Question

Consider a Carnot engine that works between thermal reservoirs with temperatures of \(1000.0 \mathrm{~K}\) and \(300.0 \mathrm{~K}\). The average power of the engine is \(1.00 \mathrm{~kJ}\) per cycle. a) What is the efficiency of this engine? b) How much energy is extracted from the warmer reservoir per cycle? c) How much energy is delivered to the cooler reservoir?

Short answer

Answer: The efficiency of the Carnot engine is 70%, the energy extracted from the warmer reservoir per cycle is 1.43 kJ, and the energy delivered to the cooler reservoir is 0.43 kJ.

Step-by-step solution

Calculate the Carnot engine's efficiency

The efficiency of a Carnot engine is given by the formula: \[ \text{Efficiency} = 1 - \frac{T_{cold}}{T_{hot}}, \] where \(T_{cold}\) is the temperature of the cooler reservoir and \(T_{hot}\) is the temperature of the warmer reservoir. In this case, we have \(T_{cold} = 300.0 \mathrm{~K}\) and \(T_{hot} = 1000.0 \mathrm{~K}\). Plugging these numbers into the formula, we get: \[ \text{Efficiency} = 1 - \frac{300.0}{1000.0} = 1 - 0.3 = 0.7. \] Therefore, the efficiency of this engine is 0.7 or 70%.

Energy extracted from the warmer reservoir per cycle

We have the average power of the engine already given as 1.00 kJ per cycle. To find the energy extracted from the warmer reservoir, let's denote it by \(Q_{in}\). Since the efficiency of the engine is 70%, it means only 70% of the extracted energy from the warmer reservoir is converted to work, which is the power of the engine. We can write this as: \[ 1.00 \mathrm{~kJ} = 0.7 \times Q_{in}. \] From this equation, we can solve for the energy extracted from the warmer reservoir: \[ Q_{in} = \frac{1.00}{0.7} = 1.43 \mathrm{~kJ}. \] So, the energy extracted from the warmer reservoir per cycle is 1.43 kJ.

Energy delivered to the cooler reservoir

Since a part of the energy extracted from the warmer reservoir is converted to work and the rest is delivered to the cooler reservoir, we can find the energy delivered to the cooler reservoir by subtracting the work done from the total energy extracted from the warmer reservoir. Let the energy delivered to the cooler reservoir be denoted by \(Q_{out}\). We can write this as: \[ Q_{out} = Q_{in} - \text{Work}. \] Now we have \(Q_{in} = 1.43 \mathrm{~kJ}\) and the work done which is the power, 1.00 kJ. Plugging these numbers in the equation, \[ Q_{out} = 1.43 - 1.00 = 0.43 \mathrm{~kJ}. \] Thus, the energy delivered to the cooler reservoir is 0.43 kJ. In conclusion: a) The efficiency of the engine is 70%. b) The energy extracted from the warmer reservoir per cycle is 1.43 kJ. c) The energy delivered to the cooler reservoir is 0.43 kJ.

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with heat, work, and temperature, and their relation to energy, radiation, and physical properties of matter. The fundamental principles of thermodynamics are encapsulated in four laws. These laws describe how energy is transferred, transformed, and conserved in a system. In our exercise with the Carnot engine, these principles help us understand how energy is converted from heat drawn from a hot reservoir to work, and how efficiency is a measure of this conversion.

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed from one form to another. This concept is particularly important when calculating the energy extracted from the warm reservoir because the engine converts this heat energy into work. The second law of thermodynamics introduces the concept of entropy and effectively defines the direction of heat flow and the limitations on the efficiency of heat engines. A Carnot engine operates on a reversible cycle, which means it is theoretically the most efficient heat engine possible, as it operates with no increase in entropy during the cycle.

Thermal Reservoirs

A thermal reservoir refers to a hypothetical body with a large thermal energy capacity that can supply or absorb finite amounts of heat without undergoing any change in temperature. In the case of a Carnot engine, we often refer to both the 'hot' and 'cold' sources as thermal reservoirs because they represent the high and low temperature limits between which the engine operates.

The hot reservoir provides thermal energy, part of which the engine converts to work, while the rest is transferred to the cold reservoir. This setup in our exercise demonstrates how thermal energy from the hot reservoir (1000 K) is partially used to generate power with the remaining energy dissipated to the cold reservoir (300 K). Understanding the behavior of thermal reservoirs and the flow of heat between them is essential for solving thermodynamic problems and illustrates the broader concepts of heat transfer and energy distribution in physical systems.

Energy Conversion

Energy conversion is the process of changing one form of energy into another. In the context of thermodynamic cycles, such as the one that the Carnot engine undergoes, the conversion of interest is from heat energy to mechanical energy or work. A key aspect of this conversion process is efficiency, which is a measure of how well the engine converts the input heat into output work.

The efficiency of any heat engine, as described by the Carnot efficiency formula, arises from the temperatures of the thermal reservoirs and is fundamentally limited by the second law of thermodynamics. The efficiency equation used in our exercise, \( \text{Efficiency} = 1 - \frac{T_{cold}}{T_{hot}} \), succinctly expresses the maximum possible efficiency that any engine operating between two temperatures can achieve. This ideal efficiency is always less than 100%, reflecting the reality that some input energy is always lost, primarily as heat to the cold reservoir. Our exercise applied this concept to determine the energy flows within the Carnot cycle, where 70% of the energy from the hot reservoir was converted to work, while 30% was expelled to the cold one.