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A \(1.00-\mathrm{kg}\) iron horseshoe is taken from a forge at \(900^{\circ} \mathrm{C}\) and dropped into \(4.00 \mathrm{kg}\) of water at \(10.0^{\circ} \mathrm{C} .\) Assuming that no energy is lost by heat to the surroundings, determine the total entropy change of the horseshoe-plus-water system.

Short Answer

Expert verified
The total entropy change of the horseshoe-plus-water system can be determined by first finding the final temperature of the system, then calculating the entropy change for the horseshoe and the water, and finally adding them together.

Step by step solution

01

Find the Final Temperature

First, we need to set up an equation using Q (Quantity of heat) = mcΔT (mass x specific heat capacity x change in temperature), where Q_horseshoe + Q_water = 0, i.e., heat gained by cool body = heat lost by hot body. Solve this equation for final temperature.
02

Calculate the Entropy Change for Each Component

Next, we need to calculate the entropy change (ΔS) for each component using the formula ΔS = Q_reversible / T. The heat Q is the same magnitude as calculated but with opposite signs for each component (one loses heat, the other gains it). The temperature T used in the equation should be calculated in Kelvin.
03

Calculate the Total Entropy Change

The total entropy change of the horseshoe and water system is the sum of the entropy change of the horseshoe and the entropy change of the water.

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Key Concepts

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

Understanding Specific Heat Capacity
The specific heat capacity is a fundamental concept in thermodynamics that describes the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). This property varies for different substances, which is why some materials heat up or cool down faster than others. Specific heat capacity is denoted usually by the symbol \(c\). It helps predict how much heat energy is required to alter the temperature of a given material under consideration. For example, let's say an iron horseshoe with a mass of \(1.00 \text{ kg}\) needs to be cooled down. Knowing the specific heat capacity of iron lets us calculate how much heat it needs to lose to reach a certain cooler temperature. In our case of thermal interaction between a hot horseshoe and water, both the horseshoe and the water have different specific heat capacities, affecting how each changes temperature over time when they are in contact. Considering these capacities is fundamental to solving problems involving heat exchange because it allows for accurate computation of the heat transferred between the substances. To put it succinctly, you can think of specific heat capacity as the resistance a substance has to changing its temperature when heat is added or removed. In practical terms, large specific heat capacities mean a substance takes more heat to change its temperature significantly, while small capacities mean it heats up or cools down quickly.
Exploring Heat Transfer
Heat transfer is the process of thermal energy moving from a warmer object or substance to a cooler one. This movement of energy can happen in three main ways: conduction, convection, and radiation. In the context of the exercise we're looking at, conduction is the most relevant, as it involves heat transfer through direct contact between the iron horseshoe and the water. When two objects at different temperatures come into contact, such as the horseshoe at 900°C and water at 10°C, heat will naturally flow from the hotter horseshoe to the cooler water. This transfer continues until both reach thermal equilibrium, meaning they share the same temperature. The equation \(Q = mc\Delta T\) is used to quantify this transferred heat. Here, \(Q\) is the heat energy transferred, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. Understanding this equation is crucial because it helps calculate how much energy is exchanged between substances during heating or cooling processes. Learning about heat transfer allows us to understand how energy conservation in closed systems leads to changes in temperature and, consequently, changes in entropy, as seen in the system of the horseshoe and water.
Thermodynamic System Simplified
A thermodynamic system refers to a defined region in space or a set of matter on which we focus to study its thermal interactions and energy exchanges. It is crucial to understand how this system behaves when analyzing any thermodynamic process, such as heat exchanges, work done, or changes in state functions like entropy. In the given exercise, the horseshoe and the water together make up the thermodynamic system. We assume that there is no heat interchange with the environment, making it an isolated system for the purpose of the calculation. This isolation is a simplifying assumption that ensures the total heat gained by water equals the total heat lost by the horseshoe. Understanding the concept of a thermodynamic system helps in visualizing how energy is conserved within closed systems. It allows us to calculate changes in quantities such as heat and entropy effectively. When you account for the system as a whole, it's easier to apply the conservation of energy principle, often seen in the first law of thermodynamics, which states that energy is neither created nor destroyed but merely transformed or transferred. Observing the horseshoe-water system demonstrates how thermal equilibrium is achieved in an isolated environment and how these closed systems naturally progress towards a state of maximum entropy or disorder.

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Most popular questions from this chapter

A sample consisting of \(n\) mol of an ideal gas undergoes a reversible isobaric expansion from volume \(V_{i}\) to volume \(3 V_{i} .\) Find the change in entropy of the gas by calculating \(\int_{i}^{t} d Q / T\) where \(d Q=n C_{P} d T\)

A firebox is at \(750 \mathrm{K},\) and the ambient temperature is \(300 \mathrm{K}\) The efficiency of a Carnot engine doing \(150 \mathrm{J}\) of work as it transports energy between these constant-temperature baths is \(60.0 \% .\) The Carnot engine must take in energy \(150 \mathrm{J} / 0.600=250 \mathrm{J}\) from the hot reservoir and must put out \(100 \mathrm{J}\) of energy by heat into the environment. To follow Carnot's reasoning, suppose that some other heat engine \(S\) could have efficiency \(70.0 \% .\) (a) Find the energy input and wasted energy output of engine \(\mathrm{S}\) as it does \(150 \mathrm{J}\) of work. (b) Let engine S operate as in part (a) and run the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, and the total energy transferred to the environment. Show that the Clausius statement of the second law of thermodynamics is violated. (c) Find the energy input and work output of engine \(S\) as it puts out exhaust energy of \(100 \mathrm{J} .\) (d) Let engine S operate as in (c) and contribute \(150 \mathrm{J}\) of its work output to running the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, the total work output, and the total energy transferred to the environment. Show that the Kelvin-Planck statement of the second law is violated. Thus our assumption about the efficiency of engine \(\mathrm{S}\) must be false. (e) Let the engines operate together through one cycle as in part (d). Find the change in entropy of the Universe. Show that the entropy statement of the second law is violated.

One of the most efficient heat engines ever built is a steam turbine in the Ohio valley, operating between \(430^{\circ} \mathrm{C}\) and \(1870^{\circ} \mathrm{C}\) on energy from West Virginia coal to produce electricity for the Midwest. (a) What is its maximum theoretical efficiency? (b) The actual efficiency of the engine is \(42.0 \% .\) How much useful power does the engine deliver if it takes in \(1.40 \times 10^{5} \mathrm{J}\) of energy each second from its hot reservoir?

At a pressure of 1 atm, liquid helium boils at \(4.20 \mathrm{K}\). The latent heat of vaporization is \(20.5 \mathrm{kJ} / \mathrm{kg} .\) Determine the entropy change (per kilogram) of the helium resulting from vaporization.

An ideal gas is taken through a Carnot cycle. The isothermal expansion occurs at \(250^{\circ} \mathrm{C},\) and the isothermal compression takes place at \(50.0^{\circ} \mathrm{C}\). The gas takes in \(1200 \mathrm{J}\) of energy from the hot reservoir during the isothermal expansion. Find (a) the energy expelled to the cold reservoir in each cycle and (b) the net work done by the gas in each cycle.

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