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A heat engine that pumps water out of an underground mine accepts \(700 \mathrm{kJ}\) of heat and produces \(250 \mathrm{kJ}\) of work. How much heat does it reject, in kJ?

Short Answer

Expert verified
Answer: The heat engine rejects 450 kJ of heat.

Step by step solution

01

Write the energy conservation equation for a heat engine

The energy conservation equation for a heat engine states that the total energy input (Q_in) must equal the total energy output (W_out + Q_out). Mathematically, this is expressed as: Q_in = W_out + Q_out
02

Substitute the given values in the equation

We are given that the heat input (Q_in) is 700 kJ and the work output (W_out) is 250 kJ. We can substitute these values into the equation from Step 1: 700 = 250 + Q_out
03

Solve for the heat rejected (Q_out)

To find the heat rejected (Q_out), we can simply rearrange the equation from Step 2 and solve for Q_out: Q_out = Q_in - W_out Q_out = 700 - 250
04

Calculate the final answer

Calculate the heat rejected (Q_out) by subtracting the work output from the heat input: Q_out = 450 kJ So, the heat engine rejects 450 kJ of heat.

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Key Concepts

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

Energy Conservation in Thermodynamics
The principle of energy conservation is foundational in thermodynamics, asserting that energy cannot be created or destroyed, but it can change forms. This principle applies to heat engines, which are devices designed to convert heat energy into mechanical work. Let's delve into the energy dynamics of a typical heat engine process.

The engine receives a certain amount of heat energy, denoted as \( Q_{in} \), from a hot reservoir. Throughout the engine cycle, it then converts a portion of this heat into work (\( W_{out} \)). However, not all the input energy can be transformed into useful work due to inherent inefficiencies and the physical limitations dictated by the second law of thermodynamics. As a result, some of the energy is released back into a cooler reservoir as waste heat, denoted as \( Q_{out} \).

Understanding the Energy Flow

In clear terms, energy enters the system as heat, part of it does the actual intended work, like moving pistons or pumping water, and the remainder of the energy is expelled as less useful heat energy. The relationship between the incoming heat, work done, and waste heat is mathematically represented by the equation: \( Q_{in} = W_{out} + Q_{out} \). This equation is a concise statement of energy conservation, tailored for the function of heat engines.
Work Output in Heat Engines
In heat engines, work output represents the useful energy that is extracted from the heat energy provided. It's this work output that drives mechanical processes, such as pistons in an engine or, in the context of our exercise, the pumping of water from an underground mine.

The efficiency of a heat engine is a measure of its capability to convert the energy input (the absorbed heat) into the desired work output. This efficiency is crucial since it indicates how much of the heat energy is being effectively used and determines how much energy is essentially 'lost' as waste heat. Mathematically, the efficiency (\( \text{Eff} \)) can be calculated using the formula: \( \text{Eff} = \frac{W_{out}}{Q_{in}} \times 100\text{%} \).

Practical Implications

For example, if a heat engine takes in 700 kJ of heat and does 250 kJ of work, the work output is utilized to perform the engine's intended function. The rest of the heat is considered excess and is not part of the work output. Understanding the work output helps engineers and scientists design more efficient engines by reducing energy waste.
Heat Rejection Calculation
The calculation of heat rejection in a heat engine is tied up with the core concepts of energy balance and efficiency. Heat rejection is the amount of heat energy expelled by the heat engine into the cooler reservoir, often symbolized as \( Q_{out} \). It's an integral part of the heat engine cycle since it represents the non-usable energy that must be disposed of for the cycle to continue.

To calculate the heat rejection, one must know the initial heat input (\( Q_{in} \)) and the work output (\( W_{out} \)). As derived from our energy conservation equation, the heat rejected can be calculated using \( Q_{out} = Q_{in} - W_{out} \).

Real-World Application

In our textbook exercise, the engine accepts 700 kJ and produces 250 kJ of work. From this, we can easily compute the amount of heat rejected to be 450 kJ, indicating this is the energy not converted into work, either due to thermodynamic inefficiencies or limitations. This calculation is critical in assessing the performance of a heat engine and planning for the necessary cooling systems to manage the waste heat.

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Most popular questions from this chapter

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at \(1400 \mathrm{K}\) and rejects the waste heat to another reservoir at temperature \(T\) The second engine receives this energy rejected by the first one, converts some of it to work, and rejects the rest to a reservoir at \(300 \mathrm{K}\). If the thermal efficiencies of both engines are the same, determine the temperature \(T .\)

A \(2.4-\mathrm{m}\) high \(200-\mathrm{m}^{2}\) house is maintained at \(22^{\circ} \mathrm{C}\) by an air-conditioning system whose COP is \(3.2 .\) It is estimated that the kitchen, bath, and other ventilating fans of the house discharge a houseful of conditioned air once every hour. If the average outdoor temperature is \(32^{\circ} \mathrm{C},\) the density of air is \(1.20 \mathrm{kg} / \mathrm{m}^{3},\) and the unit cost of electricity is \(\$ 0.10 / \mathrm{kWh}\) the amount of money "vented out" by the fans in 10 hours is \((a) \$ 0.50\) \((b) \$ 1.60\) \((c) \$ 5.00\) \((d) \$ 11.00\) \((e) \$ 16.00\)

Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid-vapor mixture region using \(0.96 \mathrm{kg}\) of refrigerant-134a as the working fluid. It is known that the maximum absolute temperature in the cycle is 1.2 times the minimum absolute temperature, and the net work input to the cycle is \(22 \mathrm{kJ}\). If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine the minimum pressure in the cycle.

Show that processes that use work for mixing are irreversible by considering an adiabatic system whose contents are stirred by turning a paddle wheel inside the system (e.g., stirring a cake mix with an electric mixer).

A heat pump cycle is executed with \(R-134 a\) under the saturation dome between the pressure limits of 1.4 and \(0.16 \mathrm{MPa}\) The maximum coefficient of performance of this heat pump is \((a) 1.1\) (b) 3.8 \((c) 4.8\) \((d) 5.3\) \((e) 2.9\)

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