/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 In making raspberry jelly, \(900... [FREE SOLUTION] | 91Ó°ÊÓ

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In making raspberry jelly, \(900 \mathrm{g}\) of raspberry juice is combined with \(930 \mathrm{g}\) of sugar. The mixture starts at room temperature, \(23.0^{\circ} \mathrm{C},\) and is slowly heated on a stove until it reaches \(220^{\circ} \mathrm{F}\). It is then poured into heated jars and allowed to cool. Assume that the juice has the same specific heat as water. The specific heat of sucrose is \(0.299 \mathrm{cal} / \mathrm{g} \cdot^{\circ} \mathrm{C}\) Consider the heating process. (a) Which of the following terms describe(s) this process: adiabatic, isobaric, isothermal, isovolumetric, cyclic, reversible, isentropic? (b) How much energy does the mixture absorb? (c) What is the minimum change in entropy of the jelly while it is heated?

Short Answer

Expert verified
The thermal process is isobaric. The mixture absorbs 88357.25cal. The minimum change in entropy of the jelly while it is heated is 253.91 cal/K.

Step by step solution

01

Identify the Thermal Process

Since the jelly mixes and heats up at constant pressure at the surface of the earth, there is no insulation to prevent heat exchange with the environment and the volume certainly changes as it gets hot and becomes a liquid, the heating process of making jelly is isobaric.
02

Calculate The Heat Absorbed By The Raspberry Juice

The formula to calculate the heat \(\Delta Q\) absorbed or released in an isobaric process is \(\Delta Q = m c \Delta T\), where \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature. Let's first convert the final temperature from Fahrenheit to Celsius, which gives \(\Delta T = (220°F - 32) * (5/9) - 23.0°C = 75°C\). The specific heat of the juice is the same as that of water, i.e. \(c = 1cal/g°C\). Replacing these values into the formula gives \(\Delta Q_{juice} = 900g * 1 cal/g°C * 75°C = 67500cal\).
03

Calculate The Heat Absorbed By The Sugar

Using the same formula and the specific heat of sugar, \(c = 0.299cal/g°C\), we get \(\Delta Q_{sugar} = 930g * 0.299cal/g°C * 75°C = 20857.25cal\).
04

Calculate The Total Heat Absorbed By The Mixture

The total heat absorbed is the sum of the heats absorbed by the juice and the sugar, which gives \(\Delta Q = \Delta Q_{juice} + \Delta Q_{sugar} = 67500cal + 20857.25cal = 88357.25cal\).
05

Calculate The Minimum Change In Entropy

The entropy change \(\Delta S\) for an isobaric process, which is irreversible in this case as we're looking for the minimum change, is calculated by using the formula \(\Delta S = \Delta Q/T\) . Dividing the total heat absorbed by the final temperature in absolute scale (Kelvin), we get \(\Delta S = \Delta Q /(75°C + 273) = 88357.25cal / 348K = 253.91 cal/K\).

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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 a fundamental concept in thermodynamics that involves the movement of thermal energy from one object or substance to another. It can occur through several mechanisms such as conduction, convection, and radiation. In the context of making raspberry jelly, heat transfer is essential as it allows the mixture of raspberry juice and sugar to reach the desired consistency and temperature.

During this process, the mixture is heated from its initial room temperature of 23°C to a final temperature of 220°F. Since no insulation is used to prevent heat exchange, the jelly undergoes an isobaric process, meaning it happens at constant pressure. The heat transferred to the raspberry juice and sugar was calculated through the formula for heat absorption:
  • To calculate the heat absorbed by the juice, we use: \( \Delta Q = m \cdot c \cdot \Delta T \), where \( m \) is the mass, \( c \) is specific heat, and \( \Delta T \) is the temperature change.
  • For the juice: \( 900g \times 1 \text{ cal/g°C} \times 75°C = 67500 \text{ cal} \)
  • For the sugar: \( 930g \times 0.299 \text{ cal/g°C} \times 75°C = 20857.25 \text{ cal} \)
  • Total heat absorbed: \( 67500 \text{ cal} + 20857.25 \text{ cal} = 88357.25 \text{ cal} \)
This total heat absorbed by the mixture illustrates the significant energy requirement to prepare jelly, showcasing heat transfer in action.
Isentropic Process
An isentropic process is an idealized thermodynamic process in which the system undergoes reversible adiabatic transformations. In such a process, there is no transfer of heat or matter into or out of a system, and entropy remains constant.

In the jelly-making exercise, we do not have an isentropic process since heat is continuously added to the mixture, and entropy changes as the system is not insulated from its surroundings. Instead, the process is isobaric and more closely aligned with real-world conditions where energy transfer occurs at constant pressure.

In comparison to an isentropic process, where all transformations are perfectly efficient, real-world processes like jelly heating are not perfectly efficient and involve entropy changes. It's crucial for students to recognize that isentropic processes are an idealization used to simplify analyses and are useful for understanding efficiencies of systems like engines and compressors in a controlled environment. However, everyday cooking processes like our raspberry jelly scenario don't fit this ideal model due to real-world inefficiencies.
Entropy Change
Entropy is a measure of the disorder or randomness in a system. It quantifies the amount of energy in a system that is unavailable to do work. In thermodynamics, when energy is transferred, especially in heating processes, the entropy of a system usually increases.

For the raspberry jelly process, calculating the change in entropy gives insight into how energy has been dispersed or spread out in the process. Since the process is primarily isobaric and irreversible in realistic cooking settings, the change in entropy can be found using:
  • \( \Delta S = \frac{\Delta Q}{T} \)
  • Given the total heat absorbed \( \Delta Q = 88357.25 \text{ cal} \) and the final temperature in Kelvin is \( 348 \text{ K} \)
  • The change in entropy calculation: \( \Delta S = \frac{88357.25 \text{ cal}}{348 \text{ K}} = 253.91 \text{ cal/K} \)
This increase in entropy highlights the greater spread of energy within the jelly mixture as it absorbs heat and aligns with the second law of thermodynamics, which suggests that entropy in an isolated system will naturally increase over time. Understanding this concept helps elucidate why energy transformation processes never proceed with perfect efficiency, and some energy is always "lost" as entropy increases.

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Most popular questions from this chapter

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.

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.

Suppose a heat engine is connected to two energy reservoirs, one a pool of molten aluminum \(\left(660^{\circ} \mathrm{C}\right)\) and the other a block of solid mercury \(\left(-38.9^{\circ} \mathrm{C}\right) .\) The engine runs by freezing \(1.00 \mathrm{g}\) of aluminum and melting \(15.0 \mathrm{g}\) of mercury during each cycle. The heat of fusion of aluminum is \(3.97 \times 10^{5} \mathrm{J} / \mathrm{kg}\); the heat of fusion of mercury is \(1.18 \times 10^{4} \mathrm{J} / \mathrm{kg} .\) What is the efficiency of this engine?

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 \(d Q=n C_{P} d T\)

An electric power plant that would make use of the temperature gradient in the ocean has been proposed. The system is to operate between \(20.0^{\circ} \mathrm{C}\) (surface water temperature) and \(5.00^{\circ} \mathrm{C}\) (water temperature at a depth of about 1 \(\mathrm{km}\) ). (a) What is the maximum efficiency of such a system? (b) If the useful power output of the plant is \(75.0 \mathrm{MW},\) how much energy is taken in from the warm reservoir per hour? (c) In view of your answer to part (a), do you think such a system is worthwhile? Note that the "fuel" is free.

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