Question

A heat engine operates with an efficiency of \(0.5 .\) What can the temperatures of the high-temperature and low-temperature reservoirs be? a) \(T_{\mathrm{H}}=600 \mathrm{~K}\) and \(T_{\mathrm{L}}=100 \mathrm{~K}\) b) \(T_{H}=600 \mathrm{~K}\) and \(T_{\mathrm{L}}=200 \mathrm{~K}\) c) \(T_{H}=500 \mathrm{~K}\) and \(T_{\mathrm{L}}=200 \mathrm{~K}\) d) \(T_{\mathrm{H}}=500 \mathrm{~K}\) and \(T_{\mathrm{L}}=300 \mathrm{~K}\) e) \(T_{\mathrm{H}}=600 \mathrm{~K}\) and \(T_{\mathrm{J}}=300 \mathrm{~K}\)

Short answer

Answer: The pair with \(T_{H}=600 \mathrm{~K}\) and \(T_{\mathrm{L}}=200 \mathrm{~K}\) has an efficiency of 0.5.

Step-by-step solution

Use the efficiency to set up equation

Use the given efficiency of 0.5 and the formula for the efficiency of a heat engine: $$0.5 = 1 - \frac{T_{low}}{T_{high}}$$

Rearrange the equation

Solving for T_{low}, we get: $$T_{low} = T_{high} \times (1-0.5)$$ $$T_{low} = \frac{T_{high}}{2}$$

Test each option to find which one(s) satisfy the equation

Now, we will check every option provided in the exercise and verify which one satisfies the equation obtained in Step 2: a) \(T_{H}=600 K\) and \(T_{L}=100 K\) $$T_{low} = \frac{T_{high}}{2}$$ $$100 = \frac{600}{2}$$ This option does not satisfy the equation. b) \(T_{H}=600 K\) and \(T_{L}=200 K\) $$T_{low} = \frac{T_{high}}{2}$$ $$200 = \frac{600}{2}$$ This option satisfies the equation. c) \(T_{H}=500 K\) and \(T_{L}=200 K\) $$T_{low} = \frac{T_{high}}{2}$$ $$200 = \frac{500}{2}$$ This option does not satisfy the equation. d) \(T_{H}=500 K\) and \(T_{L}=300 K\) $$T_{low} = \frac{T_{high}}{2}$$ $$300 = \frac{500}{2}$$ This option does not satisfy the equation. e) \(T_{H}=600 K\) and \(T_{L}=300 K\) $$T_{low} = \frac{T_{high}}{2}$$ $$300 = \frac{600}{2}$$ This option does not satisfy the equation.

Report the correct option

After testing all the given pairs of temperatures, we see that Option b) \(T_{H}=600 \mathrm{~K}\) and \(T_{\mathrm{L}}=200 \mathrm{~K}\) is the only pair that satisfies the equation and has the given efficiency of 0.5.

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationship between heat and other forms of energy. In essence, it's the study of how energy is converted from one form to another, especially in terms of heat and work. A key element of thermodynamics is the concept of a thermodynamic system, which is a defined quantity of matter or a space in which energy exchanges take place. There are laws that govern these exchanges, most notably the first law of thermodynamics which states that energy cannot be created or destroyed, only transformed.

Thermodynamics also involves the study of entropy, a measure of disorder within a system, which increases as energy is converted to heat. This idea is central to understanding why certain energy transformations are irreversible. Thermodynamics plays a crucial role in determining efficiency, especially in engineered systems like heat engines, refrigerators, and air conditioners. A good starting point for students to comprehend thermodynamics is to recognize that it's about energy, its forms, transformations, and the fundamental ways in which those are governed by natural laws.

Carnot Efficiency

Carnot efficiency is a principle that sets the maximum possible efficiency for any heat engine. This concept is named after the French physicist Sadi Carnot who deduced that the efficiency of a heat engine depends on the temperatures of the hot and cold reservoirs. Carnot efficiency serves as an ideal that real engines can aspire to but never fully achieve, due to practical limitations and friction.

The formula for Carnot efficiency is:\[\begin{equation}\eta_{Carnot} = 1 - \frac{T_{L}}{T_{H}}\end{equation}\]where \(\eta_{Carnot}\) is the Carnot efficiency, \(T_{L}\) is the absolute temperature of the cold reservoir, and \(T_{H}\) is the absolute temperature of the hot reservoir, both measured in kelvins.

In understanding the concept, students should recognize that for a heat engine operating between two temperatures, no matter how ideal, the efficiency cannot exceed the Carnot limit. This serves as a benchmark for gauging the performance of real engines and also provides insight into why enhancing energy conversion technologies is an ongoing challenge.

Temperature Reservoirs

Temperature reservoirs are key components in the study and application of thermodynamics, particularly within the context of heat engines. They are theoretical constructs imagined as bodies with a relatively infinite capacity for heat energy, such that their temperature does not change despite the exchange of heat. In reality, no perfect reservoirs exist, but for simplification in studying thermodynamics, we consider them.

A heat engine operates between two reservoirs: a hot reservoir, which provides thermal energy, and a cold reservoir, to which waste heat is expelled. The efficiency of this engine—and indeed any thermodynamic cycle—depends on the temperature difference between these two reservoirs. When analyzing problems like the one given, students should keep in mind that the greater the temperature difference, the greater the potential efficiency of the heat engine. Moreover, the exercise exemplifies that only certain combinations of temperatures will align with the given efficiency constraints, demonstrating the real-world applicability of these theoretical concepts in thermodynamics and engineering design.