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Calculate the total energy, in kilojoules, that is needed to turn a 12 g block of ice at 0 degrees C into water vapor at 135 degrees C.

Calculate the total energy, in kilojoules, that is needed to turn a 12 g block of ice at 0 degrees C into water vapor at 135 degrees C.

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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**Calculating Total Energy to Transform Ice into Water Vapor** **Problem Statement:** Calculate the total energy, in kilojoules, that is needed to turn a 12 g block of ice at 0 degrees Celsius into water vapor at 135 degrees Celsius. **Explanation:** To solve this problem, we need to perform several steps: 1. Convert the ice at 0°C to water at 0°C. 2. Heat the water from 0°C to 100°C. 3. Convert the water at 100°C to steam at 100°C. 4. Heat the steam from 100°C to 135°C. **Steps Involved:** 1. **Melting the ice (Ice to Water at 0°C):** - The latent heat of fusion for ice is 334 J/g. - Energy = mass × latent heat of fusion - Energy = 12 g × 334 J/g = 4008 J 2. **Heating the water (Water from 0°C to 100°C):** - The specific heat capacity of water is 4.18 J/g°C. - Energy = mass × specific heat capacity × temperature change - Energy = 12 g × 4.18 J/g°C × 100°C = 5016 J 3. **Boiling the water (Water at 100°C to Steam at 100°C):** - The latent heat of vaporization for water is 2260 J/g. - Energy = mass × latent heat of vaporization - Energy = 12 g × 2260 J/g = 27120 J 4. **Heating the steam (Steam from 100°C to 135°C):** - The specific heat capacity of steam is 2.02 J/g°C. - Energy = mass × specific heat capacity × temperature change - Energy = 12 g × 2.02 J/g°C × 35°C = 847.2 J **Total Energy Required:** - Total energy = Energy for melting ice + Energy for heating water + Energy for boiling water + Energy for heating steam - Total energy = 4008 J + 5016 J + 27120 J + 847.2 J = 36991.2 J **Converting Joules to Kilojoules:** - 36991.2 J = 36.9912
Transcribed Image Text:**Calculating Total Energy to Transform Ice into Water Vapor** **Problem Statement:** Calculate the total energy, in kilojoules, that is needed to turn a 12 g block of ice at 0 degrees Celsius into water vapor at 135 degrees Celsius. **Explanation:** To solve this problem, we need to perform several steps: 1. Convert the ice at 0°C to water at 0°C. 2. Heat the water from 0°C to 100°C. 3. Convert the water at 100°C to steam at 100°C. 4. Heat the steam from 100°C to 135°C. **Steps Involved:** 1. **Melting the ice (Ice to Water at 0°C):** - The latent heat of fusion for ice is 334 J/g. - Energy = mass × latent heat of fusion - Energy = 12 g × 334 J/g = 4008 J 2. **Heating the water (Water from 0°C to 100°C):** - The specific heat capacity of water is 4.18 J/g°C. - Energy = mass × specific heat capacity × temperature change - Energy = 12 g × 4.18 J/g°C × 100°C = 5016 J 3. **Boiling the water (Water at 100°C to Steam at 100°C):** - The latent heat of vaporization for water is 2260 J/g. - Energy = mass × latent heat of vaporization - Energy = 12 g × 2260 J/g = 27120 J 4. **Heating the steam (Steam from 100°C to 135°C):** - The specific heat capacity of steam is 2.02 J/g°C. - Energy = mass × specific heat capacity × temperature change - Energy = 12 g × 2.02 J/g°C × 35°C = 847.2 J **Total Energy Required:** - Total energy = Energy for melting ice + Energy for heating water + Energy for boiling water + Energy for heating steam - Total energy = 4008 J + 5016 J + 27120 J + 847.2 J = 36991.2 J **Converting Joules to Kilojoules:** - 36991.2 J = 36.9912
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