/*! 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 67 The water in a deep underground ... [FREE SOLUTION] | 91Ó°ÊÓ

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The water in a deep underground well is used as the cold reservoir of a Carnot heat pump that maintains the temperature of a house at \(301 \mathrm{K}\). To deposit 14200 J of heat in the house, the heat pump requires 800 J of work. Determine the temperature of the well water.

Short Answer

Expert verified
The temperature of the well water is approximately 284.43 K.

Step by step solution

01

Understand the Given Problem

We are dealing with a Carnot heat pump where we know the heat deposited in the house, the work done, and the temperature of the house. Our goal is to find the temperature of the well water, which acts as the cold reservoir.
02

Identify Key Formula of Carnot Cycle

The efficiency (\(\eta\)) of a Carnot heat pump is given by the relation involving the temperatures of the hot (\(T_h\)) and cold (\(T_c\)) reservoirs: \[\eta = 1 - \frac{T_c}{T_h}\].Additionally, the efficiency can also be related to work (\(W\)) and heat out (\(Q_h\)):\[\eta = \frac{W}{Q_h}\].
03

Relate Efficiency to Work and Heat

Since we know \(Q_h = 14200\) J (heat deposited in the house), and \(W = 800\) J (work done), we can use the efficiency formula:\[\eta = \frac{W}{Q_h} = \frac{800}{14200}\].
04

Calculate the Numerical Efficiency

Calculate the numerical efficiency of the heat pump:\[\eta = \frac{800}{14200} \approx 0.05634\].This is the efficiency in terms of work and heat.
05

Apply Efficiency to Temperature Relation

Use the efficiency \(\eta = 0.05634\) and the formula \(\eta = 1 - \frac{T_c}{T_h}\). Given \(T_h = 301\, \text{K}\), solve for \(T_c\):\[0.05634 = 1 - \frac{T_c}{301}\].
06

Solve for Cold Reservoir Temperature

Rearrange to find the cold temperature \(T_c\):\[T_c = 301 \times (1 - 0.05634)\].Calculate this to find \(T_c \approx 284.43\, \text{K}\).
07

Final Answer

So the temperature of the well water (cold reservoir) is approximately \(284.43\, \text{K}\).

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Key Concepts

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

Understanding Efficiency in Carnot Heat Pumps
Efficiency is a crucial term in thermodynamics, representing how well a system converts input energy into useful output energy. For a Carnot heat pump, efficiency (\(\eta\)) is defined as the ratio of the work done (\(W\)) to the heat energy provided to the hot reservoir (\(Q_h\)).
This can be expressed through the equation:
\[\eta = \frac{W}{Q_h}\].
This formula shows us the proportion of work used to deliver heat into the residence. In the given exercise, the heat pump has an efficiency calculated by dividing 800 J of work by 14200 J of heat absorbed:
\[\eta = \frac{800}{14200} \approx 0.05634\].
The result indicates the heat pump is 5.634% efficient, meaning only a small fraction of work is needed to transfer energy.
Temperature Calculation of Reservoirs
The temperature in a Carnot cycle relates directly to efficiency. When dealing with temperature, we use the absolute scale of Kelvin (K), which avoids negative numbers. In the exercise, the hot reservoir temperature (\(T_h\)) is given as 301 K.
To find the cold reservoir temperature (\(T_c\)), we use the relation between efficiency and temperatures:
\[\eta = 1 - \frac{T_c}{T_h}\].
Substituting the determined efficiency and known temperatures, the following equation can be formed:\[0.05634 = 1 - \frac{T_c}{301}\].
Rearranging gives,\[T_c = 301 \times (1 - 0.05634)\] which results in a temperature of approximately 284.43 K.
This calculation helps us understand the precise temperature needed for the cold reservoir to maintain the desired warm indoor environment.
Basics of Thermodynamics in Heat Pumps
Thermodynamics is the science of energy, heat, work, and how they are interrelated. In the context of a Carnot heat pump, it operates based on the principles of cyclical processes where work is expended to transfer heat from a cooler space to a warmer one.
Unlike typical heating systems, a heat pump can provide greater heat output than the mechanical energy it consumes. Carnot cycle's theory, developed by Sadi Carnot, outlines the ideal operational conditions for a heat engine, and by extension, heat pumps. With these guidelines, it achieves maximum possible efficiency between two energy reservoirs.
The Carnot heat pump serves to model idealized thermal movement processes, providing a benchmark against which real-world systems can be evaluated.
Cold and Hot Reservoirs
When discussing reservoirs in thermodynamics, we refer to large bodies that hold a significant amount of thermal energy. In a Carnot heat pump system, the cold reservoir is the source from which heat is absorbed, and the hot reservoir is the destination to which this heat is delivered.
In our example exercise, the underground well is the cold reservoir, while the house acts as the hot reservoir. The efficiency and practicality of the heat pump system rely on the temperature difference between these reservoirs.
Greater temperature differences tend to increase the efficiency of heat transfer.
  • Hot reservoir (\(T_h\)) at 301 K: The house that needs to be maintained at a comfortable temperature.
  • Cold reservoir (\(T_c\)) at approximately 284.43 K: The well water that acts as a source of heat.
Understanding these reservoirs helps grasp why energy characteristics such as efficiency are vital for practical heat pump implementation.

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Most popular questions from this chapter

A Carnot engine operates between temperatures of 650 and \(350 \mathrm{K}\). To improve the efficiency of the engine, it is decided either to raise the temperature of the hot reservoir by \(40 \mathrm{K}\) or to lower the temperature of the cold reservoir by \(40 \mathrm{K}\). Which change gives the greater improvement? Justify your answer by calculating the efficiency in each case.

Heat is added isothermally to \(2.5 \mathrm{mol}\) of a monatomic ideal gas. The temperature of the gas is \(430 \mathrm{K}\). How much heat must be added to make the volume of the gas double?

One mole of neon, a monatomic gas, starts out at conditions of standard temperature and pressure. The gas is heated at constant volume until its pressure is tripled, then further heated at constant pressure until its volume is doubled. Assume that neon behaves as an ideal gas. For the entire process, find the heat added to the gas.

Heat \(Q\) flows spontaneously from a reservoir at \(394 \mathrm{K}\) into a reservoir at \(298 \mathrm{K}\). Because of the spontaneous flow, \(2800 \mathrm{J}\) of energy is rendered unavailable for work when a Carnot engine operates between the rescrvoir at \(298 \mathrm{K}\) and a rescrvoir at \(248 \mathrm{K}\). Find \(Q\).

The work done by one mole of a monatomic ideal gas \(\left(\gamma=\frac{5}{3}\right)\) in expanding adiabatically is 825 J. The initial temperature and volume of the gas are \(393 \mathrm{K}\) and \(0.100 \mathrm{m}^{3}\). Obtain (a) the final temperature and (b) the final volume of the gas.

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