engineers to optimize real-world applications.
Heat Transfer
Heat transfer refers to the movement of heat from one place to another. It’s vital for understanding not only how heat pumps work, but also their efficiency. Heat can transfer in three ways: conduction, convection, and radiation.
In the context of a heat pump, it typically uses electricity to move heat, overcoming the natural flow from hot to cold. The Carnot cycle maximizes this process by theoretically minimizing energy loss.
Understanding the mechanisms of heat transfer is crucial. It enables us to grasp how a Carnot heat pump can maintain a comfortable indoor environment even when temperatures outside are frigid. This understanding also illustrates why specific designs improve the efficiency of heat pumps in various conditions.

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Most popular questions from this chapter

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 228 \(\mathrm{J}\) when 166 \(\mathrm{J}\) of work is done on the system. In the second step, the internal energy of the system increases by 115 \(\mathrm{J}\) when 177 \(\mathrm{J}\) of work is done on the system. For the overall process, find the heat. What type of process is the overall process? Explain.

A Carnot engine has an efficiency of 0.700 , and the temperatureof its cold reservoir is 378 \(\mathrm{K}\) . (a) Determine the temperature of its hot reservoir. \((\mathrm{b})\) If 5230 \(\mathrm{J}\) of heat is rejected to the cold reservoir, what amount of heat is put into the engine?

Due to design changes, the efficiency of an engine increases from 0.23 to 0.42. For the same input heat \(\left[Q_{\mathrm{H}}, \text { these changes increase }\right.\) the work done by the more efficient engine and reduce the amount of heat rejected to the cold reservoir. Find the ratio of the heat rejected to the cold reservoir for the improved engine to that for the original engine.

In a game of football outdoors on a cold day, a player will begin to feel exhausted after using approximately \(8.0 \times 10^{5} \mathrm{J}\) of internal energy. (a) One player, dressed too lightly for the weather, has to leave the game after losing \(6.8 \times 10^{5} \mathrm{J}\) of heat. How much work has he done? (b) Another player, wearing clothes that offer better protection against heat loss, is able to remain in the game long enough to do \(2.1 \times 10^{5} \mathrm{J}\) of work. What is the magnitude of the heat that he has lost?

A system gains 1500 \(\mathrm{J}\) of heat, while the internal energy of the system increases by 4500 \(\mathrm{J}\) and the volume decreases by 0.010 \(\mathrm{m}^{3}\) . Assume that the pressure is constant and find its value.

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