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Heat \(Q\) flows spontaneously from a reservoir at \(394 \mathrm{~K}\) into a reservoir that has a lower temperature \(T\). Because of the spontaneous flow, thirty percent of \(Q\) is rendered unavailable for work when a Carnot engine operates between the reservoir at temperature \(T\) and a reservoir at \(248 \mathrm{~K}\). Find the temperature \(T\)

Short Answer

Expert verified
The temperature \( T \) is approximately 826.67 K.

Step by step solution

01

Understand the Carnot Engine Efficiency Formula

The efficiency of a Carnot engine is given by the formula \( \eta = 1 - \frac{T_c}{T_h} \), where \( T_c \) is the cold reservoir temperature and \( T_h \) is the hot reservoir temperature. In this problem, the unavailable energy is 30% of \( Q \). The thermal efficiency related to this unavailable energy is 0.7, as only 70% of the energy is available for work.
02

Define the Temperatures

Here, the temperature \( T_h \) of the hot reservoir is \( T \) (since we're trying to find it), and \( T_c \) is \( 248 \mathrm{~K} \) as given in the problem. We need to use these in the efficiency formula.
03

Set Up the Equation

From the given information, we know that 70% of \( Q \) is available for work during the Carnot cycle. Therefore, \( 0.7 = 1 - \frac{248}{T} \).
04

Solve for \( T \)

Rearrange the equation from Step 3: \( 0.7 = 1 - \frac{248}{T} \). So, \( \frac{248}{T} = 0.3 \). Solving for \( T \), we get \( T = \frac{248}{0.3} \).
05

Calculate the Value

Compute the temperature \( T \) by simplifying the equation: \( T = \frac{248}{0.3} = 826.67 \).

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Key Concepts

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

Thermodynamic Efficiency
Thermodynamic efficiency is a measure of how effectively a system converts heat into work. It's one of the key concepts when analyzing heat engines like the Carnot engine. This efficiency is represented as a fraction or a percentage, showing the proportion of heat energy that can be transformed into mechanical energy.

The Carnot engine, a theoretical cycle, provides the maximum possible efficiency a heat engine operating between two temperatures can attain. According to the Carnot theorem, the efficiency \( \eta \) of an engine is given by the expression\( \eta = 1 - \frac{T_c}{T_h} \), where \( T_c \) is the absolute temperature of the cold reservoir, and \( T_h \) is the absolute temperature of the hot reservoir. This formula implies that efficiency depends entirely on the temperatures of the two reservoirs involved.

Understanding thermodynamic efficiency helps students see why real-world engines cannot be 100% efficient. This theoretical limit is due to the second law of thermodynamics, which implies that some energy in a heat transfer process is always rendered unavailable for work.
Spontaneous Heat Flow
Spontaneous heat flow refers to the natural direction of heat movement from a hot reservoir to a cold reservoir without any external work being done. This process, guided by the second law of thermodynamics, occurs because systems tend to move towards increased entropy, or disorder, and thus the distribution of thermal energy naturally flows to balance temperature differences.

In the case of the Carnot engine discussed in the exercise, heat \( Q \) flows from a hot reservoir to a cooler one. As we see, 30% of this heat is lost or rendered unavailable between the transfer, illustrating the inherent inability to convert all heat to work. This is a perfect example of spontaneous heat flow where energy disperses to reach thermal equilibrium across mediums. The Carnot cycle cannot use this portion of the heat, exemplifying the inevitable inefficiencies faced by any real or theoretical engine.

Understanding spontaneous heat flow provides clarity on why there’s always a loss in energy availability, which is fundamental to grasping why perpetual motion machines cannot exist.
Thermal Reservoir Temperatures
Thermal reservoirs play a critical role in determining the efficiency of heat engines like the Carnot engine. They essentially provide or accept thermal energy to or from the working substance. These reservoirs are assumed to have a nearly infinite capacity to supply or absorb heat without undergoing a change in temperature.

In thermodynamic problems, the temperatures of these reservoirs—denoted as \( T_h \) for the hot reservoir and \( T_c \) for the cold reservoir—are pivotal. The temperatures vary greatly, but consistent understanding is needed: higher \( T_h \) relative to \( T_c \) increases efficiency as shown by the Carnot formula: \( \eta = 1 - \frac{T_c}{T_h} \).

For example, the Carnot engine exercise discusses a hot reservoir with temperature \( T \) and a cold reservoir at \( 248 \, \mathrm{K} \). The task involves finding \( T \) using known parameters related to thermodynamic efficiency. An accurate understanding of how these reservoir temperatures influence engine efficiency is crucial to excel in thermodynamics.

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Most popular questions from this chapter

When one gallon of gasoline is burned in a car engine, \(1.19 \times 10^{8} \mathrm{~J}\) of internal energy is released. Suppose that \(1.00 \times 10^{8} \mathrm{~J}\) of this energy flows directly into the surroundings (engine block and exhaust system) in the form of heat. If \(6.0 \times 10^{5} \mathrm{~J}\) of work is required to make the car go one mile, how many miles can the car travel on one gallon of gas?

One mole of neon, a monatomic gas, starts out at conditions of standard temperature and pressure. The gas is heated at constant volume until its pressure is tripled, then further heated at constant pressure until its volume is doubled. Assume that neon behaves as an ideal gas. For the entire process, find the heat added to the gas.

A Carnot engine operates with an efficiency of \(27.0 \%\) when the temperature of its cold reservoir is \(275 \mathrm{~K}\). Assuming that the temperature of the hot reservoir remains the same, what must be the temperature of the cold reservoir in order to increase the efficiency to \(32.0 \% ?\)

In exercising, a weight lifter loses \(0.150 \mathrm{~kg}\) of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is \(1.40 \times 10^{5} \mathrm{~J}\). (a) Assuming that the latent heat of vaporization of perspiration is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg},\) find the change in the internal energy of the weight lifter. (b) Determine the minimum number of nutritional Calories of food ( 1 nutritional Calorie \(=4186 \mathrm{~J}\) ) that must be consumed to replace the loss of internal energy.

A monatomic ideal gas expands isobarically. Using the first law of thermodynamics, prove that the heat \(Q\) is positive, so that it is impossible for heat to flow out of the gas.

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