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A heat engine receives heat from a heat source at \(1200^{\circ} \mathrm{C}\) and rejects heat to a heat \(\operatorname{sink}\) at \(50^{\circ} \mathrm{C}\). The heat engine does maximum work equal to \(500 \mathrm{kJ}\). Determine the heat supplied to the heat engine by the heat source and the heat rejected to the heat sink.

Short Answer

Expert verified
Answer: The heat supplied by the heat source to the heat engine is 640.13 kJ, and the heat rejected to the heat sink is 140.13 kJ.

Step by step solution

01

Calculate the efficiency of the heat engine

To calculate the efficiency, we should first convert the given temperatures from Celsius to Kelvin. Then, using the Carnot efficiency formula, we can find the efficiency of the heat engine. The formula to convert Celsius to Kelvin is: \(K = ^{\circ}C + 273.15\) So, for the heat source temperature: \(T_H = 1200 + 273.15 = 1473.15 \mathrm{K}\) And for the heat sink temperature: \(T_C = 50 + 273.15 = 323.15 \mathrm{K}\) Now, we calculate the efficiency of the heat engine using Carnot's efficiency: \(\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{323.15}{1473.15}\)
02

Compute the efficiency

Using the temperatures we found in Step 1, let's compute the efficiency: \(\eta = 1 - \frac{323.15}{1473.15} = 1 - 0.2191 = 0.7809\)
03

Find the heat supplied from the heat source to the heat engine

We are given the maximum work done by the heat engine, which is \(500 \mathrm{kJ}\). We can use the efficiency and the work to find the heat supplied (\(Q_H\)) from the heat source to the heat engine using the following formula: \(W = \eta \cdot Q_H\) Let's rearrange the formula to find \(Q_H\): \(Q_H = \frac{W}{\eta} = \frac{500}{0.7809} = 640.13 \mathrm{kJ}\)
04

Calculate the heat rejected to the heat sink

We can now find the heat rejected to the heat sink (\(Q_C\)) by using the heat supplied and the work done: \(Q_C = Q_H - W = 640.13 - 500 = 140.13 \mathrm{kJ}\)
05

Finalize and present the findings

Now, we can provide the answers for the heat supplied to the heat engine by the heat source and the heat rejected to the heat sink: 1. The heat supplied by the heat source to the heat engine is \(640.13 \mathrm{kJ}\). 2. The heat rejected to the heat sink is \(140.13 \mathrm{kJ}\).

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Key Concepts

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

Heat Engine
A heat engine is a fascinating piece of technology that turns thermal energy into mechanical work. In simpler terms, it's like a machine that can harness the power from heat and turn it into motion that we can use to do things like driving cars or generating electricity. How does it do this magic? Well, a heat engine operates on the principle of heat flowing from a hotter place (the heat source) to a cooler place (the heat sink), and in the process, some of that heat can be converted into work.

Think of it as a clever intermediary: the heat doesn't just lazily drift from hot to cold, the engine makes it do a little dance that pumps out work we can use! This concept is crucial to industries and technologies, from the internal combustion engines in vehicles to power stations that supply electricity to our homes and workplaces.
Temperature Conversion
Temperature conversion is a bit like a language translation for science – it's all about understanding 'how hot' in multiple ways. Just as you might speak in Fahrenheit, Celsius, or Kelvin, different scientific equations and principles require temperature to be stated in the most useful unit. To chat with the laws of thermodynamics, we usually use Kelvin, as it sets absolute zero, the chilliest possible temperature, as zero Kelvin (0 K).

Converting from Celsius to Kelvin is a walk in the park: simply add 273.15 to the Celsius value. So, if your tea is steeping at a nice 100 degrees Celsius, it's actually 373.15 Kelvin. This step is super important when calculating things like Carnot efficiency because the formulas work seamlessly with Kelvin.
Heat Transfer
Heat transfer is a bit like socializing energy style! It's the process of heat moving around, spreading out from where it's warm to where it's cooler. There are a few key players in the world of heat transfer: conduction, where heat passes through solid objects that are touching; convection, where heat swirls around with fluids like air or water; and radiation, where heat beams through empty spaces like sunshine. When we talk about heat engines, we're usually interested in how much heat is transferred into the engine to do work and then how much is let go – like passing a ball in a game. Efficient engines grab a lot of heat, turn most of it into work, and then don't let too much escape unused. Understanding heat transfer is not only essential to improve heat engines but also to everyday life, like when you're cooking or keeping your house warm.

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Most popular questions from this chapter

During an experiment conducted in a room at \(25^{\circ} \mathrm{C},\) a laboratory assistant measures that a refrigerator that draws \(2 \mathrm{kW}\) of power has removed \(30,000 \mathrm{kJ}\) of heat from the refrigerated space, which is maintained at \(-30^{\circ} \mathrm{C} .\) The running time of the refrigerator during the experiment was 20 min. Determine if these measurements are reasonable.

A homeowner buys a new refrigerator with no freezer compartment and a deep freezer for the new kitchen. Which of these devices would you expect to have a lower COP? Why?

It is well established that the thermal efficiency of a heat engine increases as the temperature \(T_{L}\) at which heat is rejected from the heat engine decreases. In an effort to increase the efficiency of a power plant, somebody suggests refrigerating the cooling water before it enters the condenser, where heat rejection takes place. Would you be in favor of this idea? Why?

It is commonly recommended that hot foods be cooled first to room temperature by simply waiting a while before they are put into the refrigerator to save energy. Despite this commonsense recommendation, a person keeps cooking a large pan of stew three times a week and putting the pan into the refrigerator while it is still hot, thinking that the money saved is probably too little. But he says he can be convinced if you can show that the money saved is significant. The average mass of the pan and its contents is 5 kg. The average temperature of the kitchen is \(23^{\circ} \mathrm{C},\) and the average temperature of the food is \(95^{\circ} \mathrm{C}\) when it is taken off the stove. The refrigerated space is maintained at \(3^{\circ} \mathrm{C}\), and the average specific heat of the food and the pan can be taken to be \(3.9 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C} .\) If the refrigerator has a coefficient of performance of 1.5 and the cost of electricity is 10 cents per \(\mathrm{kWh}\) determine how much this person will save a year by waiting

Design a hydrocooling unit that can cool fruits and vegetables from 30 to \(5^{\circ} \mathrm{C}\) at a rate of \(20,000 \mathrm{kg} / \mathrm{h}\) under the following conditions: The unit will be of flood type, which will cool the products as they are conveyed into the channel filled with water. The products will be dropped into the channel filled with water at one end and be picked up at the other end. The channel can be as wide as \(3 \mathrm{m}\) and as high as \(90 \mathrm{cm} .\) The water is to be circulated and cooled by the evaporator section of a refrigeration system. The refrigerant temperature inside the coils is to be \(-2^{\circ} \mathrm{C}\), and the water temperature is not to drop below \(1^{\circ} \mathrm{C}\) and not to exceed \(6^{\circ} \mathrm{C}\) Assuming reasonable values for the average product density, specific heat, and porosity (the fraction of air volume in a box), recommend reasonable values for \((a)\) the water velocity through the channel and ( \(b\) ) the refrigeration capacity of the refrigeration system.

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