/*! 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 49 From Table \(14.4,\) we know tha... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 49

From Table \(14.4,\) we know that approximately \(2256 \mathrm{kJ}\) are needed to transform \(1.00 \mathrm{kg}\) of water at \(100^{\circ} \mathrm{C}\) to steam at \(100^{\circ} \mathrm{C}\). What is the change in entropy of \(1.00 \mathrm{kg}\) of water evaporating at \(100.0^{\circ} \mathrm{C} ?\) (Specify whether the change in entropy is an increase, \(+\), or a decrease, \(-.\) )

Short Answer

Expert verified
Answer: The change in entropy is an increase of approximately 6040.12 J/(K鈰卥g).

Step by step solution

01

Identify the given values

We are given the following information: - Energy \(Q = 2256 \mathrm{kJ/kg} \times 1000 \mathrm{J / kJ} = 2256000 \mathrm{J / kg}\) (Converted kJ to J for convenience) - Mass of water \(m = 1.00 \mathrm{kg}\) - Temperature \(T = 100^{\circ} \mathrm{C}\)
02

Convert temperature to Kelvin

We need to convert the temperature from Celsius to Kelvin before we can use it in our formula. The conversion is: \(T_\mathrm{k} = T_\mathrm{c} + 273.15\) where \(T_\mathrm{k}\) is the temperature in Kelvin and \(T_\mathrm{c}\) is the temperature in Celsius. Thus, \(T_\mathrm{k} = 100^{\circ} \mathrm{C} + 273.15 = 373.15 \mathrm{K}\)
03

Calculate the change in entropy

Now, we can use the formula to calculate the change in entropy for this process: \(\Delta S = \frac{Q}{T}\) Substituting our values: \(\Delta S = \frac{2256000 \mathrm{J / kg}}{373.15 \mathrm{K}}\) \(\Delta S \approx 6040.12 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{kg}}\) Since heat is absorbed during the evaporation process, the change in entropy will be positive: \(\Delta S = +6040.12 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{kg}}\) Thus, the change in entropy of \(1.00 \mathrm{kg}\) of water evaporating at \(100.0^{\circ} \mathrm{C}\) is an increase of approximately \(6040.12 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{kg}}\).

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Most popular questions from this chapter

What is the change in entropy of \(10 \mathrm{g}\) of steam at $100^{\circ} \mathrm{C}\( as it condenses to water at \)100^{\circ} \mathrm{C} ?$ By how much does the entropy of the universe increase in this process?
What is the efficiency of an electric generator that produces $1.17 \mathrm{kW}\(.h per \)\mathrm{kg}$ of coal burned? The heat of combustion of coal is \(6.71 \times 10^{6} \mathrm{J} / \mathrm{kg}\).

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