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Chapter 20: Problem 14

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 discard 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?

Short Answer

Expert verified
(a) 3722.85 J (b) 2727.15 J (c) 42.3% efficient

Step by step solution

01

Understanding Carnot Efficiency

The Carnot efficiency is the maximum efficiency any heat engine can achieve, operating between two temperatures. It is given by the formula \[ \eta = 1 - \frac{T_C}{T_H} \] where \( T_C \) is the temperature of the cold reservoir and \( T_H \) is the temperature of the hot reservoir. Make sure to express these temperatures in Kelvin.
02

Calculate Carnot Efficiency

Substitute the given temperatures \( T_H = 520 \, \text{K} \) and \( T_C = 300 \, \text{K} \) into the Carnot efficiency formula:\[ \eta = 1 - \frac{300}{520} \approx 0.423 \] The Carnot efficiency is approximately 42.3%.
03

Calculate Mechanical Work Output

The work output per cycle, \( W \), can be determined using the relationship between work, heat absorbed, and efficiency:\[ W = Q_H \times \eta \]where \( Q_H = 6.45 \, \text{kJ} = 6450 \, \text{J} \). Substitute the values to get:\[ W = 6450 \, \text{J} \times 0.423 \approx 2727.15 \, \text{J} \]Thus, the mechanical work performed is approximately 2727.15 J per cycle.
04

Calculate Heat Discarded

The energy conservation principle for a heat engine gives us:\[ Q_H = W + Q_C \]Rearrange to solve for \( Q_C \), the heat discarded to the cold reservoir:\[ Q_C = Q_H - W \]Substitute \( Q_H = 6450 \, \text{J} \) and \( W = 2727.15 \, \text{J} \):\[ Q_C = 6450 \, \text{J} - 2727.15 \, \text{J} \approx 3722.85 \, \text{J} \]Thus, approximately 3722.85 J per cycle is discarded to the cold reservoir.

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Key Concepts

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

Heat Reservoirs
Carnot engines operate between two heat reservoirs, which function as heat energy sources or sinks. A heat reservoir can maintain a constant temperature, even as heat is added or removed. In a Carnot engine, the high-temperature reservoir provides heat, while the low-temperature reservoir absorbs the excess heat discarded by the engine. - In our example, the hot reservoir was at 520 K, and the cold reservoir was at 300 K. - The heat energy transferred from the hot reservoir to the engine in each cycle is 6.45 kJ or 6450 J. This heat transfer enables the engine to perform work by converting a portion of this heat into mechanical energy.
Mechanical Work
Mechanical work is the energy output performed by the engine due to the heat absorbed from the hot reservoir. Not all absorbed heat is converted to work; some is inevitably lost to the cold reservoir. - Using the formula \( W = Q_H \times \eta \), we calculated the mechanical work performed by the engine per cycle.- Here, \( Q_H = 6450 \text{ J} \), and the calculated Carnot efficiency \( \eta \approx 0.423 \) results in \( W \approx 2727.15 \text{ J} \).Mechanical work here represents how efficiently the engine can convert absorbed heat into useful energy to perform tasks.
Thermal Efficiency
The thermal efficiency of a heat engine reflects how well it converts heat from the hot reservoir into useful work. It is defined by the ratio of work output to heat input. For the Carnot cycle, an idealized heat engine, thermal efficiency is determinable using the temperatures of the reservoirs:- Thermal efficiency \( \eta = 1 - \frac{T_C}{T_H} \). With \( T_H = 520 \text{ K} \) and \( T_C = 300 \text{ K} \), the efficiency is \( \eta \approx 0.423 \) or 42.3%.This means that only about 42.3% of the heat energy absorbed is converted into work; the rest is lost to the cold reservoir.
Energy Conservation
The principle of energy conservation ensures that energy cannot be created or destroyed, only transformed. For Carnot engines, this implicates the conversion between heat absorption, work output, and heat rejection.- The equation \( Q_H = W + Q_C \) shows that the heat taken from the hot reservoir equals the sum of work done and heat discarded to the cold reservoir.- Using our example, \( Q_H = 6450 \text{ J} \), \( W \approx 2727.15 \text{ J} \), and accordingly, \( Q_C \approx 3722.85 \text{ J} \).This connection ensures a balance, underscoring that no heat is lost but rather redistributed between work and heat discard.
Carnot Efficiency
Carnot efficiency represents the maximum theoretical efficiency a heat engine can attain. It establishes an ideal inefficiency ceiling based only on the temperature difference between the reservoirs.- The Carnot efficiency relies on the formula \( \eta = 1 - \frac{T_C}{T_H} \) and portrays an ideal, unattainable benchmark.- With our given temperatures, the efficiency is approximately 42.3%, indicating the maximum fraction of heat energy that can be turned into useful work without violating thermodynamic principles.This serves as a crucial standard, comparing real engine performance to the ultimate efficiency limit dictated by the second law of thermodynamics.

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Most popular questions from this chapter

A lonely party balloon with a volume of 2.40 \(\mathrm{L}\) and containing 0.100 \(\mathrm{mol}\) of air is left behind to drift in the temporarily uninhabited and depressurized International Space Station. Sunlight coming through a porthole heats and explodes the balloon, causing the air in it to undergo a free expansion into the empty station, whose total volume is 425 \(\mathrm{m}^{3} .\) Calculate the entropy change of the air during the expansion.

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation, the heat rejected by the engine melts 0.0400 \(\mathrm{kg}\) of ice. During this time, how much work \(W\) is performed by the engine?

A certain brand of freezer is advertised to use 730 \(\mathrm{kW} \cdot \mathrm{h}\) of energy per year. (a) Assuming the freezer operates for 5 hours each day, how much power does it require while operating? (b) If the freezer keeps its interior at a temperature of \(-5.0^{\circ} \mathrm{C}\) in a \(20.0^{\circ} \mathrm{C}\) room, what is its theoretical maximum performance coefficient? (c) What is the theoretical maximum amount of ice this freezer could make in an hour, starting with water at \(20.0^{\circ} \mathrm{C} ?\)

A Carnot engine has an efficiency of 59\(\%\) and performs \(2.5 \times 10^{4} \mathrm{J}\) of work in each cycle. (a) How much heat does the engine extract from its heat source in each cycle? (b) Suppose the engine exhausts heat at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) What is the temperature of its heat source?

Consider a Diesel cycle that starts (at point \(a\) in Fig. 20.7\()\) with air at temperature \(T_{a}\) The air may be treated as an ideal gas. (a) If the temperature at point \(c\) is \(T_{c}\) , derive an expression for the efficiency of the cycle in terms of the compression ratio \(r .\) (b) What is the efficiency if \(T_{a}=300 \mathrm{K}, T_{c}=950 \mathrm{K}, \gamma=1.40\) and \(r=21.0 ?\)
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