/*! 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 39 .. You are designing a Carnot en... [FREE SOLUTION] | 91Ó°ÊÓ

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.. You are designing a Carnot engine that has 2 mol of \(\mathrm{CO}_{2}\) as its working substance; the gas may be treated as ideal. The gas is to have a maximum temperature of \(527^{\circ} \mathrm{C}\) and a maximum pressure of 5.00 atm. With a heat input of 400 \(\mathrm{J}\) per cycle, you want 300 \(\mathrm{J}\) of useful work. (a) Find the temperature of the cold reservoir. (b) For how many cycles must this engine run to melt completely a 10.0 -kg block of ice originally at \(0.0^{\circ} \mathrm{C},\) using only the heat rejected by the engine?

Short Answer

Expert verified
(a) 200.04 K; (b) 33,400 cycles.

Step by step solution

01

Understanding the Carnot Cycle and Calculation of Temperatures

The Carnot cycle is an idealized thermodynamic cycle that provides the maximum possible efficiency for a heat engine. The efficiency \( \eta \) of the Carnot engine is given by \( \eta = 1 - \frac{T_c}{T_h} \) where \( T_c \) and \( T_h \) are the temperatures of the cold and hot reservoirs, respectively. First, convert the maximum temperature to Kelvin: \( T_h = 527^{\circ} \mathrm{C} + 273.15 = 800.15 \ \mathrm{K} \).
02

Calculate Efficiency of the Engine

Use the efficiency formula of the Carnot engine mentioned in Step 1. Given the heat input \( Q_h = 400 \ \mathrm{J} \) and useful work \( W = 300 \ \mathrm{J} \), the efficiency \( \eta \) is computed as \( \eta = \frac{W}{Q_h} = \frac{300}{400} = 0.75 \).
03

Solve for the Cold Reservoir Temperature

With the efficiency formula \( \eta = 1 - \frac{T_c}{T_h} \), rearrange to find the cold reservoir temperature \( T_c = T_h(1 - \eta) \). \( \eta = 0.75 \) and \( T_h = 800.15 \ \mathrm{K} \) imply \( T_c = 800.15 (1 - 0.75) = 200.0375 \ \mathrm{K} \).
04

Calculate Heat Rejected per Cycle

The heat rejected \( Q_c \) can be found using \( Q_h = Q_c + W \), thus \( Q_c = Q_h - W = 400 \ \mathrm{J} - 300 \ \mathrm{J} = 100 \ \mathrm{J} \).
05

Calculate Cycles Required to Melt Ice

The latent heat of fusion for ice is \( 334 \ \mathrm{kJ/kg} \). For a 10 kg block: \( Q_{\text{ice}} = 10 \ \mathrm{kg} \times 334 \ \mathrm{kJ/kg} = 3340 \ \mathrm{kJ} = 3,340,000 \ \mathrm{J} \). Each cycle rejects \( Q_c = 100 \ \mathrm{J} \), so the number of cycles \( n \) is given by \( n = \frac{3,340,000}{100} = 33,400 \).

