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Unreasonable Results Calculate the cold reservoir temperature of a steam engine that uses hot steam at 450ºC and has a Carnot efficiency of 0.700. (b) What is unreasonable about the temperature? (c) Which premise is unreasonable?

Short Answer

Expert verified
The cold reservoir temperature is calculated to be \( T_{cold} = 216.945 \text{ K} (\text{-56.205}^\circ\text{C}) \). The temperature appears to be below the typical ambient temperature, which is unreasonable. The premise of a Carnot efficiency of 0.700 is likely too high for a practical steam engine.

Step by step solution

01

Convert the Hot Reservoir Temperature from Celsius to Kelvin

To work with thermodynamic equations, it is standard to use temperatures in units of Kelvin. Convert the hot reservoir temperature from Celsius to Kelvin using the formula: Kelvin temperature = Celsius temperature + 273.15. For 450°C, the conversion is done as follows: \( T_{hot} = 450 + 273.15 \).
02

Use the Carnot Efficiency Formula

The Carnot efficiency (\( \text{e} \)) is given by the formula \( \text{e} = 1 - \frac{T_{cold}}{T_{hot}} \), where \( T_{cold} \) is the cold reservoir temperature, and \( T_{hot} \) is the hot reservoir temperature in Kelvin. Rearrange the formula to solve for \( T_{cold} \): \( T_{cold} = T_{hot} \times (1 - \text{e}) \).
03

Calculate the Cold Reservoir Temperature

Substitute the values of the hot reservoir temperature in Kelvin and the given efficiency into the rearranged equation: \( T_{cold} = (723.15 \text{ K}) \times (1 - 0.700) \).
04

Interpret the Result

Perform the multiplication to find the temperature of the cold reservoir. If the result appears to be below the ambient or expected temperatures, this would be considered unreasonable.
05

Address the Unreasonable Result

Discuss which premise might be unreasonable if the calculated cold reservoir temperature is significantly lower than physically expected (e.g., below absolute zero or significantly below ambient temperature).
06

Provide a Reasonable Explanation

If the calculated cold reservoir temperature is indeed unreasonable, one might argue that the assumed Carnot efficiency is too high for practical engines as real-world efficiencies are always less than those of idealized Carnot engines.

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Key Concepts

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

Thermodynamic Temperature Conversion
Understanding temperature conversion is essential when working with thermodynamics. Most thermodynamic calculations require temperatures to be in the Kelvin scale, as it is an absolute measure of temperature starting from absolute zero, where no more heat energy can be extracted from a system.

To convert from degrees Celsius to Kelvin, you add 273.15 to the Celsius temperature. This is because 0°C is equivalent to 273.15K. The conversion formula looks like this:
\[ T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \].

For example, if the temperature of the hot steam in a heat engine is given as 450°C, it gets converted to Kelvin as follows:\[ T_{\text{hot}} = 450 + 273.15 = 723.15 \text{ K} \].

This step is crucial for using the Carnot efficiency formula properly, as all temperatures must be in the same unit to ensure the accuracy of calculation and consistency in thermodynamic equations.
Carnot Engine Cycle
The Carnot engine cycle is a theoretical construct that helps us understand the maximum efficiency that a heat engine operating between two temperatures can achieve. It consists of four reversible processes:
  • Isothermal expansion (absorbing heat from the hot reservoir),
  • Adiabatic expansion (where the system does work on the surroundings),
  • Isothermal compression (releasing heat to the cold reservoir), and
  • Adiabatic compression (where work is done on the system).

During an ideal Carnot cycle, no entropy is created, which implies that the whole process is reversible and thus represents a perfect engine model.

Although no actual heat engine can achieve the efficiency of a Carnot engine due to friction and other irreversible processes, exploring the Carnot cycle gives insights into how factors such as temperature limits and heat transfer affect the efficiency of real-world engines.
Carnot Theorem
The Carnot theorem is a principle that defines the efficiency for a reversible heat engine. It states that the efficiency of a reversible heat engine depends only on the temperatures of the hot and cold reservoirs and is the same for all working substances. It is given by the formula:\[ \text{e} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \].

In the context of our exercise, the given Carnot efficiency is 0.700 (or 70%), which is unusually high for real-world engines. If we compute a cold reservoir temperature that seems physically nonviable, it may indicate that the assumed Carnot efficiency is beyond what's realistically achievable.

This can lead to the appreciation that ideal engines, as described by the Carnot theorem, provide targets for efficiency, but practical machines always fall short due to various inefficiencies such as those caused by friction, heat loss, and material limitations.

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Most popular questions from this chapter

(a) The Sun orbits the Milky Way galaxy once each \(2.60 \times 10^{8} \mathrm{y},\) with a roughly circular orbit averaging \(3.00 \times 10^{4}\) light years in radius. (A light year is the distance traveled by light in 1 y.) Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun? (b) Calculate the average speed of the Sun in its galactic orbit. Does the answer surprise you?

(a) How much heat transfer occurs from 20.0 kg of 90.0º C water placed in contact with 20.0 kg of 10.0º C water, producing a final temperature of 50.0º C? (b) How much work could a Carnot engine do with this heat transfer, assuming it operates between two reservoirs at constant temperatures of 90.0º C and 10.0º C? (c) What increase in entropy is produced by mixing 20.0 kg of 90.0º C water with 20.0 kg of 10.0º C water? (d) Calculate the amount of work made unavailable by this mixing using a low temperature of 10.0º C, and compare it with the work done by the Carnot engine. Explicitly show how you follow the steps in the Problem-Solving Strategies for Entropy. (e) Discuss how everyday processes make increasingly more energy unavailable to do work, as implied by this problem.

(a) A woman climbing the Washington Monument metabolizes 6.00×102kJ of food energy. If her efficiency is 18.0%, how much heat transfer occurs to the environment to keep her temperature constant? (b) Discuss the amount of heat transfer found in (a). Is it consistent with the fact that you quickly warm up when exercising?

What is the change in internal energy of a system which does 4.50×105J of work while 3.00×106J of heat transfer occurs into the system, and 8.00×106J of heat transfer occurs to the environment?

A coal-fired electrical power station has an efficiency of 38%. The temperature of the steam leaving the boiler is 550ºC. What percentage of the maximum efficiency does this station obtain? (Assume the temperature of the environment is 20ºC.)

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