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An air conditioner keeps the inside of a house at a temperature of \(19.0^{\circ} \mathrm{C}\) when the outdoor temperature is \(33.0^{\circ} \mathrm{C} .\) Heat, leaking into the house at the rate of 10500 joules per second, is removed by the air conditioner. Assuming that the air conditioner is a Carnot air conditioner, what is the work per second that must be done by the electrical energy in order to keep the inside temperature constant?

Short Answer

Expert verified
The air conditioner requires approximately 503.84 joules of work per second to operate.

Step by step solution

01

Understand the Carnot Cycle

The Carnot cycle is an idealized thermodynamic cycle that provides insights into the efficiency of heat engines and refrigerators. For an air conditioner, the coefficient of performance (COP) is defined as the ratio of heat removed from the cooler reservoir to the work input needed to achieve it.
02

Determine the Coefficient of Performance (COP) for a Carnot Air Conditioner

The formula for the coefficient of performance of a Carnot refrigerator or air conditioner is given by \[ \text{COP} = \frac{T_{cold}}{T_{hot} - T_{cold}} \] where \( T_{cold} \) and \( T_{hot} \) are the absolute temperatures of the indoor and outdoor environments, respectively.
03

Convert Temperatures to Kelvin

First, we need to convert the temperatures from Celsius to Kelvin. The conversion is done using the formula \[ T(K) = T(^{\circ}C) + 273.15 \] Thus, the inside temperature in Kelvin is \[ T_{cold} = 19.0 + 273.15 = 292.15 \, \text{K} \] and the outside temperature in Kelvin is \[ T_{hot} = 33.0 + 273.15 = 306.15 \, \text{K} \]
04

Calculate the Coefficient of Performance

Substitute the temperature values into the COP formula: \[ \text{COP} = \frac{292.15}{306.15 - 292.15} = \frac{292.15}{14} \approx 20.86 \]
05

Relate Heat Leak with Work Done

For a Carnot air conditioner, the relationship between the heat removed \( Q_c \), work done \( W \), and the coefficient of performance (COP) is given by: \[ \text{COP} = \frac{Q_c}{W} \] We are given that \( Q_c = 10500 \, \text{J/s} \), so substitute \( \text{COP} \) to find \( W \): \[ 20.86 = \frac{10500}{W} \]
06

Solve for the Work Done per Second

Rearrange the equation from Step 5 to solve for \( W \): \[ W = \frac{10500}{20.86} \approx 503.84 \, \text{J/s} \] Therefore, the work per second required to keep the inside temperature constant is approximately 503.84 joules.

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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, often abbreviated as COP, is a measure of the efficiency of devices like refrigerators and air conditioners. It helps us understand how well these systems work in transferring heat in relation to the energy they consume. For a Carnot air conditioner, the COP is determined by the temperatures at which it operates. The formula is given as:\[ \text{COP} = \frac{T_{\text{cold}}}{T_{\text{hot}} - T_{\text{cold}}} \]Here, \( T_{\text{cold}} \) is the temperature inside the house (where you want the cooling effect), and \( T_{\text{hot}} \) is the temperature outside the house. You need to express these temperatures in Kelvin to use the formula correctly. A higher COP indicates a more efficient air conditioner, as it can remove more heat for each unit of work done.
Thermodynamic Cycle
The thermodynamic cycle, specifically the Carnot cycle, is a theoretical model that represents the most efficient sequence of processes for heat engines. For air conditioners, the Carnot cycle demonstrates the ideal performance that the device could achieve under perfect conditions. It involves cycles of compression, expansion, and heat transfer between two thermal reservoirs. - **Compression and Expansion:** These phases involve changes in the pressure and volume of the refrigerant. - **Heat Absorption and Rejection:** The refrigerant absorbs heat from the indoor air and expels it outside. In practice, actual air conditioners are less efficient than a real Carnot device due to practical limitations, but understanding the Carnot cycle is crucial for grasping the principles of thermodynamics.
Heat Transfer
Heat transfer is the movement of thermal energy from one place to another and is a fundamental principle in the operation of air conditioners. It occurs in three ways: conduction, convection, and radiation. In air conditioners, the primary focus is on conduction and convection. - **Conduction:** This is the direct transfer of heat through a material without moving the material itself. - **Convection:** This involves the movement of fluids (like air) transferring heat as they move and mix. An air conditioner works by transferring heat from inside the house to the outside environment, effectively cooling the indoor space. The rate of heat transfer reflects how quickly the air conditioner can cool the interior, and this is often measured in joules per second or watts.
Work and Energy in Thermodynamics
In thermodynamics, work and energy are crucial in understanding how devices such as air conditioners operate. The device does mechanical work to move heat against its natural direction—from a cooler space inside to a warmer space outside. - **Work Done:** In the context of the air conditioner, work is the energy required to remove heat from inside the house and expel it outside. It is measured in joules (J) or watts (J/s). - **Energy Input:** This is typically supplied by electricity, which performs the necessary work to pump the refrigerant around the system. According to the solution for the exercise, this "work per second" is approximately 503.84 joules. Efficient air conditioners require less work for a given amount of heat removal, resulting in better energy conservation and lower electricity costs.

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Most popular questions from this chapter

A Carnot engine uses hot and cold reservoirs that have temperatures of 1684 and 842 \(\mathrm{K}\) , respectively. The input heat for this engine is \(\left|Q_{\mathrm{H}}\right|\) The work delivered by the engine is used to operate a Carnot heat pump. The pump removes heat from the \(842-\mathrm{K}\) reservoir and puts it into a hot reservoir at a temperature \(T^{\prime}\) . The amount of heat removed from the \(842-\mathrm{K}\) reservoir is also \(\left|Q_{\mathrm{H}}\right| .\) Find the temperature \(T^{\prime}\)

A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignite the fuel occurs because the pistons compress the air in the cylinders. Suppose that air at an initial temperature of \(21^{\circ} \mathrm{C}\) is compressed adiabatically to a temperature of \(688^{\circ} \mathrm{C}\) . Assume the air to be an ideal gas for which \(\gamma=\frac{7}{5} .\) Find the compression ratio, which is the ratio of the initial volume to the final volume.

A Carnot engine has an efficiency of \(0.40 .\) The Kelvin temperature of its hot reservoir is quadrupled, and the Kelvin temperature of its cold reservoir is doubled. What is the efficiency that results from these changes?

In exercising, a weight lifter loses 0.150 \(\mathrm{kg}\) of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is \(1.40 \times 10^{5} \mathrm{J}\) (a) Assuming that the latent heat of vaporization of perspiration is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\) find the change in the internal energy of the weight lifter. \((\mathbf{b})\) Determine the minimum number of nutritional Calories of food \((1 \text { nutritional Calorie }=4186 \mathrm{J})\) that must be consumed to replace the loss of internal energy.

The hot reservoir for a Carnot engine has a temperature of 890 \(\mathrm{K}\) while the cold reservoir has a temperature of 670 \(\mathrm{K}\) . The heat input for this engine is 4800 \(\mathrm{J}\) . The \(670-\mathrm{K}\) reservoir also serves as the hot reservoir for a second Carnot engine. This second engine uses the rejected heat of the first engine as input and extracts additional work from it. The rejected heat from the second engine goes into a reservoir that has a temperature of 420 \(\mathrm{K}\) . Find the total work delivered by thetwo engines.

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