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Key Concepts

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

Thermodynamic Cycle
A thermodynamic cycle refers to a series of thermodynamic processes that return a system to its initial state. Each cycle involves the transfer of heat and work into and out of the system. The Carnot cycle is one of the most important concepts in thermodynamics. It is an idealized cycle that represents the maximum possible efficiency a heat engine can achieve. Understanding this cycle is crucial because it sets a benchmark against which real engines are compared, even though no practical engine can be perfectly Carnot. During the Carnot cycle, the working substance, such as an ideal gas, undergoes a set of processes involving isothermal and adiabatic changes, which are ideal conditions that help us analyze real processes more accurately.
Efficiency
Efficiency in thermodynamics is the measure of how well a heat engine converts heat into work. For a Carnot engine, the efficiency is calculated using the temperatures of the hot and cold reservoirs. The formula is given by \[ \eta = 1 - \frac{T_c}{T_h} \] where \( T_c \) and \( T_h \) represent the absolute temperatures of the cold and hot reservoirs, respectively. This formula tells us that efficiency depends solely on these temperatures: the greater the difference, the higher the potential efficiency. For example, if the Carnot engine in question had an efficiency of 0.75, it converts 75% of the heat energy into work. This high efficiency is a theoretical maximum, highlighting the importance of operating between large temperature differences to achieve significant work output.
Ideal Gas Law
The ideal gas law is a crucial equation for gases that behave ideally, often used in calculating the states of the working substance in a Carnot cycle. It is represented as \[ PV = nRT \] where \( P \) is the pressure, \( V \) is the volume, \( n \) is the amount of substance in moles, \( R \) is the ideal gas constant, and \( T \) is the absolute temperature. This law helps predict how the gas will behave under different conditions, such as changes in temperature or pressure. In the context of a Carnot engine, it ensures calculations considering the gas's behavior during each part of the cycle are accurate, thus allowing a better design of engines owing to precise predictions of the state changes.
Temperature Conversion
Temperature conversions are essential for thermodynamic calculations to ensure all data is in the same units. In many physics problems, temperatures are required in Kelvin for compatibility with thermodynamic equations. To convert from Celsius to Kelvin, you add 273.15 to the Celsius temperature. For example, a temperature of \(527^{\circ} \mathrm{C}\) converts to \(800.15 \, \mathrm{K}\). This step is crucial because Kelvin is an absolute scale based on the theoretical lowest possible temperature (absolute zero), providing a consistent basis for calculating efficiencies and processes in thermodynamic systems.
Latent Heat of Fusion
Latent heat of fusion refers to the amount of heat needed to convert a unit mass of a solid into a liquid at a constant temperature. For ice, this is a significant concept because melting requires a considerable amount of energy without changing the temperature of the ice. The latent heat of fusion for ice is \(334 \, \mathrm{kJ/kg}\). Understanding this is essential in problems where you calculate the energy needed to melt ice using the heat rejected by another process, such as the Carnot cycle. When calculating how the energy from an engine cycle contributes to melting ice, the latent heat helps determine how many cycles are needed to provide the necessary energy for phase changes. In the situation given, knowing this value allows you to compute that the energy needed to melt a 10 kg ice block is \(3,340,000 \, \mathrm{J}\).

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Most popular questions from this chapter

A Carnot engine is operated between two heat reservoirs at temperatures of 520 \(\mathrm{K}\) and 300 \(\mathrm{K}\) . (a) If the engine receives 6.45 \(\mathrm{kJ}\) of heat energy from the reservoir at 520 \(\mathrm{K}\) in each cycle, how many joules per cycle does it reject to the reservoir at 300 \(\mathrm{K} ?\) (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?

. A human engine. You decide to use your body as a Carnot heat engine. The operating gas is in a tube with one end in your mouth (where the temperature is \(37.0^{\circ} \mathrm{C}\) ) and the other end at the surface of your skin, at \(30.0^{\circ} \mathrm{C}\) . (a) What is the maximum efficiency of such a heat engine? Would it be a very use- ful engine? (b) Suppose you want to use this human engine to lift a 2.50 \(\mathrm{kg}\) box from the floor to a tabletop 1.20 \(\mathrm{m}\) above the floor. How much must you increase the gravitational potential energy and how much heat input is needed to accomplish this? (c) How many 350 calorie (those are food calories, remember) candy bars must you eat to lift the box in this way? Recall that 80\(\%\) of the food energy goes into heat.

\(\cdot\) Fach cycle, a certain heat engine expels 250 \(\mathrm{J}\) of heat when you put in 325 \(\mathrm{J}\) of heat. Find the efficiency of this engine and the amount of work you get out of the 325 \(\mathrm{J}\) heat input.

(. A large factory furnace maintained at \(175^{\circ} \mathrm{C}\) at its outer surface is wrapped in an insulating blanket of thermal conduc- tivity 0.055 \(\mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) which is thick enough that the outer surface of the insulation is at \(42^{\circ} \mathrm{C}\) while heat escapes from the furnace at a steady rate of 125 W for each square meter of surface area. By how much does each square meter of the furnace change the entropy of the factory every second?

\(\cdot\) A gasoline engine. A gasoline engine takes in \(1.61 \times 10^{4} \mathrm{J}\) of heat and delivers 3700 \(\mathrm{J}\) of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of \(4.60 \times 10^{4} \mathrm{J} / \mathrm{g}\) . (a) What is the thermal efficiency of the engine? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?

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