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A Carnot engine has an efficiency of 66% and performs 2.5 \(\times\) 10\(^4\) 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 (20.0\(^\circ\)C). What is the temperature of its heat source?

Short Answer

Expert verified
(a) Heat extracted is approximately 3.79 × 10⁴ J. (b) Temperature of the heat source is about 862.21 K.

Step by step solution

01

Understanding Efficiency

The efficiency of a Carnot engine is given by the formula \( \eta = 1 - \frac{T_c}{T_h} \), where \( T_c \) and \( T_h \) are the temperatures of the cold and hot reservoirs, respectively. Given the efficiency \( \eta = 0.66 \), which means 66%, we'll use this information later to find the temperature of the heat source.
02

Calculating Heat Energy Input

Efficiency \( \eta \) is also defined as the ratio of work done \( W \) to the heat absorbed from the heat source \( Q_h \). The equation is \( \eta = \frac{W}{Q_h} \). Given \( W = 2.5 \times 10^4 \) J and \( \eta = 0.66 \), we solve for \( Q_h \) using \( 0.66 = \frac{2.5 \times 10^4}{Q_h} \). Rearranging gives \( Q_h = \frac{2.5 \times 10^4}{0.66} \approx 3.79 \times 10^4 \) J.
03

Understanding Temperatures in Kelvin

Convert the exhaust temperature from Celsius to Kelvin since thermodynamic calculations require the temperature in Kelvin. The conversion is given by \( T(K) = T(^{\circ}C) + 273.15 \). For a room temperature of 20.0°C, we have \( T_c = 20 + 273.15 = 293.15 \) K.
04

Finding the Temperature of the Heat Source

Using the Carnot efficiency equation \( \eta = 1 - \frac{T_c}{T_h} \), where \( \eta = 0.66 \) and \( T_c = 293.15 \) K, we can find \( T_h \). Solving \( 0.66 = 1 - \frac{293.15}{T_h} \) yields \( \frac{293.15}{T_h} = 0.34 \), and rearranging gives \( T_h = \frac{293.15}{0.34} \approx 862.21 \) K.

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Key Concepts

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

Efficiency of Heat Engines
In the world of thermodynamics, the efficiency of a heat engine is a crucial aspect to understand. A heat engine extracts heat from a source, performs work, and exhausts some of the heat to a sink. The efficiency (\(\eta\)) measures how well an engine converts the absorbed heat (\(Q_h\)) into work (\(W\)). It can be calculated using the formula:
\(\eta = \frac{W}{Q_h}\).
For a Carnot engine, which is a theoretical perfect engine, the efficiency is also given by the expression:
\(\eta = 1 - \frac{T_c}{T_h}\).
Here, \(T_c\) is the temperature of the cold reservoir and \(T_h\) is the temperature of the hot source. All these temperatures must be in Kelvin.
A Carnot engine represents an ideal because, in reality, 100% efficiency is unattainable due to the second law of thermodynamics. This means that no engine can convert all the heat it absorbs into work without losing some to the surroundings.
Thermodynamic Temperature Conversion
When dealing with thermodynamic processes, it's essential to use the Kelvin scale for temperatures. The Kelvin scale begins at absolute zero, the coldest possible temperature, which helps eliminate negative values that can complicate calculations.
To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, converting the room temperature of 20.0°C, we do:
  • \(T(K) = T(^{\circ}C) + 273.15\)
  • \(T(K) = 20 + 273.15 = 293.15\) K
In thermodynamic calculations involving engines, using Kelvin ensures consistency in calculating formulas like the Carnot efficiency. This temperature conversion is important because avoiding negative values allows for smoother and more precise calculations when using scientific equations.
Work and Heat Energy Relationship
In the study of heat engines, understanding how work and heat energy are interconnected is indispensable. The energy absorbed as heat from the heat source (\(Q_h\)) and the energy lost to the cold sink influence the amount of work (\(W\)) the engine can perform.
An important relationship here is: \(W = Q_h - Q_c\), where \(Q_c\) is the heat expelled to the cold sink. This equation demonstrates that the work done by the engine is the difference between the heat absorbed from the source and the heat lost to the sink.
Moreover, we explored the efficiency calculation \(\eta = \frac{W}{Q_h}\), concluding that efficiency helps determine how much of the absorbed heat is converted to useful work. By improving efficiency, an engine can perform more work for the same amount of heat input, but natural limitations prevent reaching the theoretical 100% efficiency.

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Most popular questions from this chapter

A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 K and 270 K. (a) If in each cycle the refrigerator receives 415 J of heat energy from the reservoir at 270 K, how many joules of heat energy does it deliver to the reservoir at 320 K? (b) If the refrigerator completes 165 cycles each minute, what power input is required to operate it? (c) What is the coefficient of performance of the refrigerator?

As a budding mechanical engineer, you are called upon to design a Carnot engine that has 2.00 mol of a monatomic ideal gas as its working substance and operates from a high temperature reservoir at 500\(^\circ\)C. The engine is to lift a 15.0-kg weight 2.00 m per cycle, using 500 J of heat input. The gas in the engine chamber can have a minimum volume of 5.00 \(L\) during the cycle. (a) Draw a \(pV\)-diagram for this cycle. Show in your diagram where heat enters and leaves the gas. (b) What must be the temperature of the cold reservoir? (c) What is the thermal efficiency of the engine? (d) How much heat energy does this engine waste per cycle? (e) What is the maximum pressure that the gas chamber will have to withstand?

CP A certain heat engine operating on a Carnot cycle absorbs 410 J of heat per cycle at its hot reservoir at 135\(^\circ\)C and has a thermal efficiency of 22.0%. (a) How much work does this engine do per cycle? (b) How much heat does the engine waste each cycle? (c) What is the temperature of the cold reservoir? (d) By how much does the engine change the entropy of the world each cycle? (e) What mass of water could this engine pump per cycle from a well 35.0 m deep?

An ideal Carnot engine operates between 500\(^\circ\)C and 100\(^\circ\)C with a heat input of 250 J per cycle. (a) How much heat is delivered to the cold reservoir in each cycle? (b) What minimum number of cycles is necessary for the engine to lift a 500-kg rock through a height of 100 m?

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\(^\circ\)C) and the other end at the surface of your skin, at 30.0\(^\circ\)C. (a) What is the maximum efficiency of such a heat engine? Would it be a very useful engine? (b) Suppose you want to use this human engine to lift a 2.50-kg box from the floor to a tabletop 1.20 m above the floor. How much must you increase the gravitational potential energy, and how much heat input is needed to accomplish this? (c) If your favorite candy bar has 350 food calories (1 food calorie = 4186 J) and 80% of the food energy goes into heat, how many of these candy bars must you eat to lift the box in this way?

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