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A carnot engine absorbs \(250 \mathrm{~J}\) of heat from reservoir at \(100^{\circ} \mathrm{C}\) and rejects heat to a reservoir at \(20^{\circ} \mathrm{C}\). Find (a) the heat rejected, (b) work done by the engine, and (c) thermal efficiency.

Short Answer

Expert verified
The heat rejected is 177.7 J, the work done by the engine is 72.3 J, and the thermal efficiency is 28.92%.

Step by step solution

01

- Understand the Temperatures

Convert both temperatures from Celsius to Kelvin by adding 273.15. For the hot reservoir: For the cold reservoir:
02

- Calculate Thermal Efficiency

Thermal efficiency (
03

- Calculate Heat Rejected

Use the thermal efficiency to find the heat rejected ( 250
04

- Calculate Work Done

The work done by the engine is the difference between the heat absorbed and the heat rejected:

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Key Concepts

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

heat transfer
Heat transfer is the movement of thermal energy from one place to another. This can occur in a variety of ways: conduction, convection, and radiation. In a Carnot engine, heat transfer happens between two reservoirs: one hot and one cold. The engine absorbs heat from the hot reservoir and releases some of this heat to the cold reservoir. The goal is to convert the maximum possible amount of absorbed heat into useful work. The heat transfer in a Carnot engine can be better understood by looking at the temperatures involved and the amount of heat absorbed and rejected.
thermodynamic cycles
Thermodynamic cycles are processes where a system goes through a series of states and eventually returns to its initial state. The Carnot cycle, named after Nicolas Léonard Sadi Carnot, is an idealized cycle that provides the maximum possible efficiency for a heat engine. The process involves four stages: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. During these cycles, the engine absorbs heat from a high-temperature reservoir and rejects heat to a low-temperature reservoir while performing work. Understanding these cycles is key to grasping how heat engines work and how thermal efficiency is achieved.
thermal efficiency
Thermal efficiency is a measure of how well a heat engine converts heat into work. It is defined as the ratio of the work done by the engine to the heat absorbed from the hot reservoir. Mathematically, it can be represented by: \[ \text{Thermal Efficiency} = \frac{\text{Work Done}}{\text{Heat Absorbed}} \times 100\text{\text{%}} \] For a Carnot engine, the maximum possible efficiency can be calculated using the temperatures of the hot and cold reservoirs: \[ \text{Carnot Efficiency} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \] The temperatures must be in Kelvin for this calculation. The key point here is that the higher the temperature difference between the two reservoirs, the higher the efficiency.
work done by a heat engine
The work done by a heat engine is derived from the heat absorbed from the hot reservoir minus the heat rejected to the cold reservoir. For a Carnot engine, this can be expressed as: \[ W = Q_{\text{abs}} - Q_{\text{rej}} \] Where \( W \) is the work done, \( Q_{\text{abs}} \) is the heat absorbed, and \( Q_{\text{rej}} \) is the heat rejected. In our example, the engine absorbs 250 J of heat. Using the thermal efficiency calculated from the temperatures, we can find the heat rejected and thus determine the work done by the engine. Understanding the amount of work done helps fulfill the purpose of a heat engine: converting heat energy into mechanical work.

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Most popular questions from this chapter

Two reversible heat pump cycles operate in series. The first cycle receives energy by heat transfer from a cold reservoir at \(260 \mathrm{~K}\) and rejects energy by heat transfer to a reservoir at an intermediate temperature \(T\) greater than 260 K. The second cycle receives energy by heat transfer from the reservoir at temperature \(T\) and rejects energy by heat transfer to a higher- temperature reservoir at \(1200 \mathrm{~K}\). If the heat pump cycles have the same coefficient of performance, determine (a) \(T\), in \(\mathrm{K}\), and (b) the value of each coefficient of performance.

At steady state, a power cycle develops a power output of \(20 \mathrm{~kW}\) while receiving energy by heat transfer at the rate of \(20 \mathrm{~kJ}\) per cycle of operation from a source at temperature \(T\). The cycle rejects energy by heat transfer to cooling water at a lower temperature of \(500 \mathrm{~K}\). If there are 150 cycles per minute, what is the minimum theoretical value for \(T\), in \(\mathrm{K} ?\)

A reversible heat pump cycle operating at steady state receives energy at the rate \(Q_{\mathrm{C}}=7.5 \mathrm{~kW}\) from a cold reservoir at temperature \(T_{\mathrm{C}}=0^{\circ} \mathrm{C}\) and discharges energy at the rate \(\dot{Q}_{\mathrm{H}}=9.5 \mathrm{~kW}\) to a hot reservoir at temperature \(T_{\mathrm{H}}\). Determine the value of \(T_{\mathrm{H}}\) and the co efficient of performance of heat pump.

At steady state, a reversible refrigeration cycle operates between hot and cold reservoirs at \(300 \mathrm{~K}\) and \(270 \mathrm{~K}\), respectively. Determine the minimum theoretical net power input required, in \(\mathrm{kW}\) per \(\mathrm{kW}\) of heat transfer from the cold reservoir.

Two reversible refrigeration cycles are arranged in series. The first cycle receives energy by heat transfer from a cold reservoir at temperature \(T_{C}\) and rejects energy by heat transfer to a reservoir at an intermediate temperature \(T\), greater than \(T_{C .}\) The second cycle receives energy by heat transfer from the reservoir at temperature \(T\) and rejects energy by heat transfer to a higher-temperature reservoir at \(T_{\mathrm{H}}\). Obtain an expression for the coefficient of performance of a single reversible refrigeration cycle operating directly between cold and hot reservoirs at \(T_{\mathrm{C}}\) and \(T_{\mathrm{H}}\), respectively, in terms of the coefficients of performance of the two cycles.

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